Sherman–Morrison for Integral Equations

July 11, 2022 by Ethan N. Epperly Leave a comment

In this post, I want to discuss two of my favorite topics in applied mathematics: the Sherman–Morrison formula and integral equations. As a bridge between these ideas, we’ll use an integral equation analog of the Sherman–Morrison formula to derive the solution for the Laplace equation with Dirichlet boundary conditions in the 2D disk.

Laplace’s Equation

Suppose we have a thin, flat (two-dimensional) plate of homogeneous material and we measure the temperature at the border. What is the temperature inside the material? The solution to this problem is described by Laplace’s equation, one of the most ubiquitous partial differential equations in physics. Let $u(x,y)$ denote the temperature of the material at point $(x,y)$ . Laplace’s equation states that, at any point $(x,y)$ on the interior of the material,

(1) $\begin{equation*} \frac{\partial^2}{\partial x^2} u(x,y) + \frac{\partial^2}{\partial y^2} u(x,y) = 0. \end{equation*}$

Laplace’s equation (1) and the specification of the temperature on the boundary form a well-posed mathematical problem in the sense that the temperature is uniquely determined at each point $(x,y)$ .¹ We call this problem the Laplace Dirichlet problem since the boundary conditions

$\begin{equation*} u(x,y) \quad \text{is specified for $(x,y)$ on the boundary} \end{equation*}$

are known as Dirichlet boundary conditions.

The Double Layer Potential

Another area of physics where the Laplace equation (1) appears is the study of electrostatics. In this case, $u(x,y)$ represents the electric potential at the point $(x,y)$ . The Laplace Dirichlet problem is to find the electric potential in the interior of the region with knowledge of the potential on the boundary.

The electrostatic application motivates a different way of thinking about the Laplace equation. Consider the following question:

How would I place electric charges on the boundary to produce the electric potential $u(x,y)$ at each point $(x,y)$ on the boundary?

This is a deliciously clever question. If I were able to find an arrangement of charges answering the question, then I could calculate the potential $u(x,y)$ at each point $(x,y)$ in the interior by adding up the contribution to the electric potential of each element of charge on the boundary. Thus, I can reduce the problem of finding the electric potential at each point $(x,y)$ in the 2D region to finding a charge distribution on the 1D boundary to that region.

We shall actually use a slight variant of this charge distribution idea which differs in two ways:

Rather than placing simple charges on the boundary of the region, we place charge dipoles.²
Since we are considering a two-dimensional problem, we use a different formula for the electric potential than given by Coulomb’s law for charges in 3D. Also, since we are interested in solving the Laplace Dirichlet problem in general, we can choose a convenient dimensionless system of units. We say that the potential at a point $(x,y)$ induced by a unit “charge” at the origin is given by $1/2\pi \cdot \ln \sqrt{x^2+y^2}$ .

With these modifications, our new question is as follows:

How would I place a density of “charge” dipoles $\phi(x,y)$ on the boundary to produce the electric potential $u(x,y)$ at each point $(x,y)$ on the boundary?

We call this function $\phi(x,y)$ the double layer potential for the Laplace Dirichlet problem. One can show the double layer potential satisfies a certain integral equation. To write down this integral equation, let’s introduce some more notation. Let $R$ be the region of interest and $\partial R$ its boundary. Denote points $(x,y)$ concisely as vectors $\mathbf{r} = (x,y)$ , with the length of $\mathbf{r}$ denoted $\|\mathbf{r}\| = \sqrt{x^2+y^2}$ . The double layer potential satisfies

(2) $\begin{equation*} \frac{1}{2} \phi(\mathbf{r}) + \frac{1}{2\pi} \int_{\partial R} \phi(\mathbf{r}') \frac{\partial}{\partial \nu_{\mathbf{r}'}} \ln \|\mathbf{r}-\mathbf{r}'\| \, dS(\mathbf{r}') = u(\mathbf{r}), \end{equation*}$

where the integral is taken over the surface $\partial R$ of the region $R$ ; $\partial / \partial \nu_{\mathbf{r}'}$ denotes the directional derivative taken in the direction normal (perpendicular) to the surface at the point $\mathbf{r}'$ . Note we choose a unit system for $\phi$ which hides physical constants particular to the electrostatic context, since we are interested in applying this methodology to the Laplace Dirichlet problem in general (possibly non-electrostatic) applications.

There’s one last ingredient: How do we compute the electric potential $u(\mathbf{r})$ at points $\mathbf{r}$ in the interior of the region? This is answered by the following formula:

(3) $\begin{equation*} u(\mathbf{r}) = \frac{1}{2\pi} \int_{\partial R} \phi(\mathbf{r}') \frac{\partial}{\partial \nu_{\mathbf{r}'}} \ln \|\mathbf{r}-\mathbf{r}'\| \, dS(\mathbf{r}'). \end{equation*}$

The integral equation (2) is certainly nothing to sneeze at. Rather than trying to comprehend it in its full glory, we shall focus on a special case for the rest of our discussion. Suppose the region $R$ is a circular disk with radius $r$ centered at $0$ . The the partial derivative in the integrand in (2) then is readily computed for points $\mathbf{r}$ and $\mathbf{r}'$ both on the boundary $\partial R$ of the circle:

$\begin{equation*} \frac{\partial}{\partial \nu_{\mathbf{r}'}} \ln \|\mathbf{r}-\mathbf{r}'\| = \frac{1}{2r}. \end{equation*}$

Substituting in (2) then gives

(4) $\begin{equation*} \frac{1}{2} \phi(\mathbf{r}) + \frac{1}{4\pi r} \int_{\partial R} \phi(\mathbf{r}') \, dS(\mathbf{r}') = u(\mathbf{r}). \end{equation*}$

The Sherman–Morrison Formula

We are interested in solving the integral equation (4) to obtain an expression for the double-layer potential $\phi$ , as this will give us a solution formula $u(\mathbf{r})$ for the Laplace Dirichlet problem. Ultimately, we accomplish this by using a clever trick. In an effort to make this trick seem more self-evident and less of a “rabbit out of a hat”, I want to draw an analogy to a seemingly unrelated problem: rank-one updates to linear systems of equations and the Sherman–Morrison formula.³ For a more general perspective on the Sherman–Morrison formula and its generalization to the Sherman–Morrison–Woodbury formula, you may be interested in the following post of mine on Schur complements and block Gaussian elimination.

Suppose we want to solve the system of linear equations

(5) $\begin{equation*} (A + uv^\top) x = b, \end{equation*}$

where $A$ is an $n\times n$ square matrix and $u$ , $v$ , and $b$ are length- $n$ vectors. We are ultimately interested in finding $x$ from $b$ . To gain insight into this problem, it will be helpful to first carefully considered the problem in reverse: computing $b$ from $x$ . We could, of course, perform this computation by forming the matrix $A + uv^\top$ in memory and multiplying it with $x$ , but there is a more economical way:

Form $\alpha = v^\top x$ .
Compute $b = \alpha u + Ax$ .

Standing back, observe that we now have a system of $n+1$ equations for unknowns $x$ and $\alpha$ . Specifically, our first equation can be rewritten as

$\begin{equation*} -\alpha + v^\top x = 0 \end{equation*}$

which combined with the second equation

$\begin{equation*} \alpha u + Ax = b \end{equation*}$

gives the $n+1$ by $n+1$ system⁴

(6) $\begin{equation*} \begin{bmatrix} - 1 & v^\top \\ u & A \end{bmatrix} \begin{bmatrix} \alpha \\ x \end{bmatrix} = \begin{bmatrix} 0 \\ b \end{bmatrix}. \end{equation*}$

The original equation for $x$ (5) can be derived from the “lifted” equation (6) by applying Gaussian elimination and eliminating the first row of the linear system (6). But now that we have the lifted equation (6), one can naturally wonder what would happen if we instead used Gaussian elimination to eliminate the last $n$ rows of (6); this will give us an equation for $\alpha = v^\top x$ which we can solved without first computing $x$ . Doing this so-called block Gaussian elimination yields

$\begin{equation*} (-1-v^\top A^{-1}u)\alpha = -v^\top A^{-1} b. \end{equation*}$

Solving this, we deduce that

$\begin{equation*} \alpha = \frac{v^\top A^{-1} b}{1 + v^\top A^{-1}u}. \end{equation*}$

From the equation $\alpha u + Ax = b$ , we have that

$\begin{equation*} x = (A+uv^\top)^{-1}b = A^{-1}b - \alpha A^{-1} u = A^{-1}b - \frac{A^{-1}uv^\top A^{-1} b}{1 + v^\top A^{-1}u}. \end{equation*}$

Since this formula holds for every vector $b$ , we deduce the famous Sherman–Morrison formula

$\begin{equation*} (A+uv^\top)^{-1}= A^{-1} - \frac{A^{-1}uv^\top A^{-1}}{1 + v^\top A^{-1}u}. \end{equation*}$

This example shows how it can be conceptually useful to lift a linear system of equations by adding additional variables and equations and then “do Gaussian elimination in a different order”. The same insight shall be useful in solving integral equations like (4).

Solving for the Double Layer Potential

Let’s try repeating the playbook we executed for the rank-one-updated linear system (5) and apply it to the integral equation (4). We are ultimately interested in computing $\phi(\cdot)$ from $u(\cdot)$ but, as we did last section, let’s first consider the reverse. To compute $u(\cdot)$ from $\phi(\cdot)$ , we first evaluate the integral

$\begin{equation*} \alpha = \int_{\partial R} \phi(\mathbf{r}') \, dS(\mathbf{r}'). \end{equation*}$

Substituting this into (4) gives the system of equations

(7) $\begin{equation*} \frac{1}{4\pi r} \alpha + \frac{1}{2} \phi(\mathbf{r}) = u(\mathbf{r}), \end{equation*}$

(8) $\begin{equation*} -\alpha + \int_{\partial R} \phi(\mathbf{r}') \, dS(\mathbf{r}') = 0. \end{equation*}$

In order to obtain (4) from (7) and (8), we add $1/4\pi r$ times equation (8) to equation (7). Following last section, we now instead eliminate $\phi(\mathbf{r})$ from equation (8) using equation (7). To do this, we need to integrate equation (7) in order to cancel the integral in equation (8):

$\begin{equation*} \frac{1}{4\pi r} \alpha \underbrace{\int_{\partial R} \, dS(\mathbf{r}')}_{=2\pi r} + \frac{1}{2} \int_{\partial R} \phi(\mathbf{r}') \, dS(\mathbf{r}') = \frac{1}{2} \alpha + \frac{1}{2} \int_{\partial R} \phi(\mathbf{r}') \, dS(\mathbf{r}') = \int_{\partial R} u(\mathbf{r}') \, dS(\mathbf{r}'). \end{equation*}$

Adding $2$ times this integrated equation to equation (8) yields

$\begin{equation*} 2\alpha = 2\int_{\partial R} u(\mathbf{r}') \, dS(\mathbf{r}') \implies \alpha = \int_{\partial R} u(\mathbf{r}') \, dS(\mathbf{r}'). \end{equation*}$

Thus plugging this expression for $\alpha$ into equation (7) yields

$\begin{equation*} \phi(\mathbf{r}) = 2u(\mathbf{r}) - \frac{1}{2\pi r} \alpha = 2u(\mathbf{r}) - \frac{1}{2\pi r} \int_{\partial R} u(\mathbf{r}') \, dS(\mathbf{r}'). \end{equation*}$

We’ve solved for our double layer potential!

As promised, the double layer potential can be used to give a solution formula (known as the Poisson integral formula) for the Laplace Dirichet problem. The details are a mechanical, but also somewhat technical, exercise in vector calculus identities. We plug through the details in the following extra section.

Poisson Integral Formula

Let’s finish this up by using the double layer to derive a solution formula for the electric potential $u(\mathbf{r})$ at a point $\mathbf{r}$ in the interior of the region. To do this, we use equation (3):

(9) $\begin{align*} u(\mathbf{r}) &= \frac{1}{2\pi} \int_{\partial R} \left( 2u(\mathbf{r}') - \frac{1}{2\pi r} \int_{\partial R} u(\mathbf{r}'') \, dS(\mathbf{r}'') \right) \frac{\partial}{\partial \nu_{\mathbf{r}'}} \ln \|\mathbf{r}-\mathbf{r}'\| \, dS(\mathbf{r}') \\ &= \frac{1}{\pi} \int_{\partial R} u(\mathbf{r}')\frac{\partial}{\partial \nu_{\mathbf{r}'}} \ln \|\mathbf{r}-\mathbf{r}'\| \, dS(\mathbf{r}') - \frac{1}{4\pi^2 r} \int_{\partial R} \frac{\partial}{\partial \nu_{\mathbf{r}'}} \ln\|\mathbf{r}-\mathbf{r}'\| \, dS(\mathbf{r}')\int_{\partial R} u(\mathbf{r}') \, dS(\mathbf{r}'). \end{align*}$

We now need to do a quick calculation which is somewhat technical and not particularly enlightening. We evaluate $-\frac{1}{4\pi}\int_{\partial R} \frac{\partial}{\partial \nu_{\mathbf{r}'}} \frac{1}{|\mathbf{r}-\mathbf{r}'|} \, dS(\mathbf{r}')$ using the divergence theorem:

$\begin{align*} \frac{1}{2\pi}\int_{\partial R} \frac{\partial}{\partial \nu_{\mathbf{r}'}} \ln \|\mathbf{r}-\mathbf{r}'\| \, dS(\mathbf{r}') &= \frac{1}{2\pi}\int_{\partial R} \nu_{\mathbf{r}'}\cdot \nabla_{\mathbf{r}'} \ln \|\mathbf{r}-\mathbf{r}'\| \, dS(\mathbf{r}') = \frac{1}{2\pi}\int_R \nabla_{\mathbf{r}'}^2 \ln\|\mathbf{r}-\mathbf{r}'\| \, d\mathbf{r}' \\ &= \int_R \delta(\mathbf{r}'-\mathbf{r}) \, d\mathbf{r}' = 1. \end{align*}$

We denote $\nabla$ for the gradient, $\nu_{\mathbf{r}'}$ for the normal vector to $\partial R$ at the point $\mathbf{r}'$ , $\cdot$ for the dot product, $\nabla^2$ for the Laplace operator, and $\delta$ for the Dirac delta “function”. The last equality holds because the function $v(\mathbf{r}') = \tfrac{1}{2\pi} \ln \| \mathbf{r} - \mathbf{r}'\|$ is a so-called fundamental solution for Laplace equation in the sense that $\nabla_{\mathbf{r}'}^2 v(\mathbf{r}') = \delta(\mathbf{r}'-\mathbf{r})$ . Therefore, (9) simplifies to

$\begin{equation*} \mathbf{u}(\mathbf{r}) = \frac{1}{\pi} \int_{\partial R} u(\mathbf{r}')\frac{\partial}{\partial \nu_{\mathbf{r}'}} \ln \|\mathbf{r}-\mathbf{r}'\| \, dS(\mathbf{r}') - \frac{1}{2\pi r} \int_{\partial R} u(\mathbf{r}') \, dS(\mathbf{r}'). \end{equation*}$

Computing the boundary derivative for the spherical region centered at the origin with radius $r$ , we obtain the formula

$\begin{equation*} \mathbf{u}(\mathbf{r}) = \frac{1}{2\pi r} \int_{\partial R} \left( 2\frac{r^2 - \mathbf{r} \cdot \mathbf{r}'}{\|\mathbf{r} - \mathbf{r}'\|^2} - 1 \right) u(\mathbf{r}') \, dS(\mathbf{r}') = \frac{r^2 - \| \mathbf{r} \|^2}{2\pi r} \int_{\partial R} \frac{u(\mathbf{r}')}{\|\mathbf{r} - \mathbf{r}'\|^2} \, dS(\mathbf{r}'). \end{equation*}$

We’ve succeeded at deriving a solution formula for $u(\mathbf{r})$ for points $\mathbf{r}$ in the interior of the disk in terms of $u(\mathbf{r}')$ for points $\mathbf{r}'$ on the boundary of the disk. This is known as the Poisson integral formula for the disk in two dimensions. This formula can be generalized to balls in higher dimensions, though this proof technique using “Sherman–Morrison” fails to work in more than two dimensions.

Sherman–Morrison for Integral Equations

Having achieved our main goal of deriving a solution formula for the 2D Laplace Dirichlet problem for a circular domain, I want to take a step back to present the approach from two sections ago in more generality. Consider a more general integral equation of the form

(10) $\begin{equation*} a \, \phi(\mathbf{x}) + \int_\Omega K(\mathbf{x},\mathbf{y}) \phi(\mathbf{y}) \, d\mathbf{y} = f(\mathbf{x}), \quad \mathbf{x} \in \Omega, \end{equation*}$

where $\Omega$ is some region in space, $K(\cdot,\cdot)$ , $f(\cdot)$ , and $\phi(\cdot)$ are functions of one or two arguments on $\Omega$ , and $a\ne 0$ is a nonzero constant. Such an integral equation is said to be of the second kind. The integral equation for the Laplace Dirichlet problem (2) is of this form with $\Omega = \partial R$ , $a = 1/2$ , $K(\mathbf{x},\mathbf{y}) = \tfrac{\partial}{\partial \nu_{\mathbf{y}}} \ln \|\mathbf{x} - \mathbf{y}\|$ , and $f(\mathbf{x}) = u(\mathbf{x})$ . We say the kernel $K(\cdot,\cdot)$ is separable with rank $k$ if $K(\cdot,\cdot)$ can be expressed in the form

$\begin{equation*} K(\mathbf{x},\mathbf{y}) = g_1(\mathbf{x})h_1(\mathbf{y}) + g_2(\mathbf{x})h_2(\mathbf{y}) + \cdots + g_k(\mathbf{x})h_k(\mathbf{y}). \end{equation*}$

With the circular domain, the Laplace Dirichlet integral equation (2) is separable with rank $k = 1$ .⁵ We shall focus on the second kind integral equation (10) assuming the kernel is separable with rank $1$ (for simplicity, we set $a = 1$ ):

(11) $\begin{equation*} \phi(\mathbf{x}) + g(\mathbf{x}) \int_\Omega h(\mathbf{y}) \phi(\mathbf{y}) \, d\mathbf{y} = f(\mathbf{x}), \quad \mathbf{x} \in \Omega. \end{equation*}$

Let’s try and write this equation in a way that’s more similar to the linear system of equation (5). To do this, we make use of linear operators defined on functions:

Let $\operatorname{Id}$ denote the identity operator on functions: It takes as inputs function $\phi(\cdot)$ and outputs the function $\phi(\cdot)$ unchanged.
Let $I_h$ denote the “integration against $h$ operator”: It takes as input a function $\phi(\cdot)$ and outputs the number $\int_\Omega h(\mathbf{y})\phi(\mathbf{y}) \, d\mathbf{y}$ .

With these notations, equation (11) can be written as

$\begin{equation*} (\operatorname{Id} + g I_h) \phi = f. \end{equation*}$

Using the same derivation which led to the Sherman–Morrison formula for linear systems of equations, we can apply the Sherman–Morrison formula to this integral equation in operator form, yielding

$\begin{equation*} \phi = \left( \operatorname{Id}^{-1} - \frac{\operatorname{Id}^{-1}g I_h \operatorname{Id}^{-1}}{1 + I_h \operatorname{Id}^{-1} g} \right)f = f - \frac{g (I_h f)}{1+I_h g}. \end{equation*}$

Therefore, the solution to the integral equation (11) is

$\begin{equation*} \phi(\mathbf{x}) = f(\mathbf{x}) - \frac{\int_\Omega h(\mathbf{y}) f(\mathbf{y}) \, d\mathbf{y}}{1 + \int_\Omega h(\mathbf{y})g(\mathbf{y}) \, d\mathbf{y} } g(\mathbf{x}). \end{equation*}$

This can be interpreted as a kind of Sherman–Morrison formula for the integral equation (11).

One can also generalize to provide a solution formula for the second-kind integral equation (10) for a separable kernel with rank $k$ ; in this case, the natural matrix analog is now the Sherman–Morrison–Woodbury identity rather than the Sherman–Morrison formula. Note that this solution formula requires the solution of a $k\times k$ system of linear equations. One can use this as a numerical method to solve second-kind integral equations: First, we approximate $K$ by a separable kernel of a modest rank $k$ and then compute the exact solution of the resulting integral equation with the approximate kernel.⁶

My goal in writing this post was to discuss two topics which are both near and dear to my heart, integral equations and the Sherman–Morrison formula. I find the interplay of these two ideas to be highly suggestive. It illustrates the power of the analogy between infinite-dimensional linear equations, like differential and integral equations, and finite-dimensional ones, which are described by matrices. Infinite dimensions certainly do have their peculiarities and technical challenges, but it can be very useful to first pretend infinite-dimensional linear operators (like integral operators) are matrices, do calculations to derive some result, and then justify these computations rigorously post hoc.⁷

The Vandermonde Decomposition

June 27, 2022 by Ethan N. Epperly Leave a comment

In this post, I want to discuss a beautiful and somewhat subtle matrix factorization known as the Vandermonde decomposition that appears frequently in signal processing and control theory. We’ll begin from the very basics, introducing the controls-and-signals context, how the Vandermonde decomposition comes about, and why it’s useful. By the end, I’ll briefly share how we can push the Vandermonde decomposition beyond matrices to the realm of tensors, which will can allow us to separate mixed signals from multiple measurements. This tensorial generalization plays an important role in my paper $(L_r,L_r,1)$ -decompositions, sparse component analysis, and the blind separation of sums of exponentials, joint work with Nithin Govindajaran and Lieven De Lathauwer, which recently appeared in the SIAM Journal of Matrix Analysis and Applications.

Finding the Frequencies

Suppose I give you a short recording of a musical chord consisting of three notes. How could you determine which three notes they were? Mathematically, we can represent such a three-note chord as a combination of scaled and shifted cosine functions

(1) $f(t) = a_0 \cos(\omega_0 t - \phi_0) + a_1 \cos(\omega_1 t - \phi_1) + a_2 \cos(\omega_2 t - \phi_2).$

We are interested in obtaining the (angular) frequencies $\omega_1$ , $\omega_2$ , and $\omega_3$ .

In the extreme limit, when we are given the values of the signal for all $t$ , both positive and negative, the frequencies are immediately given by taking a Fourier transform of the function $f(\cdot)$ . In practice, we only have access to the function $f$ at certain times $t_0,\ldots,t_{n-1}$ which we assume are equally spaced

$t_j = j\Delta t \quad \textnormal{for} \quad j = 0,1,2,\ldots,n-1.$

Given the samples

$f_j = f(t_j) \quad \textnormal{for} \quad j = 0,1,2,\ldots,n-1$

we could try to identify $\omega_1$ , $\omega_2$ , and $\omega_3$ using a discrete Fourier transform.¹ Unfortunately, this generally requires a large number of samples to identify $\omega_1$ , $\omega_2$ , and $\omega_3$ accurately. (The accuracy scales roughly like $1/n$ , where $n$ is the number of samples.) We are interested in finding a better way to identify the frequencies.

Now that we’ve moved from the function $f(\cdot)$ , defined for any real input $t$ , to a set of samples $f_0,f_1,\ldots,f_{n-1}$ it will be helpful to rewrite our formula (1) for $f$ in a different way. By Euler’s identity, the cosines can be rewritten as

$\cos \alpha = \frac{\mathrm{e}^{\mathrm{i} \alpha}+\mathrm{e}^{-\mathrm{i} \alpha}}{2}.$

As a consequence, we can rewrite one of the frequency components in (1) as

$a_0 \cos(\omega_0 t - \phi_0) = d_0 \mathrm{e}^{\mathrm{i} \omega_0t} + d_1 \mathrm{e}^{-\mathrm{i} \omega_0t}.$

Here, $d_0$ and $d_1$ are complex coefficients $d_0 = a_0 \mathrm{e}^{-\mathrm{i} \phi_0}/2$ and $d_1 = a_0 \mathrm{e}^{\mathrm{i} \phi_0}/2$ which contain the same information as the original parameters $a_0$ and $\phi_0$ . Now notice that we are only interest in values $t_j = j\, \Delta t$ which are multiples of the spacing $\Delta t$ . Thus, our frequency component can be further rewritten as

$a_0 \cos(\omega_0 t_j - \phi_0) = d_0 z_0^j + d_1 z_1^j$

where $z_0 := \mathrm{e}^{\mathrm{i} \omega_0\, \Delta t}$ and $z_1 := \mathrm{e}^{-\mathrm{i}\omega_0\, \Delta t}$ . Performing these reductions, our samples $f_j$ take the form

(2) $f_j = d_0 z_0^j + d_1 z_1^j + \cdots + d_5 z_5^j.$

We’ve now reformulated our frequency problems in identifying the parameters $d_0,\ldots,d_5$ and $z_0,\ldots,z_5$ in the relation (2) from a small number of measurements $f_0,f_1,\ldots,f_{n-1}$ .

Frequency Finding as a Matrix Factorization

We will return to the algorithmic problem of identifying the parameters in the relation (2) from measurements in a little bit. First, we will see that (2) can actually be written as a matrix factorization. Understanding computations by matrix factorization has been an extremely successful paradigm in applied mathematics, and we will see in this post how viewing (2) as a matrix factorization can be very useful.

While it may seem odd at first,² it will be illuminating to repackage the measurements $f_0,f_1,\ldots,f_{n-1}$ as a matrix:

(3) $H = \begin{bmatrix} f_0 & f_1 & f_2 & \cdots & f_{(n-1)/2} \\f_1 & f_2 & f_3 & \cdots & f_{(n-1)/2+1} \\f_2 & f_3 & f_4 & \cdots & f_{(n-1)/2+2} \\\vdots & \vdots & \vdots & \ddots & \vdots \\f_{(n-1)/2} & f{(n-1)/2+1} & f{(n-1)/2+2} & \cdots & f_{n-1} \end{bmatrix}.$

Here, we have assumed $n$ is odd. The matrix $H$ is known as the Hankel matrix associated with the sequence $f_0,\ldots,f_{n-1}$ . Observe that the entry in position $ij$ of $H$ depends only on the sum of the indices $i$ and $j$ , $H_{ij} = f_{i+j}$ . (We use a zero-indexing system to label the rows and columns of $H$ where, for instance, the first row of $H$ is row $0$ .)

Let’s see how we can interpret the frequency decomposition (2) as a factorization of the Hankel matrix $H$ . We first write out $H_{ij}$ using (2):

(4) $H_{ij} = f_{i+j} = \sum_{k=0}^5 d_k z_k^{i+j} = \sum_{k=0}^5 d_k z_k^i \cdot z_k^j.$

The power $z_k^{i+j}$ was just begging to be factorized as $z_k^i\cdot z_k^j$ , which we did. Equation (4) almost looks like the formula for the product of two matrices with entries $z_k^i$ , so it makes sense to introduce the $6\times (n-1)/2$ matrix $V$ with entry $V_{ki} = z_k^i$ . This is a so-called Vandermonde matrix associated with $z_0,\ldots,z_5$ and has the form

$V = \begin{bmatrix}z_0^0 & z_0^1 & z_0^2 & \cdots & z_0^{(n-1)/2} \\z_1^0 & z_1^1 & z_1^2 & \cdots & z_1^{(n-1)/2} \\\vdots & \vdots & \vdots & \ddots & \vdots \\z_5^0 & z_5^1 & z_5^2 & \cdots & z_5^{(n-1)/2}\end{bmatrix}.$

If we also introduce the $6\times 6$ diagonal matrix $D = \operatorname{diag}(d_0,d_1,\ldots,d_5)$ , the formula (4) for $H$ can be written as the matrix factorization³

(5) $H = V^\top D V.$

This is the Vandermonde decomposition of the Hankel matrix $H$ , a factorization of $H$ as a product of the transpose of a Vandermonde matrix, a diagonal matrix, and that same Vandermonde matrix.

The Vandermonde decomposition immediately tells us all the information $d_0,\ldots,d_5$ and $z_0,z_1,\ldots,z_5$ describing our sampled recording $f_0,\ldots,f_{n-1}$ via (2). Thus, the problem of determining $d_0,\ldots,d_5$ and $z_0,z_1,\ldots,z_5$ is equivalent to finding the Vandermonde decomposition (5) of the Hankel matrix $H$ .

Computing the Vandermonde Decomposition: Prony’s Method

Computing the Vandermonde decomposition accurately can be a surprisingly hard task, particularly if the measurements $f_0,f_1,\ldots,f_{n-1}$ are corrupted by even a small amount of measurement error. In view of this, I want to present a very classical way of computing this decomposition (dating back to 1795!) known as Prony’s method. This method is conceptually simple and will be a vehicle to continue exploring frequency finding and its connection with Hankel matrices. It remains in use, though it’s accuracy may be significantly worse compared to other methods.

As a first step to deriving Prony’s method, let’s reformulate the frequency finding problem in a different way. Sums of cosines like the ones in our expression (1) for the function $f(\cdot)$ often appear as the solution to a (linear) ordinary differential equation (ODE). This means that one way we could find the frequencies comprising $f(\cdot)$ would be to find a differential equation which $f(\cdot)$ satisfies. Together with the initial condition $f(0)$ , determining all the frequencies would be very straightforward.

Since we only have access to samples $f_0, f_1,\ldots,f_{n-1}$ of $f(\cdot)$ at regular time intervals, we will instead look for the “discrete-time” analog of a linear ODE, a linear recurrence relation. This is an expression of the form

(6) $f_m = c_{k-1} f_{m-1} + \cdots + c_1f_{m-k+1}+ c_0 f_{m-k} \quad \textnormal{for every} \quad m = k,\,{k+1},\,{k+2},\cdots.$

In our case, we’ll have $k = 6$ because there are six terms in the formula (2) for $f_j$ . Together with initial conditions $f_0,f_1,\ldots,f_{k-1}$ , such a recurrence will allow us to determine the parameters $z_0,\ldots,z_5$ and $d_0,\ldots,d_5$ in our formula (2) for our sampled recordings $f_0,\ldots,f_{n-1}$ and hence also allow us to compute the Vandermonde decomposition (5).

Observe that the recurrence (6) is a linear equation in the variables $c_0,\ldots,c_5$ . A very good rule of thumb in applied mathematics is to always write down linear equations in matrix–vector notation in see how it looks. Doing this, we obtain

(7) $\begin{bmatrix} f_6 \\ f_7 \\ \vdots \\ f_{n-1} \end{bmatrix} = \underbrace{\begin{bmatrix} f_0 & f_1 & f_2 & f_3 & f_4 & f_5 \\ f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ f_{n-7} & f_{n-6} & f_{n-5} & f_{n-4} & f_{n-3} & f_{n-2} \end{bmatrix}}_{=F}\begin{bmatrix} c_0 \\ c_1 \\ \vdots \\ c_5 \end{bmatrix}.$

Observe that the matrix on the right-hand side of this equation is also a Hankel matrix (like $H$ in (3)) formed from the samples $f_0,\ldots,f_{n-1}$ . Call this Hankel matrix $F$ . Unlike $H$ in (3), $F$ is rectangular. If $n$ is much larger than $6$ , $F$ will be tall, possessing many more rows than columns. We assume $n > 12$ going forward.⁴

Let’s write (7) a little more compactly as

(8) $f_{6 \, :\, n-1} = F c,$

where we’ve introduced $f_{6\,:\, n-1}$ for the vector on the left-hand side of (7) and collected the recurrence coefficients $c_0,\ldots,c_5$ into a vector $c$ . For a typical system of linear equations like (8), we would predict the system to have no solution $c$ : Because $F$ has more rows than columns (if $n > 12$ ), the system equations (8) has more equations than unknowns. Fortunately, we are not in the typical case. Despite the fact that we have more equations than unknowns, the linear equations (8) have a unique solution $c$ .⁵ The existence of a unique solution is a consequence of the fact that the samples $f_0,\ldots,f_{n-1}$ satisfy the formula (2). As a fun exercise, you might want to verify the existence of a unique $c$ satisfying (8)!

As a quick aside, if the measurements $f_0,\ldots,f_{n-1}$ are corrupted by small measurement errors, then the equations (8) will usually not possess a solution. In this case, it would be appropriate to find the least squares solution to equation (8) as a way of mitigating these errors.

Hurrah! We’ve found the coefficients $c_0,\ldots,c_5$ providing a recurrence relation (6) for our measurements $f_0,\ldots,f_{n-1}$ . All that is left is to find the parameters $z_0,\ldots,z_5$ and $d_0,\ldots,d_5$ in our signal formula (2) and the Vandermonde decomposition (5). Fortunately, this is just a standard computation for linear recurrence relations, nicely paralleling the solution of (homogenous) linear ODEs by means of the so-called “characteristic equation”. I’ll go through fairly quickly since this material is well-explained elsewhere on the internet (like Wikipedia). Let’s guess that our recurrence (6) has a solution of the form $f_j = z^j$ ; we seek to find all complex numbers $z$ for which this is a bonafide solution. Plugging this solution into the formula (6) for $f_6$ gives

(9) $z^6 = c_0 z^5 + c_1 z^4 + \cdots + c_6.$

This is the so-called characteristic equation for the recurrence (6). As a single-variable polynomial equation of degree six, it has six complex solutions $z_0,z_1,\ldots,z_5$ . These numbers $z_0,z_1,\ldots,z_5$ are precisely those numbers which appear in the sequence formula (2) and the Vandermonde decomposition (5).

Finally, we need to compute the coefficients $d_0,\ldots,d_5$ . But this is easy. Observe that the formula (2) provides the following system of linear equations for $d_0,\ldots,d_5$ :

(10) $\begin{bmatrix}f_0 \\ f_1 \\ \vdots \\ f_{n-1}\end{bmatrix} = \begin{bmatrix}1 & 1 & \cdots & 1 \\ z_0 & z_1 & \cdots & z_5 \\ \vdots & \vdots & \ddots & \vdots \\ z_0^{n-1} & z_1^{n-1} & \cdots & z_5^{n-1}\end{bmatrix} \begin{bmatrix}d_0 \\ d_1 \\ \vdots \\ d_{n-1}.\end{bmatrix}.$

Again, this system of equations will have a unique solution if the measurements $f_0,\ldots,f_{n-1}$ are uncorrupted by errors (and can be solved in the least squares sense if corrupted). This gives $d_0,\ldots,d_5$ , completing our goal of computing the parameters in the formula (2) or, equivalently, finding the Vandermonde decomposition (5).

We have accomplished our goal of computing the Vandermonde decomposition. The approach by which we did so is known as Prony’s method, as mentioned in the introduction to this section. As suggested, this method may not always give high-accuracy results. There are two obvious culprits that jump out about why this is the case. Prony’s method requires solving for the roots of the polynomial equation (9) expressed “in the monomial basis” and solving a system of linear equations (10) with a (transposed) Vandermonde matrix. Both of these problems can be notoriously ill-conditioned and thus challenging to solve accurately and may require the measurements $f_0,\ldots,f_{n-1}$ to be done to very high accuracy. Notwithstanding this, Prony’s method does useful results in some cases and forms the basis for potentially more accurate methods, such as those involving generalized eigenvalue problems.

Separating Signals: Extending the Vandermonde Decomposition to Tensors

In our discussion of the frequency identification problem, the Vandermonde decomposition (5) has effectively been an equivalent way of showing the samples $f_j$ are a combination of exponentials $z^j$ . So far, the benefits of the matrix factorization perspective have yet to really reveal themselves.

So what are the benefits of the Vandermonde decompostions? A couple of nice observations related to the Vandermonde decomposition and the “Hankelization” of the signals $H$ have already been lurking in the background. For instance, the rank of the Hankel matrix $H$ is the number of frequency components $z_k$ needed to describe the samples and the representation of the samples as a mixture of exponentials is uniquely determined only if the matrix $H$ does not have full rank; I have a little more to say about this at the very end. There are also benefits to certain computational problems; one can use Vandermonde decompositions to compute super high accuracy singular value decompositions of Hankel matrices.

The power of the Vandermonde decomposition really starts to shine when we go beyond the basic frequency finding problem we discussed by introducing more signals. Suppose now there are three short recordings $f^{(1)}(\cdot)$ , $f^{(2)}(\cdot)$ , and $f^{(3)}(\cdot)$ . (Here, the superscript denotes an index rather than differentiation.) Each signal is a weighted mixture of three sources $s^{(1)}(\cdot)$ , $s^{(2)}(\cdot)$ , and $s^{(3)}(\cdot)$ , each of which plays a musical chord of three notes (thus representable as a sum of cosines as in (1)). One can think of the sources of being produced three different musical instruments at different places in a room and the recordings $f^{(1)}(\cdot)$ , $f^{(2)}(\cdot)$ , and $f^{(3)}(\cdot)$ being taken from different microphones in the room.⁶ Our goal is now not just to identify the musical notes in the recordings but also to identify how to assign those notes to reconstruct the source signals $s^{(1)}(\cdot)$ , $s^{(2)}(\cdot)$ , and $s^{(3)}(\cdot)$ .

Taking inspiration from earlier, we record samples $f_0^{(\ell)},\ldots,f_{n-1}^{(\ell)}$ for each recording $\ell = 1,2,3$ and form each collection of samples into a Hankel matrix

$H^{(\ell)} = \begin{bmatrix} f_0^{(\ell)} & f_1^{(\ell)} & f_2^{(\ell)} & \cdots & f_{(n-1)/2}^{(\ell)} \\f_1^{(\ell)} & f_2^{(\ell)} & f_3^{(\ell)} & \cdots & f_{(n-1)/2+1}^{(\ell)} \\f_2^{(\ell)} & f_3^{(\ell)} & f_4^{(\ell)} & \cdots & f_{(n-1)/2+2}^{(\ell)} \\\vdots & \vdots & \vdots & \ddots & \vdots \\f_{(n-1)/2}^{(\ell)} & f_{(n-1)/2+1}^{(\ell)} & f_{(n-1)/2+2}^{(\ell)} & \cdots & f_{n-1}^{(\ell)} \end{bmatrix}.$

Here comes the crazy part: Stack the Hankelized recordings $H^{(1)}$ , $H^{(2)}$ , and $H^{(3)}$ as slices of a tensor $\mathcal{H}$ . A tensor, in this context, just means a multidimensional array of numbers. Just as a vector is a one-dimensional array and a matrix is a two-dimensional array, a tensor could have any number of dimensions. In our case, we need just three. If we use a MATLAB-esque indexing notation, $\mathcal{H}$ is a $(n-1)/2\times (n-1)/2 \times 3$ array given by

$\mathcal{H}(:,:,\ell) = H^{(\ell)} \quad \textnormal{for} \quad \ell=1,2,3.$

The remarkable thing is that the source signals can be determined (under appropriate conditions) by computing a special kind of Vandermonde decomposition of the tensor $\mathcal{H}$ ! (Specifically, the required decomposition is a Vandermonde-structured $(L_r,L_r,1)$ -block term decomposition of the tensor $\mathcal{H}$ .) Even more cool, this decomposition can be computed using general-purpose software like Tensorlab.

If this sounds interesting, I would encourage you to check out my recently published paper $(L_r,L_r,1)$ -decompositions, sparse component analysis, and the blind separation of sums of exponentials, joint work with Nithin Govindajaran and Lieven De Lathauwer and recently published in the SIAM Journal on Matrix Analysis and Applications. In the paper, we explain what this $(L_r,L_r,1)$ -decomposition is and how applying it to $\mathcal{H}$ can be used to separate mixtures of exponentials signals from the resulting Vandermonde structure, an idea originating in the work of De Lathauewer. A very important question for these signal separation problems is that of uniqueness. Given the three sampled recordings (comprising the tensor $\mathcal{H}$ ), is there just one way of unscrambling the mixtures into different sources or multiple? If there are multiple, then we might have possibly computed the wrong one. If there is just a single unscrambling, though, then we’ve done our job and unmixed the scrambled signals. The uniqueness of these tensor decompositions is fairly complicated math, and we survey existing results and prove new ones in this paper.⁷

Conclusions, Loose Ends, and Extensions

The central idea that we’ve been discussing is how it can be useful to convert between a sequence of observations $f_0,f_1,\ldots,f_{n-1}$ and a special matricization of this sequence into a Hankel matrix (either square, as in (3), or rectangular, as in (7)). By manipulating the Hankel matrix, say, by computing its Vandermonde decomposition (5), we learn something about the original signal, namely a representation of the form (2).

This is a powerful idea which appears implicitly or explicitly throughout various subfields of mathematics, engineering, and computation. As with many other useful ideas, this paradigm admits many natural generalizations and extensions. We saw one already in this post, where we extended the Vandermonde decomposition to the realm of tensors to solve signal separation problems. To end this post, I’ll place a few breadcrumbs at the beginning of a few of the trails of these generalizations for any curious to learn more, wrapping up a few loose ends on the way.

Is the Vandermonde Decomposition Unique?

A natural question is whether the Vandermonde decomposition (5) is unique. That is, is it possible that there exists two Vandermonde decompositions

$H = V^\top DV = \tilde{V}^\top \tilde{D} \tilde{V}$

of the same (square) Hankel matrix $H$ ? This is equivalent to whether the frequency components $z_0,z_1,\ldots$ can be uniquely determined from the measurements $f_0,f_1,\ldots,f_{n-1}$ .

Fortunately, the Vandermonde decomposition is unique if (and only if) the matrix $H$ is a rank-deficient matrix. Let’s unpack this a little bit. (For those who could use a refresher on rank, I have a blog post on precisely this topic.) Note that the Vandermonde decomposition is a rank factorization⁸ since $V$ has $\rank H$ rows, $V$ has full (row) rank, and $D$ is invertible. This means that if take enough samples $f_0,\ldots,f_{n-1}$ of a function $f(\cdot)$ which is a (finite) combinations of exponentials, the matrix $H$ will be rank-deficient and the Vandermonde decomposition unique.⁹ If too few samples are taken, then $H$ does not contain enough information to determine the frequency components of the signal $f(\cdot)$ and thus the Vandermonde decomposition is non-unique.

Does Every Hankel Matrix Have a Vandermonde Decomposition?

This post has exclusively focused on a situation where we are provided with sequence we know to be represented as a mixture of exponentials (i.e., taking the form (2)) from which the existence of the Vandermonde decomposition (5) follows immediately. What if we didn’t know this were the case, and we were just given a (square) Hankel matrix $H$ . Is $H$ guaranteed to possess a Vandermonde decomposition of the form (5)?

Unfortunately, the answer is no; there exist Hankel matrices which do not possess a Vandermonde decomposition. The issue is related to the fact that the appropriate characteristic equation (analogous to (9)) might possess repeated roots, making the solutions to the recurrence (6) not just take the form $z^j$ but also $jz^j$ and perhaps $j^2z^j$ , $j^3z^j$ , etc.

Are There Cases When the Vandermonde Decomposition is Guaranteed To Exist?

There is one natural case when a (square) Hankel matrix is guaranteed to possess a Vandermonde decomposition, namely when the matrix is nonsingular/invertible/full-rank. Despite this being a widely circulated fact, I am unaware of a simple proof for why this is the case. Unfortunately, there is not just one but infinitely many Vandermonde decompositions for a nonsingular Hankel matrix, suggesting these decompositions are not useful for the frequency finding problem that motivated this post.

What If My Hankel Matrix Does Not Possess a Vandermonde Decomposition?

As discussed above, a Hankel matrix may fail to have a Vandermonde decomposition if the characteristic equation (a la (9)) has repeated roots. This is very much analogous to the case of a non-diagonalizable matrix for which the characteristic polynomial has repeated roots. In this case, while diagonalization is not possible, one can “almost-diagonalize” the matrix by reducing it to its Jordan normal form. In total analogy, every Hankel matrix can be “almost Vandermonde decomposed” into a confluent Vandermonde decomposition (a discovery that appears to have been made independently several times). I will leave these links to discuss the exact nature of this decomposition, though I warn any potential reader that these resources introduce the decomposition first for Hankel matrices with infinitely many rows and columns before considering the finite case as we have. One is warned that while the Vandermonde decomposition is always a rank decomposition, the confluent Vandermonde decomposition is not guaranteed to be one.¹⁰

The Toeplitz Vandermonde Decomposition

Just as it proved useful to arrange samples $f_0,\ldots, f_{n-1}$ into a Hankel matrix, it can also be useful to form them into a Toeplitz matrix

$T = \begin{bmatrix} f_{(n-1)/2} & f_{(n-1)/2+1} & f_{(n-1)/2+2} & \cdots & f_{n-1} \\\\f_{(n-1)/2-1} & f_{(n-1)/2} & f_{(n-1)/2+1} & \cdots & f_{n-2} \\\\f_{(n-1)/2-2} & f_{(n-1)/2-1} & f_{(n-1)/2} & \cdots & f_{n-3} \\\\\vdots & \vdots & \vdots & \ddots & \vdots \\\\ f_0 & f_1 & f_2 & \cdots & f_{(n-1)/2} \end{bmatrix}.$

The Toeplitz matrix $T$ has the appealing propery that the matrix–vector product $Tx$ computes a (discrete) convolution of the sampled signal $f$ with the sampled signal $x$ which has all sorts of uses in signal processing and related fields.¹¹

One can interconvert between Hankel and Toeplitz matrices by reversing the order of the rows. As such, to the extent to which Hankel matrices possess Vandermonde decompositions (with all the asterisks and fine print just discussed), Toeplitz matrices do as well but with the rows of the first factor reversed:

$T = \operatorname{ReversedRows}(V^\top) \cdot DV.$

There is a special and important case where more is true. If a Toeplitz matrix is (Hermitian) positive semidefinite, then $T$ always possesses a Vandermonde decomposition of the form

$T = V^* D V,$

where $V$ is a Vandermonde matrix associated with parameters $z_0,z_1,\ldots,z_{k-1}$ which are complex numbers of absolute value one and $D$ is a diagonal matrix with real positive entries.¹² This Vandermonde decomposition is unique if and only if $T$ is rank-deficient. Positive semidefinite Toeplitz matrices are important as they occur as autocorrelation matrices which effectively describe the similarity between a signal and different shifts of itself in time. Autocorrelation matrices appear under different names in everything from signal processing to random processes to near-term quantum algorithms (a topic near and dear to my heart). A delightfully simple and linear algebraic derivation of this result is given by Yang and Xie (see Theorem 1).¹³ This decomposition can be generalized to infinite positive semidefinite Toeplitz matrices (appropriately defined).¹⁴

The Elegant Geometry of Generalized Eigenvalue Perturbation Theory

December 14, 2021 by Ethan N. Epperly 1 Comment

In this post, I want to discuss a beautiful and simple geometric picture of the perturbation theory of definite generalized eigenvalue problems. As a culmination, we’ll see a taste of the beautiful perturbation theory of Mathias and Li, which appears to be not widely known in virtue of only being explained in a technical report. Perturbation theory for the generalized eigenvalue problem is a bit of a niche subject, but I hope you will stick around for some elegant arguments. In addition to explaining the mathematics, I hope this post serves as an allegory for the importance of having the right way of thinking about a problem; often, the solution to a seemingly unsolvable problem becomes almost self-evident when one has the right perspective.

What is a Generalized Eigenvalue Problem?

This post is about the definite generalized eigenvalue problem, so it’s probably worth spending a few words talking about what generalized eigenvalue problems are and why you would want to solve them. Slightly simplifying some technicalities, a generalized eigenvalue problem consists of finding nonzero vectors $x$ and a (possibly complex) numbers $\lambda$ such that $Ax = \lambda \, Bx$ .¹ The vector $x$ is called an eigenvector and $\lambda$ its eigenvalue. For our purposes, $A$ and $B$ will be real symmetric (or even complex Hermitian) matrices; one can also consider generalized eigenvalue problemss for nonsymmetric and even non-square matrices $A$ and $B$ , but the symmetric case covers many applications of practical interest. The generalized eigenvalue problem is so-named because it generalizes the standard eigenvalue problem $Ax = \lambda x$ , which is a special case of the generalized eigenvalue problem with $B = I$ .²

Why might we want so solve a generalized eigenvalue problem? The applications are numerous (e.g., in chemistry, quantum computation, systems and control theory, etc.). My interest in perturbation theory for generalized eigenvalue problems arose in the analysis of a quantum algorithm for eigenvalue problems in chemistry, and the theory discussed in this article played a big role in that analysis. To give an application which is more easy to communicate than the quantum computation application which motivated my own interest, let’s discuss an application in classical mechanics.

The Lagrangian formalism is a way of reformulating Newton’s laws of motion in a general coordinate system.³ If $q$ denotes a vector of generalized coordinates describing our system and $\dot{q}$ denotes $q$ ‘s time derivative, then the time evolution of a system with Lagrangian functional $L(q,\dot{q})$ are given by the Euler–Lagrange equations $\tfrac{d}{dt} \nabla_{\dot{q}} L = \nabla_q L$ . If we choose $q$ to represent the deviation of our system from equilibrium,⁴ then our Lagrangian is well-approximated by it’s second order Taylor series:

$\begin{equation*} L(q,\dot{q}) \approx L_0 + \frac{1}{2} q^\top A q - \frac{1}{2} \dot{q}^\top B\dot{q}. \end{equation*}$

By the Euler–Lagrange equations, the equations of motion for small deviations from this equillibrium point are described by

$\begin{equation*} B\ddot{q} = -Aq. \end{equation*}$

A fundamental set of solutions of this system of differential equations is given by $\mathrm{e}^{\pm \sqrt{-\lambda} \, t} x$ , where $\lambda$ and $x$ are the generalized eigenvalues and eigenvectors of the pair $(A,B)$ .⁵ In particular, if all the generalized eigenvalues are positive, then the equillibrium is stable and the square roots of the eigenvalues represent the modes of vibration. In the most simple mechanical systems, such as masses in one dimension connected by springs with the natural coordinate system, the matrix $B$ is diagonal with diagonal entries equal to the different masses. In even slightly more complicated “freshman physics” problems, it is quite easy to find examples where, in the natural coordinate system, the matrix $B$ is nondiagonal.⁶ As this example shows, generalized eigenvalue problems aren’t crazy weird things since they emerge as natural descriptions of simple mechanical systems like coupled pendulums.

One reason generalized eigenvalue problems aren’t more well-known is that one can easily reduce a generalized eigenvalue problem into a standard one. If the matrix $B$ is invertible, then the generalized eigenvalues of $(A,B)$ are just the eigenvalues of the matrix $B^{-1}A$ . For several reasons, this is a less-than-appealing way of reducing a generalized eigenvalue problem to a standard eigenvalue problem. A better way, appropriate when $A$ and $B$ are both symmetric and $B$ is positive definite, is to reduce the generalized eigenvalue problem for $(A,B)$ to the symmetrically reduced matrix $B^{-1/2}AB^{-1/2}$ , which also possesses the same eigenvalues as $(A,B)$ . In particular, the matrix $B^{-1/2}AB^{-1/2}$ remains symmetric, which shows that $(A,B)$ has real eigenvalues by the spectral theorem. In the mechanical context, one can think of this reformulation as a change of coordinate system in which the “mass matrix” $B$ becomes the identity matrix $I$ .

There are several good reasons to not simply reduce a generalized eigenvalue problem to a standard one, and perturbation theory gives a particular good reason. In order for us to change coordinates to change the $B$ matrix into an identity matrix, we must first know the $B$ matrix. If I presented you with an elaborate mechanical system which you wanted to study, you would need to perform measurements to determine the $A$ and $B$ matrices. But all measurements are imperfect and the entries of $A$ and $B$ are inevitably corrupted by measurement errors. In the presence of these measurement errors, we must give up on computing the normal modes of vibration perfectly; we must content ourselves with computing the normal modes of vibration plus-or-minus some error term we hope we can be assured is small if our measurement errors are small. In this setting, reducing the problem to $B^{-1/2}AB^{-1/2}$ seems less appealing, as I have to understand how the measurement errors in $A$ and $B$ are amplified in computing the triple product $B^{-1/2}AB^{-1/2}$ . This also suggests that computing $B^{-1/2}AB^{-1/2}$ may be a poor algorithmic strategy in practice: if the matrix $B$ is ill-conditioned, there might be a great deal of error amplification in the computed product $B^{-1/2}AB^{-1/2}$ . One might hope that one might be able to devise algorithms with better numerical stability properties if we don’t reduce the matrix pair $(A,B)$ to a single matrix. This is not to say that reducing a generalized eigenvalue problem to a standard one isn’t a useful tool—it definitely is. However, it is not something one should do reflexively. Sometimes, a generalized eigenvalue problem is best left as is and analyzed in its native form.

The rest of this post will focus on the question if $A$ and $B$ are real symmetric matrices (satisfying a definiteness condition, to be elaborated upon below), how do the eigenvalues of the pair $(A+E,B+F)$ compare to those of $(A,B)$ , where $E$ and $F$ are small real symmetric perturbations? In fact, it shall be no additional toil to handle the complex Hermitian case as well while we’re at it, so we shall do so. (Recall that a Hermitian matrix $A$ satisfies $A^* = A$ , where $(\cdot)^*$ is the conjugate transpose. Since the complex conjugate does not change a real number, a real Hermitian matrix is necesarily symmetric $A^* = A^\top = A$ .) For the remainder of this post, let $A$ , $B$ , $E$ , and $F$ be Hermitian matrices of the same size. Let $\widetilde{A} := A+E$ and $\widetilde{B} := B + F$ denote the perturbations.

Symmetric Treatment

As I mentioned at the top of this post, our mission will really be to find the right way of thinking about perturbation theory for the generalized eigenvalue problem, after which the theory will follow much more directly than if we were to take a frontal assault on the problem. As we go, we shall collect nuggets of insight, each of which I hope will follow quite naturally from the last. When we find such an insight, we shall display it on its own line.

The first insight is that we can think of the pair $A$ and $B$ interchangeably. If $\lambda$ is a nonzero eigenvalue of the pair $(A,B)$ , satisfying $Ax = \lambda\, Bx$ , then $Bx = \lambda^{-1} \, Ax$ . That is, $\lambda^{-1}$ is an eigenvalue of the pair $(B,A)$ . Lack of symmetry is an ugly feature in a mathematical theory, so we seek to smooth it out. After thinking a little bit, notice that we can phrase the generalized eigenvalue condition symmetrically as $\beta \, Ax = \alpha \, Bx$ with the associated eigenvalue being given by $\lambda = \alpha/\beta$ . This observation may seem trivial at first, but let us collect it for good measure.

Treat $A$ and $B$ symmetrically by writing the eigenvalue as $\lambda = \alpha/\beta$ with $\beta\, Ax = \alpha\, Bx$ .

Before proceeding, let’s ask a question that, in our new framing, becomes quite natural: what happens when $\beta = 0$ ? The case $\beta = 0$ is problematic because it leads to a division by zero in the expression $\lambda = \alpha/\beta$ . However, if we have $\alpha \ne 0$ , this expression still makes sense: we’ve found a vector $x$ for which $Ax \ne 0$ but $Bx = 0$ . It makes sense to consider $x$ still an eigenvector of $(A,B)$ with eigenvalue $\alpha / 0 = \infty$ ! Dividing by zero should justifiably make one squeemish, but it really is quite natural in the case to treat $x$ as a genuine eigenvector with eigenvalue $\infty$ .

Things get even worse if we find a vector $x$ for which $Ax = Bx = 0$ . Then, any $(\alpha,\beta)$ can reasonably considered an eigenvalue of $(A,B)$ since $\alpha \, Bx = 0 = \beta\, Ax$ . In such a case, all complex numbers are simultaneously eigenvalues of $(A,B)$ , in which case we call $(A,B)$ singular.⁷ For the generalized eigenvalue problem to make sense for a pair $(A,B)$ , it is natural to require that $(A,B)$ not be singular. In fact, we shall assume an even stronger “definiteness” condition which ensures that $(A,B)$ has only real (or infinite) eigenvalues. Let us return to this question of definiteness in a moment and for now assume that $(A,B)$ is not singular and possesses real eigenvalues.

With this small aside taken care of, let us return to the main thread. By modeling eigenvalues as pairs $(\alpha,\beta)$ , we’ve traded one ugliness for another. While reformulating the eigenvalue as a pair $(\alpha,\beta)$ treats $A$ and $B$ symmetrically, it also adds an additional indeterminacy, scale. For instance, if $(\alpha,\beta)$ is an eigenvalue of $(A,B)$ , then so is $(10\alpha,10\beta)$ . Thus, it’s better not to think of $(\alpha,\beta)$ so much as a pair of numbers together with all of its possible scalings.⁸ For reasons that shall hopefully become more clear as we go forward, it will be helpful to only consider all the possible positive scalings of $(\alpha,\beta)$ —e.g., all $(t\alpha,t\beta)$ for $t > 0$ . Geometrically, the set of all positive scalings of a point in two-dimensional space is precisely just a ray emanating from the origin.

Represent eigenvalue pairs $(\alpha,\beta)$ as rays emanating from the origin to account for scale ambiguity.

Now comes a standard eigenvalue trick. It’s something you would never think to do originally, but once you see it once or twice you learn to try it as a matter of habit. The trick: multiply the eigenvalue-eigenvector relation by the (conjugate) transpose of $x$ :⁹

$\begin{equation*} \beta \, Ax = \alpha \, Bx \implies \beta \, x^*Ax = \alpha \, x^*Bx \implies \frac{\alpha}{\beta} = \frac{x^*Ax}{x^*Bx}. \end{equation*}$

The above equation is highly suggestive: since $\alpha$ and $\beta$ are only determined up to a scaling factor, it shows we can take $\alpha = x^*Ax$ and $\beta = x^*Bx$ . And by different scalings of the eigenvector $x$ , we can scale $x^*Ax = \alpha$ and $x^*Bx = \beta$ by any positive factor we want. (This retroactively shows why it makes sense to only consider positive scalings of $\alpha$ and $\beta$ .¹⁰) The expression $x^*Ax$ is so important that we give it a name: the quadratic form (associated with $A$ and evaluated at $x$ ).

The eigenvalue pair $(\alpha,\beta)$ can be taken equal to the pair of quadratic forms $(x^*Ax,x^*Bx)$ .

Complexifying Things

Now comes another standard mathematical trick: represent points in two-dimensional space by complex numbers. In particular, we identify the pair $(\alpha,\beta)$ with the complex number $\alpha + \mathrm{i}\beta$ .¹¹ Similar to the previous trick, it’s not fully clear why this will pay off, but let’s note it as an insight.

Identify the pair $(\alpha,\beta)$ with the complex number $\alpha + \mathrm{i}\beta$ .

Now, we combine all the previous observations. The eigenvalue $\lambda = x^*Ax / x^*Bx$ is best thought of as a pair $(\alpha,\beta)$ which, up to scale, can be taken to be $\alpha = x^*Ax$ and $\beta = x^*Bx$ . But then we represent $(\alpha,\beta)$ as the complex number

$\begin{equation*} \alpha + \mathrm{i} \beta = x^*Ax + \mathrm{i} x^*Bx = x^*(A + \mathrm{i} B) x. \end{equation*}$

Let’s stop for a moment and appreciate how far we’ve come. The generalized eigenvalue problem $Ax = \lambda\, Bx$ is associated with the expression $x^*(A+\mathrm{i} B)x$ .If we just went straight from one to the other, this reduction would appear like some crazy stroke of inspiration: why would I ever think to write down $x^*(A+\mathrm{i} B)x$ ? However, just following our nose lead by a desire to treat $A$ and $B$ symmetrically and applying a couple standard tricks, this expression appears naturally. The expression $x^*(A+\mathrm{i} B)x$ will be very useful to us because it is linear in $A$ and $B$ , and thus for the perturbed problem $(\widetilde{A},\widetilde{B}) = (A+E,B+F)$ , we have that $x^*(\widetilde{A}+\mathrm{i} \widetilde{B})x = x^*(A+\mathrm{i} B)x + x^*(E+\mathrm{i} F)x$ : consequently, $x^*(\widetilde{A}+\mathrm{i} \widetilde{B})x$ is a small perturbation of $x^*(A+\mathrm{i} B)x$ . This observation will be very useful to us.

If $x$ is the eigenvector, then the complex number $\alpha + \mathrm{i} \beta$ is $x^*(A+\mathrm{i} B)x$ .

Definiteness and the Crawford Number

With these insights in hand, we can now return to the point we left earlier about what it means for a generalized eigenvalue problem to be “definite”. We know that if there exists a vector $x$ for which $Ax = Bx = 0$ , then the problem is singular. If we multiply by $x^*$ , we see that this means that $x^*Ax = x^*Bx = 0$ as well and thus $x^*(A+\mathrm{i}B)x = 0$ . It is thus quite natural to assume the following definiteness condition:

The pair $(A,B)$ is said to be definite if $x^*(A+\mathrm{i}B)x \ne 0$ for all complex nonzero vectors $x$ .

A definite problem is guaranteed to be not singular, but the reverse is not necessarily true; one can easily find pairs $(A,B)$ which are not definite and also not singular.¹² (Note $x^*Ax = x^*Bx = 0$ does not imply $Ax = Bx = 0$ unless $A$ and $B$ are both positive (or negative) semidefinite.)

The “natural” symmetric condition for $(A,B)$ to be “definite” is for $x^*(A+\mathrm{i} B)x \ne 0$ for all vectors $x$ .

Since the expression $x^*(A+\mathrm{i}B)x$ is just scaled by a positive factor by scaling the vector $x$ , it is sufficient to check the definiteness condition $x^*(A+\mathrm{i}B)x \ne 0$ for only complex unit vectors $x$ . This leads naturally to a quantitative characterization of the degree of definiteness of a pair $(A,B)$ :

The Crawford number¹³ $c(A,B)$ of a pair $(A,B)$ is the minimum value of $|x^*(A+\mathrm{i}B)x| = \sqrt{(x^*Ax)^2 + (x^*Bx)^2}$ over all complex unit vectors $x$ .

The Crawford number naturally quantifies the degree of definiteness.¹⁴ A problem which has a large Crawford number (relative to a perturbation) will remain definite after perturbation, whereas the pair may become indefinite if the size of the perturbation exceeds the Crawford number. Geometrically, the Crawford number has the following interpretation: $x^*(A+\mathrm{i}B)x$ must lie on or outside the circle of radius $c(A,B)$ centered at $0$ for all (complex) unit vectors $x$ .

The “degree of definiteness” can be quantified by the Crawford number $c(A,B) := \min_{\|x\|=1} x^*(A+iB)x$ .

Now comes another step in our journey which is a bit more challenging. For a matrix $C$ (in our case $C = A+\mathrm{i}B$ ), the set of complex numbers $x^*Cx$ for all unit vectors $x$ has been the subject of considerable study. In fact, this set has a name

The field of values of a matrix $C$ is the set $W(C) := \{ x^*Cx : x\in\mathbb{C}^n, \: \|x\| = 1\}$ .

In particular, the Crawford number is just the absolute value of the closest complex number in the field of values $W(A+iB)$ to zero.

It is a very cool and highly nontrivial fact (called the Toeplitz–Hausdorff Theorem) that the field of values is always a convex set, with every two points in the field of values containing the line segment connecting them. Thus, as a consequence, the field of values $W(A+\mathrm{i}B)$ for a definite matrix pair $(A,B)$ is always on “one side” of the complex plane (in the sense that there exists a line through zero which $W(A+\mathrm{i}B)$ lies strictly on one side of¹⁵).

The numbers $x^*(A+iB)x$ for unit vectors $x$ lie on one half of the complex plane.

The field of values $W(A+\mathrm{i}B)$ lies outside the circle of radius $c(A,B)$ centered at $0$ and thus on one side of the complex plane.

The field of values $W(A+\mathrm{i}B)$ lies outside the circle of radius $c(A,B)$ centered at $0$ and thus on one side of the complex plane.

From Eigenvalues to Eigenangles

All of this geometry is lovely, but we need some way of relating it to the eigenvalues. As we observed a while ago, each eigenvalue is best thought of as a ray emanating from the origin, owing to the fact that the pair $(\alpha,\beta)$ can be scaled by an arbitrary positive factor. A ray is naturally associated with an angle, so it is natural to characterize an eigenvalue pair $(\alpha,\beta)$ by the angle describing its arc.

But the angle of a ray is only defined up additions by full rotations ( $2\pi$ radians). As such, to associate each ray a unique angle we need to pin down this indeterminacy in some fixed way. Moreover, this indeterminacy should play nice with the field of values $W(A+\mathrm{i}B)$ and the field of values $W(\widetilde{A}+\mathrm{i}\widetilde{B})$ of the perturbation. But in the last section, we say that each of these field of angles lies (strictly) on one half of the complex plane. Thus, we can find a ray $R$ which does not intersect either field of values!

One possible choice is to measure the angle from this ray. We shall make a slightly different choice which plays better when we treat $(\alpha,\beta)$ as a complex number $\alpha + \mathrm{i}\beta$ . Recall that a number $\theta$ is an argument for $\alpha + \mathrm{i}\beta$ if $\alpha + \mathrm{i}\beta = r\mathrm{e}^{i\theta}$ for some real number $r \ge 0$ . The argument is multi-valued since $\theta + 2\pi n$ is an argument for $\alpha+\mathrm{i}\beta$ as long as $\theta$ is (for all integers $n$ ). However, once we exclude our ray $R$ , we can assign each complex number $\alpha+\mathrm{i}\beta$ not on this ray a unique argument which depends continuously on $(\alpha,\beta)$ . Denote this “branch” of the argument by $\operatorname{arg}$ . If $(\alpha,\beta)$ represents an eigenvalue $\lambda = \alpha/\beta$ , we call $\theta = \arg(\alpha+\mathrm{i}\beta)$ an eigenangle.

Represent an eigenvalue pair $(\alpha,\beta)$ by its associated eigenangle $\theta = \arg(\alpha+i\beta)$ .

How are these eigenangles related to the eigenvalues? It’s now a trigonometry problem:

$\begin{equation*} \lambda = \frac{\alpha}{\beta} = \frac{\mbox{adjacent}}{\mbox{opposite}} = \cot \left( \operatorname{arg}(\alpha+i\beta)). \end{equation*}$

The eigenvalues are the cotangents of the eigenangles!

The eigenvalue $\lambda = \alpha/\beta$ is the cotangent of the eigenangle $\theta = \arg(\alpha+\mathrm{i}\beta)$ .

Variational Characterization

Now comes another difficulty spike in our line of reasoning, perhaps the largest in our whole deduction. To properly motivate things, let us first review some facts about the standard Hermitian/symmetric eigenvalue problem. The big idea is that eigenvalues can be thought of as the solution to a certain optimization problem. The largest eigenvalue of a Hermitian/symmetric matrix $A$ is given by the maximization problem

$\begin{equation*} \lambda_{\rm max}(A) = \max_{\|x\| = 1} x^*Ax. \end{equation*}$

The largest eigenvalue is the maximum of the quadratic form over unit vectors $x$ . What about the other eigenvalues? The answer is not obvious, but the famous Courant–Fischer Theorem shows that the $j$ th largest eigenvalue $\lambda_j(A)$ can be written as the following minimax optimization problem

$\begin{equation*} \lambda_j(A) = \min_{\dim \mathcal{X} = n-j+1} \max_{\substack{x \in \mathcal{X} \\ \|x\| = 1}} x^*Ax. \end{equation*}$

The minimum is taken over all subspaces $\mathcal{X}$ of dimension $n-j+1$ whereas the maximum is taken over all unit vectors $x$ within the subspace $\mathcal{X}$ . Symmetrically, one can also formulate the eigenvalues as a max-min optimization problem

$\begin{equation*} \lambda_j(A) = \max_{\dim \mathcal{X} = j} \min_{\substack{x \in \mathcal{X} \\ \|x\| = 1}} x^*Ax. \end{equation*}$

These variational/minimax characterizations of the eigenvalues of a Hermitian/symmetric matrix are essential to perturbation theory for Hermitian/symmetric eigenvalue problems, so it is only natural to go looking for a variational characterization of the generalized eigenvalue problem. There is one natural way of doing this that works for $B$ positive definite: specifically, one can show that

$\begin{equation*} \lambda_j(A,B) = \min_{\dim \mathcal{X} = n-j+1} \max_{\substack{x \in \mathcal{X} \\ \|x\| = 1}} \frac{x^*Ax}{x^*Bx}. \end{equation*}$

This characterization, while useful in its own right, is tricky to deal with because it is nonlinear in $A$ and $B$ . It also treats $A$ and $B$ non-symmetrically, which should set off our alarm bells that there might be a better way. Indeed, the ingenious idea, due to G. W. Stewart in 1979, is to instead provide a variational characterization of the eigenangles! Specifically, Stewart was able to show¹⁶

(1) $\begin{align*} \theta_j &= \min_{\dim \mathcal{X} = n-j+1} \max_{\substack{x\in\mathcal{X} \\ \|x\|=1}} \arg(x^*(A+\mathrm{i}B)x), \\ \theta_j &= \max_{\dim \mathcal{X} = j} \min_{\substack{x\in\mathcal{X} \\ \|x\|=1}} \arg(x^*(A+\mathrm{i}B)x), \end{align*}$

for the eigenangles $\theta_1\ge \theta_2\ge\cdots\ge \theta_n$ .¹⁷ This shows, in particular, that the field of values is subtended by the smallest and largest eigenangles.

The eigenangles satisfy a minimax variational characterization.

How Big is the Perturbation?

We’re tantalizingly close to our objective. The final piece in our jigsaw puzzle before we’re able to start proving perturbation theorems is to quantify the size of the perturbing matrices $E$ and $F$ . Based on what we’ve done so far, we see that the eigenvalues are natural associated with the complex number $x^*(A+\mathrm{i}B)x$ , so it is natural to characterize the size of the perturbing pair $(E,F)$ by the distance between $x^*(A+\mathrm{i}B)x$ and $x^*(\widetilde{A}+\mathrm{i}\widetilde{B})x$ . But the difference between these two quantities is just

$\begin{equation*} x^*(\widetilde{A}+\mathrm{i}\widetilde{B})x - x^*(A+\mathrm{i}B)x = x^*(E+\mathrm{i}F)x. \end{equation*}$

We’re naturally led to the question: how big can $x^*(E+\mathrm{i}F)x$ be? If the vector $x$ has a large norm, then quite large, so let’s fix $x$ to be a unit vector. With this assumption in place, the maximum size of $x^*(E+\mathrm{i}F)x$ is simple the distance of the farthest point in the field of values $E+iF$ from zero. This quantity has a name:

The numerical radius of a matrix $G$ (in our case $=E+\mathrm{i}F$ ) is $r(G) := \max_{\|x\|=1} |x^*Gx|$ .¹⁸

The size of the perturbation $(E,F)$ is the numerical radius $r(E+\mathrm{i}F) = \max_{\|x\|=1} |x^*(E+\mathrm{i}F)x|$ .

It is easy to upper-bound the numerical radius $r(E+\mathrm{i}F)$ by more familiar quantities. For instance, once can straightforwardly show the bound $r(E+\mathrm{i}F) \le \sqrt{\|E\|^2+\|F\|^2}$ , where $\|\cdot\|$ is the spectral norm. We prefer to state results using the numerical radius because of its elegance: it is, in some sense, the “correct” measure of the size of the pair $(E,F)$ in the context of this theory.

Stewart’s Perturbation Theory

Now, after many words of prelude, we finally get to our first perturbation theorem. With the work we’ve done in place, the result is practically effortless.

Let $\widetilde{\theta}_1\ge \widetilde{\theta}_2\ge \cdots\ge\widetilde{\theta}_n$ denote the eigenangles of the perturbed pair $(\widetilde{A},\widetilde{B})$ and consider the $j$ th eigenangle. Let $\mathcal{X}^*$ be the subspace of dimension $n-j+1$ achieving the minimum in the first equation of the variational principle (1) for the original unperturbed pair $(A,B)$ . Then we have

(2) $\begin{equation*} \widetilde{\theta}_j = \min_{\dim \mathcal{X} = n-j+1} \max_{\substack{x\in\mathcal{X} \\ \|x\|=1}} \arg(x^*(\widetilde{A}+\mathrm{i}\widetilde{B})x) \le \max_{\substack{x\in\mathcal{X}^* \\ \|x\|=1}} \arg(x^*(A+\mathrm{i}B)x + x^*(E+\mathrm{i}F)x). \end{equation*}$

This is something of a standard trick when dealing with variational problems in matrix analysis: take the solution (in this case the minimizing subspace) for the original problem and plug it in for the perturbed problem. The solution may no longer be optimal, but it at least gives an upper (or lower) bound. The complex number $x^*(A+\mathrm{i}B)x$ must lie at least a distance $c(A,B)$ from zero and $|x^*(E+\mathrm{i}F)x| \le r(E+\mathrm{i}F)$ . We’re truly toast if the perturbation is large enough to perturb $x^*(\widetilde{A}+i\widetilde{B})x$ to be equal to zero, so we should assume that $r(E+\mathrm{i}F) < c(A,B)$ .

For our perturbation theory to work, we must assume $r(E+\mathrm{i}F) < c(A,B)$ .

$x^*(A+\mathrm{i}B)x$ lies on or outside the circle centered at zero with radius $c(A,B)$ . $x^*(\tilde{A}+\mathrm{i}B)x$ might lie anywhere in a circle centered at $x^*(A+\mathrm{i}B)x$ with radius $r(E+\mathrm{i}F)$ , so one must have $r(E+\mathrm{i}F) < c(A,B)$ to ensure the perturbed problem is nonsingular (equivalently $x^*(\tilde{A}+\mathrm{i}B)x\ne 0$ for every $x$ ).

$x^*(A+\mathrm{i}B)x$ lies on or outside the circle centered at zero with radius $c(A,B)$ . $x^*(\tilde{A}+\mathrm{i}B)x$ might lie anywhere in a circle centered at $x^*(A+\mathrm{i}B)x$ with radius $r(E+\mathrm{i}F)$ , so one must have $r(E+\mathrm{i}F) < c(A,B)$ to ensure the perturbed problem is nonsingular (equivalently $x^*(\tilde{A}+\mathrm{i}B)x\ne 0$ for every $x$ ).

Making the assumption that $r(E+\mathrm{i}F) < c(A,B)$ , bounding the right-hand side of (2) requires finding the most-counterclockwise angle necessary to subtend a circle of radius $r(E+\mathrm{i}F)$ centered at $x^*(A+\mathrm{i}B)x$ , which must lie a distance $c(A,B)$ from the origin. The worst-case scenario is when $x^*(A+\mathrm{i}B)x$ is exactly a distance $c(A,B)$ from the origin, as is shown in the following diagram.

In the worst case, $x^*(A+\mathrm{i}B)x$ lies on the circle centered at zero with radius $c(A,B)$ , which is subtended above by angle $\theta_j + \sin^{-1}(r(E+iF)/c(A,B))$ .

Solving the geometry problem for the counterclockwise-most subtending angle in this worst-case sitation, we conclude the eigenangle bound $\widetilde{\theta}_j -\theta_j \le \sin^{-1}(r(E+\mathrm{i}F)/c(A,B))$ . An entirely analogous argument using the max-min variational principle (1) proves an identical lower bound, thus showing

(3) $\begin{equation*} \sin |\theta_j - \widetilde{\theta}_j| \le \frac{r(E+\mathrm{i}F)}{c(A,B)}. \end{equation*}$

In the language of eigenvalues, we have¹⁹

$\begin{equation*} |\cot^{-1}(\widetilde{\lambda}_j) - \cot^{-1}(\lambda_j)| \le \sin^{-1}\left( \frac{r(E+\mathrm{i}F)}{c(A,B)} \right). \end{equation*}$

Interpreting Stewart’s Theory

After much work, we have finally proven our first generalized eigenvalue perturbation theorem. After taking a moment to celebrate, let’s ask ourselves: what does this result tell us?

Let’s start with the good. This result shows us that if the perturbation, measured by the numerical radius $r(E+iF)$ , is much smaller than the definiteness of the original problem, measured by the Crawford number $c(A,B)$ , then the eigenangles change by a small amount. What does this mean in terms of the eigenvalues? For small eigenvalues (say, less than one in magnitude), small changes in the eigenangles also lead to small changes of the eigenvalues. However, for large eigenangles, small changes in the eigenangle are magnified into potentially large changes in the eigenvalues. One can view this result in a positive or negative framing. On the one hand, large eigenvalues could be subject to dramatic changes by small perturbations; on the other hand, the small eigenvalues aren’t “taken down with the ship” and are much more well-behaved.

Stewart’s theory is beautiful. The variational characterization of the eigenangles (1) is a master stroke and exactly the extension one would want from the standard Hermitian/symmetric theory. From the variational characterization, the perturbation theorem follows almost effortlessly from a little trigonometry. However, Stewart’s theory has one important deficit: the Crawford number. All that Stewart’s theory tells is that all of the eigenangles change by at most roughly “perturbation size over Crawford number”. If the Crawford number is quite small since the problem is nearly indefinite, this becomes a tough pill to swallow.

The Crawford number is in some ways essential: if the perturbation size exceeds the Crawford number, the problem can become indefinite or even singular. Thus, we have no hope of fully removing the Crawford number from our analysis. But might it be the case that some eigenangles change by much less than “pertrubation size over Crawford number”? Could we possibly improve to a result of the form “the eigenangles change by roughly perturbation size over something (potentially) much less than the Crawford number”? Sun improved Stewart’s analysis in 1982, but the scourge of the Crawford number remained.²⁰ The theory of Mathias and Li, published in a technical report in 2004, finally produced a bound where the Crawford number is replaced.

The Mathias–Li Insight and Reduction to Diagonal Form

Let’s go back to the Stewart theory and look for a possible improvement. Recall in the Stewart theory that we considered the point $x^*(A+\mathrm{i}B)x$ on the complex plane. We then argued that, in the worst case, this point would lie a distance $c(A,B)$ from the origin and then drew a circle around it with radius $r(E+\mathrm{i}F)$ . To improve on Stewart’s bound, we must somehow do something better than using the fact that $|x^*(A+\mathrm{i}B)x|\ge c(A,B)$ . The insight of the Mathias–Li theory is, in some sense, as simple as this: rather than using the fact that $|x^*(A+\mathrm{i}B)x| \ge c(A,B)$ (as in Stewart’s analysis), use how far $x^*(A+\mathrm{i}B)x$ actually is from zero, where $x$ is chosen to be the unit norm eigenvectors of $(A,B)$ .²¹

Before going further, let us quickly make a small reduction which will simplify our lives greatly. Letting $X$ denote a matrix whose columns are the unit-norm eigenvectors of $(A,B)$ , one can verify that $X^*AX$ and $X^*BX$ are diagonal matrices with entries $\alpha_1,\ldots,\alpha_n$ and $\beta_1,\ldots,\beta_n$ respectively. With this in mind, it can make our lives a lot easy to just do a change of variables $A \mapsto X^*AX$ and $B\mapsto X^*BX$ (which in turn sends $E\mapsto X^*EX$ and $F \mapsto X^*FX$ ). The change of variables $A \mapsto X^*AX$ is very common in linear algebra and is called a congruence transformation.

Perform a change of variables by a congruence transformation with the matrix of eigenvectors.

While this change of variables makes our lives a lot easier, we must first worry about how this change of variables might effect the size of the perturbation matrices $(E,F)$ . It turns out this change of variables is not totally benign, but it is not maximally harmful either. Specifically, the spectral radius $r(E+\mathrm{i}F)$ can grow by as much as a factor of $n$ .²² This factor of $n$ isn’t great, but it is much better than if the bound were to degrade by a factor of the condition number $\|X\|\|X^{-1}\|$ , which can be arbitrarily large.

This change of variables may increase $r(E+\mathrm{i}F)$ by at most a factor of $n$ .

From now on, we shall tacitly assume that this change of variables has taken place, with $A$ and $B$ being diagonal and $E$ and $F$ being such that $r(E+\mathrm{i}F)$ is at most a factor $n$ larger than it was previously. We denote by $\alpha_j$ and $\beta_j$ the $j$ th diagonal element of $A$ and $B$ , which are given by $\alpha_j = x_j^*Ax_j$ and $\beta_j = x_j^*Bx_j$ where $x_j$ is the $j$ th unit-norm eigenvector

Mathias and Li’s Perturbation Theory

We first assume the perturbation $(E,F)$ is smaller than the Crawford number in the sense $r(E+\mathrm{i}F) < c(A,B)$ , which is required to be assured that the perturbed problem $(\widetilde{A},\widetilde{B})$ does not lose definiteness. This will be the only place in this analysis where we use the Crawford number.

Draw a circle of radius $r(E+\mathrm{i}F)$ around $\alpha_j + \mathrm{i}\beta_j$ .

If $\theta_j$ is the associated eigenangle, then this circle is subtended by arcs with angles

$\begin{equation*} \ell_j = \theta_j - \sin^{-1}\left(\frac{r(E+\mathrm{i}F)}{|\alpha_j+\mathrm{i}\beta_j|}\right), \quad u_j = \theta_j + \sin^{-1}\left(\frac{r(E+\mathrm{i}F)}{|\alpha_j+\mathrm{i}\beta_j|}\right). \end{equation*}$

It would be nice if the perturbed eigenangles $\widetilde{\theta}_j$ were guaranteed to lie in these arcs (i.e., $\ell_j \le \widetilde{\theta}_j \le u_j$ ). Unfortunately this is not necessarily the case. If one $\alpha_j + \mathrm{i}\beta_j$ is close to the origin, it will have a large arc which may intersect with other arcs; if this happens, we can’t guarantee that each perturbed eigenangle will remain within its individual arc. We can still say something though.

What follows is somewhat technical, so let’s start with the takeaway conclusion: $\widetilde{\theta}_j$ is larger than any $j$ of the lower bounds $\ell_j$ . In particular, this means that $\widetilde{\theta}_j$ is larger than the $j$ th largest of all the lower bounds. That is, if we rearrange the lower bounds $\ell_1,\ldots,\ell_n$ in decreasing order $\ell_1^\downarrow \ge \ell_2^\downarrow \ge \cdots \ge \ell_n^\downarrow$ , we hace $\widetilde{\theta}_j \ge \ell_j^\downarrow$ . An entirely analogous argument will give an upper bound, yielding

(4) $\begin{equation*} \ell_j^\downarrow \le \widetilde{\theta}_j \le u_j^\downarrow. \end{equation*}$

For those interested in the derivation, read on the in the following optional section:

Derivation of the Mathias–Li Bounds

Since $A$ and $B$ are diagonal, the eigenvectors of the pair $(A,B)$ are just the standard basis vectors, the $j$ th of which we will denote $e_j$ . The trick will be to use the max-min characterization (1) with the subspace $\mathcal{X}$ spanned by some collection of $j$ basis vectors $e_{i_1},\ldots,e_{i_j}$ . Churning through a couple inequalities in quick fashion,²³ we obtain

$\begin{align*} \widetilde{\theta}_j &\ge \min_{\substack{x \in \mathcal{X} \\ \|x\| = 1}} \arg \left( x^*(A+\mathrm{i}B)x + x^*(E+\mathrm{i}F)x \right) \\ &\ge \min \left\{ \arg(y+z) : y \in \operatorname{conv} \{ \alpha_{i_1}+\mathrm{i}\beta_{i_1},\ldots,\alpha_{i_j}+\mathrm{i}\beta_{i_j} \}, z\in W(E+\mathrm{i}F) \} \\ &\ge \min \left\{ \arg(y+z) : y \in \operatorname{conv} \{ \alpha_{i_1}+\mathrm{i}\beta_{i_1},\ldots,\alpha_{i_j}+\mathrm{i}\beta_{i_j} \}, |z|\le r(E+\mathrm{i}F) \} \\ &\ge \min \left\{ \arg(w) : w\in\operatorname{conv} \bigcup_{k=1}^j \{ a\in\mathbb{C} : |\alpha_{i_k} + \mathrm{i}\beta_{i_k} - a| \le r(E+\mathrm{i}F) \} \right\} \\ &= \min_{k=1,\ldots,j} \ell_{i_k}. \end{align*}$

Here, $\operatorname{conv}$ denotes the convex hull. Since this holds for every set of indices $i_1,\ldots,i_j$ , it in particular holds for the set of indices which makes $\min_{k=1,\ldots,j} \ell_{i_k}$ the largest. Thus, $\widetilde{\theta}_j \ge \ell^\downarrow_j$ .

How to Use Mathias–Li’s Perturbation Theory

The eigenangle perturbation bound (4) can be instantiated in a variety of ways. We briefly sketch two. The first is to bound $|\alpha_j + \mathrm{i}\beta_j|$ by its minimum over all $j$ , which then gives a bound on $u^\downarrow_j$ (and $\ell^\downarrow_j$ )

$\begin{equation*} |\alpha_j + \mathrm{i} \beta_j| \ge \min_{1\le j\le n} |\alpha_j + \mathrm{i} \beta_j| \implies u_j^\downarrow \le \theta_j + \sin^{-1} \frac{r(E+\mathrm{i}F)}{\min_{1\le j\le n} |\alpha_j + \mathrm{i} \beta_j|}. \end{equation*}$

Plugging into (4) and simplifying gives

(5) $\begin{equation*} \sin \left| \widetilde{\theta}_j - \theta_j \right| \le \frac{r(E+\mathrm{i}F)}{\min_{1\le j\le n} |\alpha_j + \mathrm{i} \beta_j|}. \end{equation*}$

This improves on Stewart’s bound (3) by replacing the Crawford number $c(A,B)$ by $\min_{1\le j\le n} |\alpha_j + \mathrm{i} \beta_j|$ ; as Mathias and Li show $\min_{1\le j\le n} |\alpha_j + \mathrm{i} \beta_j|$ is always smaller than or equal to $c(A,B)$ and can be much much smaller.²⁴

For the second instantiation (4), we recognize that if an eigenangle $\theta_j$ is sufficiently well-separated from other eigenangles (relative to the size of the perturbation and $\min_{1\le j\le n} |\alpha_j + \mathrm{i} \beta_j|$ ), then we have $u_j^\downarrow \le u_j$ and $\ell_j^\downarrow \ge \ell_j$ . (The precise instantiation of “sufficiently well-separated” requires some tedious algebra; if you’re interested, see Footnote 7 in Mathias and Li’s paper.²⁵) Under this separation condition, (4) then reduces to

(6) $\begin{equation*} \sin \left| \widetilde{\theta}_j - \theta_j \right| \le \frac{r(E+\mathrm{i}F)}{|\alpha_j + \mathrm{i} \beta_j|}. \end{equation*}$

This result improves on Stewart’s result (4) by even more, since we have now replaced the Crawford number $c(A,B)$ by $|\alpha_j + \mathrm{i} \beta_j|$ for a sufficiently small perturbation. In fact, a result of this form is nearly as good as one could hope for.²⁶

The Elegant Geometry of Generalized Eigenvalue Perturbation Theory

As I said at the start of this post, what fascinates me about this generalized eigenvalue perturbation is the beautiful and elegant geometry. When I saw it for the first time, it felt like a magic trick: a definite generalized eigenvalue problem with real eigenvalues was transformed by sleight of hand into a geometry problem on the complex plane, with solutions involving just a little high school geometry and trigonometry. Upon studying the theory, I began to appreciate it for a different reason. Upon closer examination, the magic trick was revealed to be a sequence of deductions, each logically following naturally from the last. To the pioneers of this subject—Stewart, Sun, Mathias, Li, and others—this sequence of logical deductions was not preordained, and their discovery of this theory doubtlessly required careful thought and leaps of insight. Now that this theory has been discovered, however, we get the benefit of retrospection, and can retell a narrative of this theory where each step follows naturally from the last. When told this way, one almost imagines being able to develop this theory by oneself, where at each stage we appeal to some notion of mathematical elegance (e.g., by treating $A$ and $B$ symmetrically) or by applying a standard trick (e.g., identifying a pair $(\alpha,\beta)$ with the complex number $\alpha + \mathrm{i}\beta$ ). Since this theory took several decades to fall into place, we should not let this storytelling exercise fool us into thinking the prospective act of developing a new theory will be as straightforward and linear as this retelling, pruned of dead ends and halts in progress, might suggest.

That said, I do think the development of the perturbation theory of the generalized eigenvalue problem does have a lesson for those of us who seek to develop mathematical theories: be guided by mathematical elegance. At several points in the development of the perturbation theory, we obtained great gains by treating quantities which play a symmetric role in the problem symmetrically in the theory or by treating a pair of real numbers as a complex number and asking how to interpret that complex number. My hope is that this perturbation theory serves as a good example for how letting oneself be guided by intuition, a small array of standard tricks, and a search for elegance can lead one to conceptualize a problem in the right way which leads (after a considerable amount of effort and a few lucky breaks) to a natural solution.

Big Ideas in Applied Math: Low-rank Matrices

October 26, 2021 by Ethan N. Epperly 10 Comments

Let’s start our discussion of low-rank matrices with an application. Suppose that there are 1000 weather stations spread across the world, and we record the temperature during each of the 365 days in a year.¹ If we were to store each of the temperature measurements individually, we would need to store 365,000 numbers. However, we have reasons to believe that significant compression is possible. Temperatures are correlated across space and time: If it’s hot in Arizona today, it’s likely it was warm in Utah yesterday.

If we are particularly bold, we might conjecture that the weather approximately experiences a sinusoidal variation over the course of the year:

(1) $\begin{equation*} \mbox{temperature at station $i$ on day $j$} \approx a_i + b_i \sin\left( 2\pi \times \frac{j}{365} + \phi \right). \end{equation*}$

For a station $i$ , $a_i$ denotes the average temperature of the station and $b_i$ denotes the maximum deviation above or below this station, signed so that it is warmer than average in the Northern hemisphere during June-August and colder-than-average in the Southern hemisphere during these months. The phase shift $\phi$ is chosen so the hottest (or coldest) day in the year occurs at the appropriate time. This model is clearly grossly inexact: The weather does not satisfy a simple sinusoidal model. However, we might plausibly expect it to be fairly informative. Further, we have massively compressed our data, only needing to store the $2000 \ll 365,000$ numbers $a_1,a_2,\ldots,b_{1000}$ rather than our full data set of 365,000 temperature values.

Let us abstract this approximation procedure in a linear algebraic way. Let’s collect our weather data into a matrix $W$ with 1000 rows, one for each station, and 365 columns, one for each day of the year. The entry $W_{ij}$ corresponding to station $i$ and day $j$ is the temperature at station $i$ on day $j$ . The approximation Eq. (1) corresponds to the matrix approximation

(2) $\begin{equation*} W \approx \underbrace{\begin{bmatrix} a_1 + b_1 \sin\left( 2\pi \times \frac{1}{365} + \phi \right) & \cdots & a_1 + b_1 \sin\left( 2\pi \times \frac{365}{365} + \phi \right) \\ \vdots & \ddots & \vdots \\ a_{1000} + b_{1000} \sin\left( 2\pi \times \frac{1}{365} + \phi \right) & \cdots & a_{1000} + b_{1000} \sin\left( 2\pi \times \frac{365}{365} + \phi \right) \end{bmatrix}}_{:=\hat{W}}. \end{equation*}$

Let us call the matrix on the right-hand side of Eq. (2) $\hat{W}$ for ease of discussion. When presented in this linear algebraic form, it’s less obvious in what way $\hat{W}$ is simpler than $W$ , but we know from Eq. (1) and our previous discussion that $\hat{W}$ is much more efficient to store than $W$ . This leads us naturally to the following question: Linear algebraically, in what way is $\hat{W}$ simpler than $W$ ?

The answer is that the matrix $\hat{W}$ has low rank. The rank of the matrix $\hat{W}$ is $2$ whereas $W$ almost certainly possesses the maximum possible rank of $365$ . This example is suggestive that low-rank approximation, where we approximate a general matrix by one of much lower rank, could be a powerful tool. But there any many questions about how to use this tool and how widely applicable it is. How can we compress a low-rank matrix? Can we use this compressed matrix in computations? How good of a low-rank approximation can we find? What even is the rank of a matrix?

What is Rank?

Let’s do a quick review of the foundations of linear algebra. At the core of linear algebra is the notion of a linear combination. A linear combination of vectors $v_1,\ldots,v_k$ is a weighted sum of the form $\alpha_1 v_1 + \cdots + \alpha_k v_k$ , where $\alpha_1,\ldots,\alpha_k$ are scalars². A collection of vectors $v_1,\ldots,v_k$ is linearly independent if there is no linear combination of them which produces the zero vector, except for the trivial $0$ -weighted linear combination $0 v_1 + \cdots + 0v_k$ . If $v_1,\ldots,v_k$ are not linearly independent, then they’re linearly dependent.

The column rank of a matrix $B$ is the size of the largest possible subset of $B$ ‘s columns which are linearly independent. So if the column rank of $B$ is $r$ , then there is some sub-collection of $r$ columns of $B$ which are linearly independent. There may be some different sub-collections of $r$ columns from $B$ that are linearly dependent, but every collection of $r+1$ columns is guaranteed to be linearly dependent. Similarly, the row rank is defined to be the maximum size of any linearly independent collection of rows taken from $B$ . A remarkable and surprising fact is that the column rank and row rank are equal. Because of this, we refer to the column rank and row rank simply as the rank; we denote the rank of a matrix $B$ by $\operatorname{rank}(B)$ .

Linear algebra is famous for its multiple equivalent ways of phrasing the same underlying concept, so let’s mention one more way of thinking about the rank. Define the column space of a matrix to consist of the set of all linear combinations of its columns. A basis for the column space is a linear independent collection of elements of the column space of the largest possible size. Every element of the column space can be written uniquely as a linear combination of the elements in a basis. The size of a basis for the column space is called the dimension of the column space. With these last definitions in place, we note that the rank of $B$ is also equal to the dimension of the column space of $B$ . Likewise, if we define the row space of $B$ to consist of all linear combinations of $B$ ‘s rows, then the rank of $B$ is equal to the dimension of $B$ ‘s row space.

The upshot is that if a matrix $B$ has a small rank, its many columns (or rows) can be assembled as linear combinations from a much smaller collection of columns (or rows). It is this fact that allows a low-rank matrix to be compressed for algorithmically useful ends.

Rank Factorizations

Suppose we have an $m\times n$ matrix $B$ which is of rank $r$ much smaller than both $m$ and $n$ . As we saw in the introduction, we expect that such a matrix can be compressed to be stored with many fewer than $mn$ entries. How can this be done?

Let’s work backwards and start with the answer to this question and then see why it works. Here’s a fact: a matrix $B$ of rank $r$ can be factored as $B = LR^\top$ , where $L$ is an $m\times r$ matrix and $R$ is an $n\times r$ matrix. In other words, $B$ can be factored as a “thin” matrix $L$ with $r$ columns times a “fat” matrix $R^\top$ with $r$ rows. We use the symbols $L$ and $R$ for these factors to stand for “left” and “right”; we emphasize that $L$ and $R$ are general $m\times r$ and $n\times r$ matrices, not necessarily possessing any additional structure.³ The fact that we write the second term in this factorization as a transposed matrix “ $R^\top$ ” is unimportant: We adopt a convention where we write a fat matrix as the transpose of a thin matrix. This notational choice is convenient allows us to easily distinguish between thin and fat matrices in formulas; this choice of notation is far from universal. We call a factorization such as $B = LR^\top$ a rank factorization.⁴

Rank factorizations are useful as we can compactly store $B$ by storing its factors $L$ and $R$ . This reduces the storage requirements of $B$ to $(m+n)r$ numbers down from $mn$ numbers. For example, if we store a rank factorization of the low-rank approximation $\hat{W}$ from our weather example, we need only store 2,730 numbers rather than 365,000. In addition to compressing $B$ , we shall soon see that one can rapidly perform many calculations from the rank factorization $LR^\top = B$ without ever forming $B$ itself. For these reasons, whenever performing computations with a low-rank matrix, your first step should almost always be to express it using a rank factorization. From there, most computations can be done faster and using less storage.

Having hopefully convinced ourselves of the usefulness of rank factorizations, let us now convince ourselves that every rank- $r$ matrix $B$ does indeed possess a rank factorization $B = LR^\top$ where $L$ and $R$ have $r$ columns. As we recalled in the previous section, since $B$ has rank $r$ , there is a basis of $B$ ‘s column space consisting of $r$ vectors $\ell_1,\ldots,\ell_r$ . Collect these $r$ vectors as columns of an $m\times r$ matrix $L = \begin{bmatrix} \ell_1 & \cdots & \ell_r\end{bmatrix}$ . But since the columns of $L$ comprise a basis of the column space of $B$ , every column of $B$ can be written as a linear combination of the columns of $L$ . For example, the $j$ th column $b_j$ of $B$ can be written as a linear combination $b_j = R_{j1} \ell_1 + \cdots + R_{jr} \ell_r$ , where we suggestively use the labels $R_{j1},\ldots,R_{jr}$ for the scalar multiples in our linear combination. Collecting these coefficients into a matrix $R$ with $ij$ th entry $R_{ij}$ , we have constructed a factorization $B = LR^\top$ . (Check this!)

This construction gives us a look at what a rank factorization is doing. The columns of $L$ comprise a basis for the column space of $B$ and the rows of $R^\top$ comprise a basis for the row space of $B$ . Once we fix a “column basis” $L$ , the “row basis” $R^\top$ is comprised of linear combination coefficients telling us how to assemble the columns of $B$ as linear combinations of the columns in $L$ .⁵ Note that this means there exist many different rank factorizations of a matrix since one may pick different column bases $L$ for $B$ .⁶

Now that we’ve convinced ourselves that every matrix indeed has a rank factorization, how do we compute them in practice? In fact, pretty much any matrix factorization will work. If you can think of a matrix factorization you’re familiar with (e.g., LU, QR, eigenvalue decomposition, singular value decomposition,…), you can almost certainly use it to compute a rank factorization. In addition, many dedicated methods have been developed for the specific purpose of computing rank factorizations which can have appealing properties which make them great for certain applications.

Let’s focus on one particular example of how a classic matrix factorization, the singular value decomposition, can be used to get a rank factorization. Recall that the singular value decomposition (SVD) of a (real) matrix $B$ is a factorization $B = U\Sigma V^\top$ where $U$ and $V$ are an $m\times m$ and $n\times n$ (real) orthogonal matrices and $\Sigma$ is a (possibly rectangular) diagonal matrix with nonnegative, descending diagonal entries $\sigma_1 \ge \sigma_2 \ge \cdots \ge \sigma_{\min(m,n)}$ . These diagonal entries are referred to as the singular values of the matrix $B$ . From the definition of rank, we can see that the rank of a matrix $B$ is equal to its number of nonzero singular values. With this observation in hand, a rank factorization of $B$ can be obtained by letting $L$ be the first $r$ columns of $U$ and $R^\top$ being the first $r$ rows of $\Sigma V^\top$ (note that the remaining rows of $\Sigma V^\top$ are zero).

Computing with Rank Factorizations

Now that we have a rank factorization in hand, what is it good for? A lot, in fact. We’ve already seen that one can store a low-rank matrix expressed as a rank factorization using only $(m+n)r$ numbers, down from $mn$ numbers by storing all of its entries. Similarly, if we want to compute the matrix-vector product $Bx$ for a vector $x$ of length $n$ , we can compute this product as $Bx = L(R^\top x)$ . This reduces the operation count down from $2mn$ operations to $2(m+n)r$ operations using the rank factorization. As a general rule of thumb, when we have something expressed as a rank factorization, we can usually expect to reduce our operation count (and storage costs) from something proportional to $mn$ (or worse) down to something proportional to $m+n$ .

Let’s try something more complicated. Say we want to compute an SVD $B = U\Sigma V^\top$ of $B$ . In the previous section, we computed a rank factorization of $B$ using an SVD, but suppose now we computed $B = LR^\top$ in some other way. Our goal is to “upgrade” the general rank factorization $B = LR^\top$ into an SVD of $B$ . Computing the SVD of a general matrix $B$ requires $\mathcal{O}(mn\min(m,n))$ operations (expressed in big O notation). Can we do better? Unfortunately, there’s a big roadblock for us: We need $m^2+n^2$ operations even to write down the matrices $U$ and $V$ , which already prevents us from achieving an operation count proportional to $m+n$ like we’re hoping for. Fortunately, in most applications, only the first $r$ columns of $U$ and $V$ are important. Thus, we can change our goal to compute a so-called economy SVD of $B$ , which is a factorization $B = \hat{U}\hat{\Sigma}\hat{V}^\top$ , where $\hat{U}$ and $\hat{V}$ are $m\times r$ and $n\times r$ matrices with orthonormal columns and $\hat{\Sigma}$ is a $r\times r$ diagonal matrix listing the nonzero singular values of $B$ in decreasing order.

Let’s see how to upgrade a rank factorization $B = LR^\top$ into an economy SVD $B = \hat{U}\hat{\Sigma}\hat{V}^\top$ . Let’s break our procedure into steps:

Compute (economy⁷) QR factorizations of $L$ and $R$ : $L = Q_1T_1$ and $R = Q_2 T_2$ . Reader beware: We call the “ $R$ ” factor in the QR factorizations of $L$ and $R$ to be $T_1$ and $T_2$ , as we have already used the letter $R$ to denote the second factor in our rank factorization.
Compute the small matrix $S = T_1T_2^\top$ .
Compute an SVD of $S=\tilde{U}\hat{\Sigma}\tilde{V}^\top$ .
Set $\hat{U} := Q_1\tilde{U}$ and $\hat{V} := Q_2\tilde{V}$ .

By following the procedure line-by-line, one can check that indeed the matrices $\hat{U}$ and $\hat{V}$ have orthonormal columns and $B = \hat{U}\hat{\Sigma}\hat{V}^\top$ , so this procedure indeed computes an economy SVD of $B$ . Let’s see why this approach is also faster. Let’s count operations line-by-line:

Economy QR factorization of an $m\times r$ and $n\times r$ matrix require $\mathcal{O}(mr^2)$ and $\mathcal{O}(nr^2)$ operations.
The product of two $r\times r$ matrices requires $\mathcal{O}(r^3)$ operations.
The SVD of an $r\times r$ matrix requires $\mathcal{O}(r^3)$ operations.
The products of a $m\times r$ and a $n\times r$ matrix by $r\times r$ matrices requires $\mathcal{O}(mr^2)$ and $\mathcal{O}(nr^2)$ operations.

Accounting for all the operations, we see the operation count is $\mathcal{O}((m+n)r^2)$ , a significant improvement over the $\mathcal{O}(mn\min(m,n))$ operations for a general matrix.⁸

As the previous examples show, many (if not most) things we want to compute from a low-rank matrix $B$ can be dramatically more efficiently computed using its rank factorization. The strategy is simple in principle, but can be subtle to execute: Whatever you do, avoid explicitly computing the product $LR^\top$ at all costs. Instead, compute with the matrices $L$ and $R$ directly, only operating on $m\times r$ , $n\times r$ , and $r\times r$ matrices.

Another important type of computation one can perform with low-rank matrices are low-rank updates, where we have already solved a problem for a matrix $A$ and we want to re-solve it efficiently with the matrix $A+B$ where $B$ has low rank. If $B$ is expressed in a rank factorization, very often we can do this efficiently as well, as we discuss in the following bonus section. As this is somewhat more niche, the uninterested reader should feel free to skip this and continue to the next section.

Low-rank Updates

Suppose we’ve solved a system of linear equations $Ax = b$ by computing an LU factorization of the $n\times n$ matrix $A$ . We now wish to solve the system of linear equations $(A+B)y = c$ , where $B$ is a low-rank matrix expressed as a rank factorization $B = LR^\top$ . Our goal is to do this without recomputing a new $LU$ factorization from scratch.

The first solution uses the Sherman-Morrison-Woodbury formula, which has a nice proof via the Schur complement and block Gaussian elimination which I described here. In our case, the formula yields

(3) $\begin{equation*} (A+B)^{-1} = (I_n-(A^{-1}L)(I_r+R^\top(A^{-1}L))^{-1}R^\top)A^{-1}, \end{equation*}$

where $I_n$ and $I_r$ denote the $n\times n$ and $r\times r$ identity matrices. This formula can easily verified by multiplying with $A+B$ and confirming one indeed recovers the identity matrix. This formula suggests the following approach to solving $(A+B)y = c$ . First, use our already-computed LU factorization for $A$ to compute $S:=A^{-1}L$ . (This involves solving $r$ linear systems of the form $As = p$ to compute each column $s$ of $S$ from each column $p$ of $P$ .) We then compute an LU factorization of the much smaller $r\times r$ matrix $I_r+R^\top S$ . Finally, we use our $LU$ factorization of $A$ once more to compute $z = A^{-1}c$ , from which our solution $y = (A+B)^{-1}c$ is given by

(4) $\begin{equation*} y = (I_n-(A^{-1}L)(I_r+R^\top(A^{-1}L))^{-1}R^\top)A^{-1}c = z - S((I_r+R^\top S)^{-1}(R^\top z)). \end{equation*}$

The net result is we solved our rank- $r$ -updated linear system using $r+1$ solutions of the original linear system with no need to recompute any $LU$ factorizations of $n\times n$ matrices. We’ve reduced the solution of the system $(A+B)y=c$ to an operation count of $\mathcal{O}(n^2r)$ which is dramatically better than the $\mathcal{O}(n^3)$ operation count of recomputing the LU factorization from scratch.

This simple example demonstrates a broader pattern: Usually if a matrix problem took $\mathcal{O}(n^3)$ to solve originally, one can usually solve the problem after a rank- $r$ update in an additional time of only something like $\mathcal{O}(n^2r)$ operations.⁹ For instance, not only can we solve rank- $r$ -updated linear systems in $\mathcal{O}(n^2r)$ operations, but we can actually update the LU factorization itself in $\mathcal{O}(n^2r)$ operations. Similar updates exist for Cholesky, QR, symmetric eigenvalue, and singular value decompositions to update these factorizations in $\mathcal{O}(n^2r)$ operations.

An important caveat is that, as always with linear algebraic computations, it’s important to read the fine print. There are many algorithms for computing low-rank updates to different matrix factorizations with dramatically different accuracy properties. Just because in principle rank-updated versions of these factorizations can be computed doesn’t mean it’s always advisable. With this qualification stated, these ways of updating matrix computations with low-rank updates can be a powerful tool in practice and reinforce the computational benefits of low-rank matrices expressed via rank factorizations.

Low-rank Approximation

As we’ve seen, computing with low-rank matrices expressed as rank factorizations can yield significant computational savings. Unfortunately, many matrices in application are not low-rank. In fact, even if a matrix in an application is low-rank, the small rounding errors we incur in storing it on a computer may destroy the matrix’s low rank, increasing its rank to the maximum possible value of $\min(m,n)$ . The solution in this case is straightforward: approximate our high-rank matrix with a low-rank one, which we express in algorithmically useful form as a rank factorization.

Here’s one simple way of constructing low-rank approximations. Start with a matrix $B$ and compute a singular value decomposition of $B$ , $B = U\Sigma V^\top$ . Recall from two sections previous that the rank of the matrix $B$ is equal to its number of nonzero singular values. But what if $B$ ‘s singular values aren’t exactly zero, but they’re very small? It seems reasonable to expect that $B$ is nearly low-rank in this case. Indeed, this intuition is true. To approximate $B$ a low-rank matrix, we can truncate $B$ ‘s singular value decomposition by setting $B$ ‘s small singular values to zero. If we zero out all but the $r$ largest singular values of $B$ , this procedure results in a rank- $r$ matrix $\hat{B}$ which approximates $B$ . If the singular values that we zeroed out were tiny, then $\hat{B}$ will be very close to $B$ and the low-rank approximation is accurate. This matrix $\hat{B}$ is called an $r$ -truncated singular value decomposition of $B$ , and it is easy to represent it using a rank factorization once we have already computed an SVD of $B$ .

It is important to remember that low-rank approximations are, just as the name says, approximations. Not every matrix is well-approximated by one of small rank. A matrix may be excellently approximated by a rank-100 matrix and horribly approximated by a rank-90 matrix. If an algorithm uses a low-rank approximation as a building block, then the approximation error (the difference between $B$ and its low-rank approximation $\hat{B}$ ) and its propagations through further steps of the algorithm need to be analyzed and controlled along with other sources of error in the procedure.

Despite this caveat, low-rank approximations can be startlingly effective. Many matrices occurring in practice can be approximated to negligible error by a matrix with very modestly-sized rank. We shall return to this surprising ubiquity of approximately low-rank matrices at the end of the article.

We’ve seen one method for computing low-rank approximations, the truncated singular value decomposition. As we shall see in the next section, the truncated singular value decomposition produces excellent low-rank approximations, the best possible in a certain sense, in fact. As we mentioned above, almost every matrix factorization can be used to compute rank factorizations. Can these matrix factorizations also compute high quality low-rank approximations?

Let’s consider a specific example to see the underlying ideas. Say we want to compute a low-rank approximation to a matrix $B$ by a QR factorization. To do this, we want to compute a QR factorization $B = QR$ and then throw away all but the first $r$ columns of $Q$ and the first $r$ rows of $R$ . This will be a good approximation if the rows we discard from $R$ are “small” compared to the rows of $R$ we keep. Unfortunately, this is not always the case. As a worst case example, if the first $r$ columns of $B$ are zero, then the first $r$ rows of $R$ will definitely be zero and the low-rank approximation computed this way is worthless.

We need to modify something to give QR factorization a fighting chance for computing good low-rank approximations. The simplest way to do this is by using column pivoting, where we shuffle the columns of $B$ around to bring columns of the largest size “to the front of the line” as we computing the QR factorization. QR factorization with column pivoting produces excellent low-rank approximations in a large number of cases, but it can still give poor-quality approximations for some special examples. For this reason, numerical analysts have developed so-called strong rank-revealing QR factorizations, such as the one developed by Gu and Eisenstat, which are guaranteed to compute quite good low-rank approximations for every matrix $B$ . Similarly, there exists a strong rank-revealing LU factorizations which can compute good low-rank approximations using LU factorization.

The upshot is that most matrix factorizations you know and love can be used to compute good-quality low-rank approximations, possibly requiring extra tricks like row or column pivoting. But this simple summary, and the previous discussion, leaves open important questions: what do we mean by good-quality low-rank approximations? How good can a low-rank approximation be?

Best Low-rank Approximation

As we saw in the last section, one way to approximate a matrix by a lower rank matrix is by a truncated singular value decomposition. In fact, in some sense, this is the best way of approximating a matrix by one of lower rank. This fact is encapsulated in a theorem commonly referred to as the Eckart–Young theorem, though the essence of the result is originally due to Schmidt and the modern version of the result to Mirsky.¹⁰

But what do we mean by best approximation? One ingredient we need is a way of measuring how big the discrepancy between two matrices is. Let’s define a measure of the size of a matrix $E$ which we will call $E$ ‘s norm, which we denote as $\|E\|$ . If $B$ is a matrix and $\hat{B}$ is a low-rank approximation to it, then $\hat{B}$ is a good approximation to $B$ if the norm $\|B-\hat{B}\|$ is small. There might be many different ways of measuring the size of the error, but we have to insist on a couple of properties on our norm $\|\cdot\|$ for it to really define a sensible measure of size. For instance if the norm of a matrix $E$ is $\|E\| = 5$ , then the norm of $10E$ should be $\|10E\| = 10\|E\| = 50$ . A list of the properties we require a norm to have are listed on the Wikipedia page for norms. We shall also insist on one more property for our norm: the norm should be unitarily invariant.¹¹ What this means is the norm of a matrix $E$ remains the same if it is multiplied on the left or right by an orthogonal matrix. This property is reasonable since multiplication by orthogonal matrices geometrically represents a rotation or reflection¹² which preserves distances between points, so it makes sense that we should demand that the size of a matrix as measured by our norm does not change by such multiplications. Two important and popular matrix norms satisfy the unitarily invariant property: the Frobenius norm $\| E\|_{\rm F} = \sum_{ij} |E_{ij}|^2$ and the spectral (or operator 2-) norm $\| E \|_{\rm op} = \sigma_{\rm max}(E)$ , which measures the largest singular value.¹³

With this preliminary out of the way, the Eckart–Young theorem states that the truncated singular value decomposition of $B$ truncated to rank $r$ is the closest of all rank- $r$ matrices $B$ when distances are measured using any unitarily invariant norm $\|\cdot\|$ . If we let $B_r$ denote the $r$ -truncated singular value decomposition of $B$ , then the Eckart–Young theorem states that

(5) $\begin{equation*} \| B - B_r \| \le \|B - C\| \mbox{ for all matrices $C$ of rank $r$}. \end{equation*}$

Less precisely, the $r$ -truncated singular value decomposition is the best rank- $r$ approximation to a matrix.

Let’s unpack the Eckart–Young theorem using the spectral and Frobenius norms. In this context, a brief calculation and the Eckart–Young theorem proves that for any rank- $r$ matrix $C$ , we have

(6) $\begin{equation*} \| B - C \|_{\rm op} \ge \sigma_{r+1},\quad \| B - C\|_{\rm F} \ge \sqrt{\sum_{j>r} \sigma_j^2}, \end{equation*}$

where $\sigma_1,\sigma_2,\ldots$ are the singular values of $B$ . This bound is quite intuitive. The error in low-rank approximation will be “small” when we measure the error in the spectral norm when each singular value we zero out is “small”. When we measure error in the Frobenius norm, the error in low-rank approximation is “small” when all of the singular values we zero out are “small” in aggregate when squared and added together.

The Eckart–Young theorem shows that possessing a good low-rank approximation is equivalent to the singular values rapidly decaying.¹⁴ If a matrix does not have nice singular value decay, no good low-rank approximation exists, computed by the $r$ -truncated SVD or otherwise.

Why Are So Many Matrices (Approximately) Low-rank?

As we’ve seen, we can perform computations with low-rank matrices represented using rank factorizations much faster than general matrices. But all of this would be a moot point if low-rank matrices rarely occurred in practice. But in fact precisely the opposite is true: Approximately low-rank matrices occur all the time in practice.

Sometimes, exact low-rank matrices appear for algebraic reasons. For instance, when we perform one step Gaussian elimination to compute an $LU$ factorization, the lower right portion of the eliminated matrix, the so-called Schur complement, is a rank-one update to the original matrix. In such cases, a rank- $r$ matrix might appear in a computation when one performs $r$ steps of some algebraic process: The appearance of low-rank matrices in such cases is unsurprising.

However, often, matrices appearing in applications are (approximately) low-rank for analytic reasons instead. Consider the weather example from the start again. One might reasonably model the temperature on Earth as a smooth function $T(\cdot,\cdot)$ of position $x$ and time $t$ . If we then let $x_i$ denote the position on Earth of station $i$ and $t_j$ the time representing the $j$ th day of a given year, then the entries of the $W$ matrix are given by $W_{ij} = T(x_i,t_j)$ . As discussed in my article on smoothness and degree of approximation, a smooth function function of one variable can be excellently approximated by, say, a polynomial of low degree. Analogously, a smooth function depending on two arguments, such as our function $T(\cdot,\cdot)$ , can be excellently be approximated by a separable expansion of rank $r$ :

(7) $\begin{equation*} T(x,t) \approx \phi_1(x) \psi_1(t) + \cdots + \phi_r(x) \psi_r(t). \end{equation*}$

Similar to functions of a single variable, the degree to which a function $T(\cdot,\cdot)$ can to be approximated by a separable function of small rank depends on the degree smoothness of the function $T(\cdot,\cdot)$ . Assuming the function $T(\cdot,\cdot)$ is quite smooth, then $T(\cdot,\cdot)$ can be approximated has a separable expansion of small rank $r$ . This leads immediately to a low-rank approximation to the matrix $W$ given by the rank factorization

(8) $\begin{equation*} W = \begin{bmatrix} \phi_1(x_1) & \cdots & \phi_r(x_1) \\ \vdots & \ddots & \vdots \\ \phi_1(x_{1000}) & \cdots & \phi_r(x_{1000}) \end{bmatrix}\begin{bmatrix} \psi_1(t_1) & \cdots & \psi_r(t_1) \\ \vdots & \ddots & \vdots \\ \psi_1(t_{365}) & \cdots & \psi_r(t_{365}) \end{bmatrix}^\top. \end{equation*}$

Thus, in the context of our weather example, we see that the data matrix can be expected to be low-rank under the reasonable-sounding assumption that the temperature depends smoothly on space and time.

What does this mean in general? Let’s speak informally. Suppose that the $ij$ th entries of a matrix $B$ are samples $f(x_i,y_j)$ from a smooth function $f(\cdot,\cdot)$ for points $x_1,\ldots,x_m$ and $y_1,\ldots,y_n$ . Then we can expect that $B$ will be approximately low-rank. From a computational point of view, we don’t need to know a separable expansion for the function $f(\cdot,\cdot)$ or even the form of the function $f(\cdot,\cdot)$ itself: If the smooth function $f(\cdot,\cdot)$ exists and $B$ is sampled from it, then $B$ is approximately low-rank and we can find a low-rank approximation for $B$ using the truncated singular value decomposition.¹⁵

This “smooth function” explanation for the prevalence of low-rank matrices is the reason for the appearance of low-rank matrices in fast multipole method-type fast algorithms in computational physics and has been proposed¹⁶ as a general explanation for the prevalence of low-rank matrices in data science.

(Another explanation for low-rank structure for highly structured matrices like Hankel, Toeplitz, and Cauchy matrices¹⁷ which appear in control theory applications has a different explanation involving a certain Sylvester equation; see this lecture for a great explanation.)

Upshot: A matrix is low-rank if it has many fewer linearly independent columns than columns. Such matrices can be efficiently represented using rank-factorizations, which can be used to perform various computations rapidly. Many matrices appearing in applications which are not genuinely low-rank can be well-approximated by low-rank matrices; the best possible such approximation is given by the truncated singular value decomposition. The prevalence of low-rank matrices in diverse application areas can partially be explained by noting that matrices sampled from smooth functions are approximately low-rank.

Why Randomized Algorithms?

August 11, 2021 by Ethan N. Epperly 4 Comments

In this post, I want to answer a simple question: how can randomness help in solving a deterministic (non-random) problem?

Let’s start by defining some terminology. An algorithm is just a precisely defined procedure to solve a problem. For example, one algorithm to compute the integral of a function $f$ on the interval $[0,1]$ is to pick 100 equispaced points $0 = x_1 < \tfrac{1}{100} = x_2 < \tfrac{2}{100} = x_3 <\cdots<\tfrac{99}{100}=x_{100}$ on this interval and output the Riemann sum $\tfrac{1}{n} \sum_{j=1}^{100} f(x_j)$ . A randomized algorithm is simply an algorithm which uses, in some form, randomly chosen quantities. For instance, a randomized algorithm¹ for computing integrals would be to pick 100 random points $X_1,\ldots,X_{100}$ uniformly distributed in the interval $[0,1]$ and output the average $\tfrac{1}{n} \sum_{j=1}^{100} f(X_j)$ . To help to distinguish, an algorithm which does not use randomness is called a deterministic algorithm.

To address the premise implicit in our central question, there are problems where randomized algorithms provably outperform the best possible deterministic algorithms.² Additionally, even for problems for which randomized algorithms have not been formally proven to outperform deterministic ones, the best randomized algorithms we know for many problems dramatically outperform the best-known deterministic algorithms. For example, testing whether an $n$ -digit number is prime takes roughly $n^6$ operations using the best-known deterministic algorithm and only roughly $n^2$ operations allowing randomness. The power of randomness extends beyond discrete problems typically considered by computer scientists to continuous problems of interest to computational mathematicians and scientists. For instance, the (asymptotically) fastest known algorithms to solve Laplacian linear system of equations use random sampling as a key ingredient. The workhorse optimization routines in machine learning are mostly variants of stochastic gradient descent, a randomized algorithm. To list off a few more, randomized algorithms have proven incredibly effective in solving eigenvalue and singular value problems, approximating the trace of a very large matrix, computing low-rank approximations, evaluating integrals, and simulating subgrid phenomena in fluid problems.

This may seem puzzling and even paradoxical. There is no randomness in, say, a system of linear equations, so why should introducing randomness help solve such a problem? It can seem very unintuitive that leaving a decision in an algorithm up to chance would be better than making an informed (and non-random) choice. Why do randomized algorithms work so well; what is the randomness actually buying us?

A partial answer to this question is that it can be very costly to choose the best possible option at runtime for an algorithm. Often, a random choice is “good enough”.

A Case Study: Quicksort

Consider the example of quicksort. Given a list of $N$ integers to sort in increasing order, quicksort works by selecting an element to be the pivot and then divides the elements of the list into groups larger and smaller than the pivot. One then recurses on the two groups, ending up with a sorted list.

The best choice of pivot is the median of the list, so one might naturally think that one should use the median as the pivot for quicksort. The problem with this reasoning is that finding the median is time-consuming; even using the fastest possible median-finding algorithm,³quicksort with exact median pivot selection isn’t very quick. However, one can show quicksort with a pivot selected (uniformly) at random achieves the same $\mathcal{O}(N\log N)$ expected runtime as quicksort with optimal pivot selection and is much faster in practice.⁴

But perhaps we have given up on fast deterministic pivot selection in quicksort too soon. What if we just pick the first entry of the list as the pivot? This strategy usually works fairly well too, but it runs into an unfortunate shortfall: if one feeds in a sorted list, the first-element pivot selection is terrible, resulting in a $\mathcal{O}(N^2)$ algorithm. If one feeds in a list in a random order, the first-element pivot selection has a $\mathcal{O}(N \log N)$ expected runtime,⁵ but for certain specific orderings (in this case, e.g., the sorted ordering) the runtime is a disappointing $\mathcal{O}(N^2)$ . The first-element pivot selection is particularly bad since its nemesis ordering is the common-in-practice already-sorted ordering. But other simple deterministic pivot selections are equally bad. If one selects, for instance, the pivot to be the entry in the position $\lfloor N/2 \rfloor$ , then we can still come up with an ordering of the input list that makes the algorithm run in time $\mathcal{O}(N^2)$ . And because the algorithm is deterministic, it runs in $\mathcal{O}(N^2)$ time every time with such-ordered inputs. If the lists we need to sort in our code just happen to be bad for our deterministic pivot selection strategy, quicksort will be slow every time.

We typically analyze algorithms based on their worst-case performance. And this can often be unfair to algorithms. Some algorithms work excellently in practice but are horribly slow on manufactured examples which never occur in “real life”.⁶ But average-case analysis of algorithms, where one measures the average performance of an algorithm over a distribution of inputs, can be equally misleading. If one were swayed by the average-case analysis of the previous section for quicksort with first-element pivot selection, one would say that this should be an effective pivot selection in practice. But sorting an already-sorted or nearly-sorted list occurs all the time in practice: how many times do programmers read in a sorted list from a data source and sort it again just to be safe? In real life, we don’t encounter random inputs to our algorithms: we encounter a very specific distribution of inputs specific to the application we use the algorithm in. An algorithm with excellent average-case runtime analysis is of little use to me if it is consistently and extremely slow every time I run it on my input data on the problem I’m working on. This discussion isn’t meant to disparage average-case analysis (indeed, very often the “random input” model is not too far off), but to explain why worst-case guarantees for algorithms can still be important.

To summarize, often we have a situation where we have an algorithm which makes some choice (e.g. pivot selection). A particular choice usually works well for a randomly chosen input, but can be stymied by certain specific inputs. However, we don’t get to pick which inputs we receive, and it is fully possible the user will always enter inputs which are bad for our algorithms. Here is where randomized algorithms come in. Since we are unable to randomize the input, we instead randomize the algorithm.

A randomized algorithm can be seen as a random selection from a collection of deterministic algorithms. Each individual deterministic algorithm may be confounded by an input, but most algorithms in the collection will do well on any given input. Thus, by picking a random algorithm from our collection, the probability of poor algorithmic performance is small. And if we run our randomized algorithm on an input and it happens to do poorly by chance, if we run it again it is unlikely to do so; there isn’t a specific input we can create which consistently confounds our randomized algorithm.

The Game Theory Framework

Let’s think about algorithm design as a game. The game has two players: you and an opponent. The game is played by solving a certain algorithmic problem, and the game has a score which quantifies how effectively the problem was solved. For example, the score might be the runtime of the algorithm, the amount of space required, or the error of a computed quantity. For simplicity of discussion, let’s use runtime as an example for this discussion. You have to pay your opponent a dollar for every second of runtime, so you want to have the runtime be as low as possible.⁷

Each player makes one move. Your move is to present an algorithm $A$ which solves the problem. Your opponent’s move is to provide an input $I$ to your algorithm. Your algorithm $A$ is then run on your opponents input $I$ and you pay your opponent a dollar for every second of runtime.

This setup casts algorithm design as a two-person, zero-sum game. It is a feature of such games that a player’s optimal strategy is often mixed, picking a move at random subject to some probability distribution over all possible moves.

To see why this is the case, let’s consider a classic and very simple game: rock paper scissors. If I use a deterministic strategy, then I am forced to pick one choice, say rock, and throw it every time. But then my savvy opponent could pick paper and be assured victory. By randomizing my strategy and selecting between rock, paper, and scissors (uniformly) at random, I improve my odds to winning a third of the time, losing a third of the time, and drawing a third of the time. Further, there is no way my opponent can improve their luck by adopting a different strategy. This strategy is referred to as minimax optimal: among all (mixed) strategies I could adopt, this strategy has the best performance over all strategies provided my opponent always counters my strategy with the best response strategy they can find.

Algorithm design is totally analogous. For the pivot selection problem, if I pick a the fixed strategy of choosing the first entry to be the pivot (analogous to always picking rock), then my opponent could always give me a sorted list (analogous to always picking paper) and I would always lose. My randomizing my pivot selection, my opponent could input the list in whatever order they choose and my pivot selection will always have the excellent (expected) $\mathcal{O}(N\log N)$ runtime characteristic of quicksort. In fact, randomized pivot selection is the minimax optimal pivot selection strategy (assuming pivot selection is non-adaptive in that we don’t choose the pivot based on the values in the list).⁸

How Much Does Randomness Buy?

Hopefully, now, the utility of randomness is less opaque. Randomization, in effect, allows an algorithm designer to trade algorithm which runs fast for most inputs all of the time for an algorithm which runs fast for all inputs most of the time. They do this by introducing randomized decision-making to hedge against particular bad inputs which could confound their algorithm.

Randomness, of course, is not a panacea. Randomness does not allow us to solve every algorithmic problem arbitrarily fast. But how can we quantify this? Are there algorithmic speed limits for computational problems for which no algorithm, randomized or not, can exceed?

The game theoretic framework for randomized algorithms can shed light on this question. Let us return to the framing of last section where you choose an algorithm $A$ , your opponent chooses an input $I$ , and you pay your opponent a dollar for every second of runtime. Since this cost depends on the algorithm and the input, we can denote the cost $C(A,I)$ .

Suppose you devise a randomized algorithm for this task, which can be interpreted as selecting an algorithm $A$ from a probability distribution $P$ over the class of all algorithms $\mathcal{A}$ which solve the problem. (Less formally, we assign each possible algorithm a probability of picking it and pick one subject to these probabilities.) Once your opponent sees your randomized algorithm (equivalently, the distribution $P$ ), they can come up with the on-average slowest possible input $I_{\rm worst}$ and present it to your algorithm.

But now let’s switch places to see things from your opponent. Suppose they choose their own strategy of randomly selecting an input $I$ from among all inputs $\mathcal{I}$ with some distribution $Q$ . Once you see the distribution $Q$ , you can come up with the best possible deterministic algorithm $A_{\rm best}$ to counter their strategy.

The next part gets a bit algebraic. Suppose now we apply your randomized algorithm against your opponents strategy. Then, your randomized algorithm could only take longer on average than $A_{\rm best}$ because, by construction, $A_{\rm best}$ is the fastest possible algorithm against the input distribution $Q$ . Symbolically, we have

(1) $\begin{equation*} \mathbb{E}_{I\sim Q} \left[ C(A_{\rm best},I) \right] \le \mathbb{E}_{A\sim P} \left[ \mathbb{E}_{I\sim Q} \left[ C(A, I) \right] \right]. \end{equation*}$

Here, $\mathbb{E}_{I\sim Q}$ and $\mathbb{E}_{A\sim P}$ denote the average (expected) value of a random quantity when an input $I$ is drawn randomly with distribution $Q$ or an algorithm $A$ is drawn randomly with distribution $P$ . But we know that $I_{\rm worst}$ is the slowest input for our randomized algorithm, so, on average, our randomized algorithm will take longer on worst-case input $I_{\rm worst}$ then a random input from $Q$ . In symbols,

(2) $\begin{equation*} \mathbb{E}_{A\sim P} \left[ \mathbb{E}_{I\sim Q} \left[ C(A, I) \right] \right] \le \mathbb{E}_{A\sim P} \left[ C(A,I_{\rm worst}) \right]. \end{equation*}$

Combining Eqs. (1) and (2) gives Yao’s minimax principle⁹:

(3) $\begin{equation*} \mathbb{E}_{I\sim Q} \left[ C(A_{\rm best},I) \right] \le \mathbb{E}_{A\sim P} \left[ C(A,I_{\rm worst}) \right]. \end{equation*}$

In words, the average performance of any randomized algorithm on its worst-case input can be no better than the average performance of the best possible deterministic algorithm for a distribution of inputs.¹⁰

This, in effect, allows the algorithm analyst seeking to prove “no randomized algorithm can do better than this” to trade randomness in the algorithm to randomness in the input in their analysis. This is a great benefit because randomness in an algorithm can be used in arbitrarily complicated ways, whereas random inputs can be much easier to understand. To prove any randomized algorithm takes at least cost $c$ to solve a problem, the algorithm analyst can find a distribution $Q$ on which every deterministic algorithm takes at least cost $c$ . Note that Yao’s minimax principle is an analytical tool, not an algorithmic tool. Yao’s principle establishes speed limits on the best possible randomized algorithm: it does not imply that one can just use deterministic algorithms and assume or make the input to be random.

There is a fundamental question concerning the power of randomized algorithms not answered by Yao’s principle that is worth considering: how much better can randomized algorithms be than deterministic ones? Could infeasible problems for deterministic computations be solved tractably by randomized algorithms?

Questions such as these are considered in the field of computational complexity theory. In this context, one can think of a problem as tractably solvable if it can be solved in polynomial time—that is, in an amount of time proportional to a polynomial function of the size of the input. Very roughly, we call the class of all such problems to be $\mathsf{P}$ . If one in addition allows randomness, we call this class of problems $\mathsf{BPP}$ .¹¹

It has widely been conjectured that $\mathsf{P} = \mathsf{BPP}$ : all problems that can be tractably solved with randomness can tractably be solved without. There is some convincing evidence for this belief. Thus, provided this conjecture turns out to be true, randomness can give us reductions in operation counts by a polynomial amount in the input size for problems already in $\mathsf{P}$ , but they cannot efficiently solve $\mathsf{NP}$ -hard computational problems like the traveling salesman problem.¹²

So let’s return to the question which is also this article’s title: why randomized algorithms? Because randomized algorithms are often faster. Why, intuitively, is this the case? Randomization can us to upgrade simple algorithms that are great for most inputs to randomized algorithms which are great most of the time for all inputs. How much can randomization buy? A randomized algorithm on its worst input can be no better than a deterministic algorithm on a worst-case distribution. Assuming a widely believed and theoretically supported, but not yet proven, conjecture, randomness can’t make intractable problems into tractable ones. Still, there is great empirical evidence that randomness can be an immensely effective algorithm design tool in practice, including computational math and science problems like trace estimation and solving Laplacian linear systems.

Minimal Rank Completions

June 22, 2021 by Ethan N. Epperly Leave a comment

I’m delighted to share that my first applied mathematics paper, Minimal Rank Completions for Overlapping Blocks, coauthored with Nithin Govindarajan and Shivkumar Chandrasekaran, has been accepted for publication in Linear Algebra and its Applications. (We also have a preprint on arXiv.) In this post, I wanted to share what our paper is about and share an alternate derivation which ended up being cut from the final paper.

Our paper is concerned with the following question: given a collection of matrices which overlap in a particular way, can we choose the region of simultaneous overlap so as to minimize the rank of each of the individual matrices? This is a multi-objective optimization problem, so a priori it is not clear it has a simultaneous solution. So it was of great surprise to us when we discovered that it does. In fact, we were able to find a tractable algorithm to find all solutions to this multi-objective problem.

Our enormous thanks go to the anonymous reviewer, who went above and beyond by discovering and sketching to us a dramatically shorter and more revealing proof¹ of our main theorem. We are truly indebted to this extraordinary reviewer for the significantly streamlined and more understandable presentation in the current iteration of our paper.

For the remainder of this post, I want to discuss minimal rank completions in a little more generality and present a solution to a special case of a minimal rank completion problem which can be tractably solved. I will present a solution to this problem we discovered which I have not seen elsewhere in the literature. I hope that this perspective will be complementary to existing solutions to this problem published in the literature including the summary of this result in our paper.

Minimal Rank Completion Problems

Minimal rank completions have achieved a lot of buzz in statistics and data science. Given a matrix with unknown entries, it is often very natural to impute the missing entries as those minimizing the rank of the matrix. This is justified by a kind of Occam’s razor argument: big data matrices are very often (approximately) low-rank, so when data is missing, we may assume it takes whatever values are necessary to minimize the rank.

Finding the rank-minimizing choice of missing entries is, in general, NP-hard. However, under certain assumptions, semidefinite programming relaxations can sometimes exactly recover the missing entries.

Alternately, if the missing entries belong to a special pattern, it may be possible to find the rank-minimizing choice of the unknown entries using linear algebraic tools like matrix factorizations. These approaches have significantly more limited domains of applicability than optimization-based approaches, but have advantages in that they

can be more tractable as they involve matrix factorizations rather than solving optimization problems;
work over arbitrary fields of numbers, such as finite fields appearing in computer science;
can find all ways of choosing the entries to minimize the rank, in particular certifying or dis-certifying the uniqueness of the rank-minimizing choice; and
are guaranteed to find the rank-minimizer, without any statistical assumptions.

These algebraic methods have their roots in systems theory, integral operators, and rank-structured matrices. The last of these was the application which motivated our interest in the subject.

Our paper concerns an overlapping variant of this problem, where we simultaneously minimize the ranks of several matrices by choosing the entries in the overlap carefully. This problem emerged to us naturally in the construction of certain matrix representations, and we hope it might prove useful in tackling other problems. As it turns out, the overlapping problem can be solved very much in the spirit of a much simpler problem, the block $2\times 2$ minimal rank completion problem, which we will spend most of the remainder of this post discussing.

A Solution to the Block 2×2 Case

The block $2\times 2$ minimal rank completion problem is as follows: given a partially filled block matrix

(1) $\begin{equation*} \begin{bmatrix} A & B \\ ? & C \end{bmatrix} \end{equation*}$

how can the “?” be filled in to minimize the rank of this matrix?

A generalized version of this problem was originally solved by Kaashoek and Woerdeman. An alternate derivation using matrix factorizations is given by Eidelman, Gohberg, and Haimovici, though they only find some of the solutions to this problem. We seek to characterize all ways of choosing the “?” so that the rank is minimize the rank.

Here, I present the solution to this problem which my coauthors and I originally discovered, which is different than the one we summarize in the final version of our paper.² This solution is in the spirit of the one by Eidelman, Gohberg, and Haimovici but does produce all solutions.

Let’s start by recalling some facts about linear algebra. Given an $m\times n$ real³ matrix $K$ , one can define its column space, which I will denote $\operatorname{Col} K$ , to be the vector space consist of all linear combinations of the columns of $K$ . From this column space, or indeed any vector subspace of $\mathbb{R}^m$ , one can extract a basis for this subspace, meaning every vector in the subspace can be uniquely decomposed as a linear combination of elements from the basis. In this discussion, we shall always arrange the basis vectors as columns of a basis matrix $P$ . If we instead consider the row space $\operatorname{Row} K$ of $K$ , then we arrange the elements of a basis as rows of a matrix $Q$ .

Before we solve the problem, let’s reason about the lowest possible rank we could possibly expect for the completed matrix. Since the rank of a matrix can be no smaller than the rank of any of its submatrices, clearly we must have that, for any assignment $X$ of the “?”,

(2) $\begin{equation*} \operatorname{rank} \begin{bmatrix} A & B \\ X & C \end{bmatrix} \ge \operatorname{rank} \begin{bmatrix} A & B\end{bmatrix}. \end{equation*}$

However, in general, the rank of the completed matrix must be higher than $\operatorname{rank} \begin{bmatrix} A & B\end{bmatrix}$ , as is exemplified when $A$ and $B$ are both zero but $C$ is not. In addition to rank of $\begin{bmatrix} A & B\end{bmatrix}$ , we must also account for rows of $\begin{bmatrix} X & C\end{bmatrix}$ which cannot be written as linear combinations of the rows above, no matter how $X$ is chosen. With a little noodling, one can convince oneself that there are always at least $\operatorname{rank} \begin{bmatrix} B \\ C \end{bmatrix} - \operatorname{rank} B$ such rows, leading to the bound

(3) $\begin{equation*} \operatorname{rank} \begin{bmatrix} A & B \\ X & C \end{bmatrix} \ge \operatorname{rank} \begin{bmatrix} A & B\end{bmatrix} + \operatorname{rank} \begin{bmatrix} B \\ C \end{bmatrix} - \operatorname{rank} B =: r_{\rm opt}. \end{equation*}$

We shall show that, by judicious choice of $X$ , Eq. (3) can always be achieved with equality, making $r_{\rm opt}$ the minimal rank for this completion problem.

Let us now find such a rank-minimizing $X$ . The construction begins by considering the column spaces $\operatorname{Col} A$ and $\operatorname{Col} B$ of the matrices $A$ and $B$ . The intersection of these column spaces $\operatorname{Col} A \cap \operatorname{Col} B$ is again a vector subspace, and we choose a basis $P_{AB}$ for it. The subspace $\operatorname{Col} A \cap \operatorname{Col} B$ might have a smaller size than $\operatorname{Col} A$ . Therefore, to find a basis for $\operatorname{Col} A$ , we can extend the basis $P_{AB}$ by adding new columns $P_{\overline{A}}$ to it so that the enlarged matrix $\begin{bmatrix} P_{AB} & P_{\overline{A}} \end{bmatrix}$ forms a basis for $\operatorname{Col} A$ . Similarly, we can find new columns $P_{\overline{B}}$ to add to $P_{AB}$ so that the matrix $\begin{bmatrix} P_{AB} & P_{\overline{B}} \end{bmatrix}$ forms a basis for $\operatorname{Col} B$ .

Now comes something subtle. Since $P = \begin{bmatrix} P_{AB} & P_{\overline{A}} \end{bmatrix}$ forms a basis for $\operatorname{Col} A$ , every column in $A$ can be written as a linear combination of columns in $P$ . In matrix language, this means there exists a matrix $Q^\top$ such that $A = PQ^\top$ —in fact, this exact $Q$ forms a basis for $\operatorname{Row} A$ . Since $P$ is divided into two collections of columns, it is only natural to similarly (and conformally) divide $Q$ into two pieces as $Q = \begin{bmatrix} Q_{AB,A} & Q_{\overline{A},A} \end{bmatrix}$ . Thus, doing the same with $B$ as we did with $A$ , we obtain two matrix factorizations

(4) $\begin{equation*} A = \begin{bmatrix} P_{AB} & P_{\overline{A}} \end{bmatrix} \begin{bmatrix} Q_{AB,A}^\top \\ Q_{\overline{A},A}^\top \end{bmatrix}, \quad B = \begin{bmatrix} P_{AB} & P_{\overline{B}} \end{bmatrix} \begin{bmatrix} Q_{AB,B}^\top \\ Q_{\overline{B},B}^\top \end{bmatrix}. \end{equation*}$

Now we do the same song and dance with the row spaces of $B$ and $C$ . Let’s go through the construction somewhat quickly. We start with a basis $Q_{BC}$ for the intersection of $\operatorname{Row} B \cap \operatorname{Row} C$ . We then extend these to bases $\begin{bmatrix} Q_{BC} & Q_{\overline{B}} \end{bmatrix}$ and $\begin{bmatrix} Q_{BC} & Q_{\overline{C}} \end{bmatrix}$ for $\operatorname{Row} B$ and $\operatorname{Row} C$ . From here, we derive the existence of matrix factorizations analogous to Eq. (4):

(5) $\begin{equation*} B = \begin{bmatrix} P_{B,BC} & P_{B,\overline{B}} \end{bmatrix} \begin{bmatrix} Q_{BC}^\top \\ Q_{\overline{B}}^\top \end{bmatrix},\quad C = \begin{bmatrix} P_{C,BC} & P_{C,\overline{C}} \end{bmatrix} \begin{bmatrix} Q_{BC}^\top \\ Q_{\overline{C}}^\top \end{bmatrix}. \end{equation*}$

For the next part, we shall take advantage of a powerful fact: if I have a bases $P$ and $Q$ for the row and column spaces of a matrix $K$ , there exists a nonsingular matrix $R$ for which $K=PRQ^\top$ . Applying this fact to the bases $\begin{bmatrix} P_{B,BC} & P_{B,\overline{B}} \end{bmatrix}$ and $\begin{bmatrix} Q_{AB,B} & Q_{\overline{B},B} \end{bmatrix}$ for $B$ ‘s column and row spaces, we get the existence of a matrix $R = \begin{bmatrix} R_{BC,AB} & R_{\overline{B},AB} \\ R_{BC,\overline{B}} & R_{\overline{B},\overline{B}} \end{bmatrix}$ such that

(6) $\begin{equation*} B = \begin{bmatrix} P_{B,BC} & P_{B,\overline{B}} \end{bmatrix}\begin{bmatrix} R_{BC,AB} & R_{BC,\overline{B}} \\ R_{\overline{B},AB} & R_{\overline{B},\overline{B}} \end{bmatrix} \begin{bmatrix} Q_{AB,B}^\top \\ Q_{\overline{B},B}^\top \end{bmatrix}. \end{equation*}$

We now have all ingredients to describe the solution. Rather than trying to “derive” the solution in a rigorous way, let us try and discover the solution in a non-rigorous way and justify our work later. We’re hoping to find assignments $X$ of the “?” such that $\begin{bmatrix} A & B \\ X & C \end{bmatrix}$ achieves the minimum possible rank $r_{\rm opt}$ defined in Eq. (3). To do so, let’s try and find a rank factorization of the completed matrix and then infer what values $X$ will take. First off, let’s build a rank factorization for $\begin{bmatrix} A & B \end{bmatrix}$ using the building blocks we’ve already built

(7) $\begin{equation*} \begin{bmatrix} A & B \end{bmatrix} = \begin{bmatrix} P_{AB} & P_{\overline{B}} & P_{\overline{A} \end{bmatrix} \begin{bmatrix} Q_{AB,A}^\top & Q_{AB,B}^\top \\ 0 & Q_{\overline{B},B}^\top \\ Q_{\overline{A},A}^\top & 0 \end{bmatrix}. \end{equation*}$

Now we want to extend this to a rank factorization for the entire completed matrix. Let’s build up to this in stages, denoting by $\star$ entries whose value we have yet to determine. For us to have a hope of representing the matrix $C$ , we’ll need to somehow add $Q^\top_{\overline{C}}$ into our rank factorization. Doing exactly this, we get the partially specified rank factorization

(8) $\begin{equation*} \begin{bmatrix} A & B \\ X & C \end{bmatrix} = \begin{bmatrix} P_{AB} & P_{\overline{B}} & P_{\overline{A} & 0 \\ \star & \star & \star & \star \end{bmatrix} \begin{bmatrix} Q_{AB,A}^\top & Q_{AB,B}^\top \\ 0 & Q_{\overline{B},B}^\top \\ Q_{\overline{A},A}^\top & 0\\ \star & Q_{\overline{C}}^\top \end{bmatrix}. \end{equation*}$

Now, to fill in the second block row of the left factor, we recall that $C = P_{C,BC} Q_{BC}^\top + P_{C,\overline{C}} Q_{\overline{C}}^\top$ . From Eqs. (5) and (6), we deduce that $Q_{BC}^\top = R_{BC,AB} Q_{AB,B}^\top + R_{BC,\overline{B}} Q_{\overline{B},B}^\top$ . Thus, we can fill in more entries:

(9) $\begin{equation*} \begin{bmatrix} A & B \\ X & C \end{bmatrix} = \begin{bmatrix} P_{AB} & P_{\overline{B}} & P_{\overline{A} & 0 \\ P_{C,BC}R_{BC,AB} & P_{C,BC} R_{BC,\overline{B}} & \star & P_{C,\overline{C}} \end{bmatrix} \begin{bmatrix} Q_{AB,A}^\top & Q_{AB,B}^\top \\ 0 & Q_{\overline{B},B}^\top \\ Q_{\overline{A},A}^\top & 0\\ \star & Q_{\overline{C}}^\top \end{bmatrix}. \end{equation*}$

With these entries filled in, the remaining $\star$ ‘s can be chosen arbitrarily. Assigning names $F_{\overline{A}}^\top$ and $F_{\overline{C}}$ to these free variables, we conclude that

(10) $\begin{equation*} \begin{bmatrix} A & B \\ X & C \end{bmatrix} = \begin{bmatrix} P_{AB} & P_{\overline{B}} & P_{\overline{A} & 0 \\ P_{C,BC}R_{BC,AB} & P_{C,BC} R_{BC,\overline{B}} & F_{\overline{C}} & P_{C,\overline{C}} \end{bmatrix} \begin{bmatrix} Q_{AB,A}^\top & Q_{AB,B}^\top \\ 0 & Q_{\overline{B},B}^\top \\ Q_{\overline{A},A}^\top & 0\\ F_{\overline{A}}^\top & Q_{\overline{C}}^\top \end{bmatrix} \end{equation*}$

and

(11) $\begin{equation*} X = P_{C,BC}R_{BC,AB}Q_{AB,A}^\top + F_{\overline{A}} Q_{\overline{A},A}^\top + P_{C,\overline{C}} F_{\overline{C}}^\top. \end{equation*}$

From all the analysis previous, we know that all $X$ ‘s of the form Eq. (11) solve the minimal rank completion problem, making the completed matrix achieve the minimal rank $r_{\rm opt}$ defined in Eq. (3). With a little more elbow grease, one can also confirm that all such minimal completions $X$ are of the form Eq. (11), proving that Eq. (11) indeed characterizes all solutions to the minimal rank completion problem. This completes the characterization and construction of the complete set of minimizers to the block $2\times 2$ minimal rank completion problem.

The Overlapping Block Minimal Rank Completion Problem

If you found this post interesting, be sure to check out our paper (or here on arXiv) for an explanation of a different way of thinking about the solution of the block $2\times 2$ minimal rank completion problem and a solution to a more general “overlapping” variant. The treatment should be reasonably self-contained, and we hope the solution to this problem could prove a useful tool in tackling open problems in systems theory and the study of rank-structured matrices.

Big Ideas in Applied Math: The Fast Fourier Transform

May 10, 2021 by Ethan N. Epperly 4 Comments

The famous law of the instrument states that “when all you have is a hammer, every problem looks like a nail.” In general, this tendency is undesirable: most problems in life are not nails and could better be addressed by a more appropriate tool. However, one can also review the law of the instrument in a more positive framing: when presented with a powerful new tool, it is worth checking how many problems it can solve. The fast Fourier transform (FFT) is one of the most important hammers in an applied mathematician’s toolkit. And it has made many seemingly unrelated problems look like nails.

In this article, I want to consider three related questions:

What is the FFT—what problem is it solving and how does it solve it fast?
How can the ideas behind the FFT be used to solve other problems?
How can the FFT be used as a building block in solving a seemingly unrelated problem?

The FFT is widely considered one of the most important numerical algorithms, and as such every sub-community of applied mathematics is inclined to see the most interesting applications of the FFT as those in their particular area. I am unapologetically victim to this tendency myself, and thus will discuss an application of the FFT that I find particularly beautiful and surprising. In particular, this article won’t focus on the manifold applications of the FFT in signal processing, which I think has been far better covered by authors more familiar with that field.

The Discrete Fourier Transform

At its core, the FFT is a fast algorithm to compute $n$ complex numbers $\hat{f}_0,\ldots,\hat{f}_{n-1}$ given $n$ real or complex numbers $f_0,\ldots,f_{n-1}$ defined by the formula¹

(1) $\begin{equation*} \hat{f}_k = \frac{1}{\sqrt{n}} \sum_{j=0}^{n-1} f_j e^{-(2\pi i/n)jk} \quad \mbox{for } k = 0,1,2,\ldots,n-1. \end{equation*}$

The outputs $\hat{f} = (\hat{f}_0,\ldots,\hat{f}_{n-1})$ is called the discrete Fourier transform (DFT) of $f = (f_0,\ldots,f_{n-1})$ . The FFT is just one possible algorithm to evaluate the DFT.

The DFT has the following interpretation. Suppose that $f$ is a periodic function defined on the integers with period $n$ —that is, $f(j + n) = f(j)$ for every integer $j$ . Choose $f_0,\ldots,f_{n-1}$ to be the values of $f$ given by $f_j = f(j)$ for $j=0,1,2,\ldots,n-1$ . Then, in fact, $\hat{f}_0,\ldots,\hat{f}_{n-1}$ gives an expression for $f$ as a so-called trigonometric polynomial²:

(2) $\begin{equation*} f(j) = \frac{1}{\sqrt{n}} \sum_{k=0}^{n-1} \hat{f}_k e^{(2\pi i/n)jk} \mbox{ for every integer } j. \end{equation*}$

This shows that (1) converts function values $f_0,f_1,\ldots,f_{n-1}$ of a periodic function $f$ to coefficients $\hat{f}_0,\hat{f}_1,\ldots,\hat{f}_{n-1}$ of a trigonometric polynomial representation of $f$ , which can be called the Fourier series of $f$ . Eq. (2), referred to as the inverse discrete Fourier transform, inverts this, converting coefficients $\hat{f}_0,\hat{f}_1,\ldots,\hat{f}_{n-1}$ to function values $f_0,f_1,\ldots,f_{n-1}$ .

Fourier series are an immensely powerful tool in applied mathematics. For example again, if $f$ represents a sound wave produced by a chord on a piano, its Fourier coefficients $\hat{f}$ represents the intensity of each pitch comprising the chord. An audio engineer could, for example, compute a Fourier series for a piece of music and zero out Fourier coefficients, thus reducing the amount of data needed to store a piece of music. This idea is indeed part of the way audio compression standards like MP3 work. In addition to many more related applications in signal processing, the Fourier series is also a natural way to solve differential equations, either by pencil and paper or by computer via so-called Fourier spectral methods. As these applications (and more to follow) show, the DFT is a very useful computation to perform. The FFT allows us to perform this calculation fast.

The Fast Fourier Transform

The first observation to make is that Eq. (1) is a linear transformation: if we think of Eq. (1) as describing a transformation $f \mapsto \hat{f}$ , then we have that $\widehat{\alpha f + \beta g} = \alpha \hat{f} + \beta \hat{g}$ . Recall the crucial fact from linear algebra that every linear transformation can be represented by a matrix-vector muliplication.³ In my experience, one of the most effective algorithm design strategies in applied mathematics is, when presented with a linear transformation, to write its matrix down and poke and prod it to see if there are any patterns in the numbers which can be exploited to give a fast algorithm. Let’s try to do this with the DFT.

We have that $\hat{f} = F_n f$ for some $n\times n$ matrix $F_n$ . (We will omit the subscript $n$ when its value isn’t particularly important to the discussion.) Let us make the somewhat non-standard choice of describing rows and columns of $F$ by zero-indexing, so that the first row of $F$ is row $0$ and the last is row $n-1$ . Then we have that $\hat{f}_k = \sum_{j=0}^{n-1} F_{kj} f_j$ . Comparing with Eq. (1), we see that $F_{kj} = \tfrac{1}{\sqrt{n}} e^{(-2\pi i/n) jk}$ . Let us define $\omega_n = e^{-2\pi i/n}$ . Thus, we can write the matrix $F$ out as

(3) $\begin{equation*} F_n = \frac{1}{\sqrt{n}} \begin{bmatrix}\omega_n^0 & \omega_n^0 & \omega_n^0 & \cdots & \omega_n^0 \\\omega_n^{0} & \omega_n^1 & \omega_n^2 & \cdots & \omega_n^{n-1} \\\omega_n^0 & \omega_n^2 & \omega_n^4 & \cdots & \omega_n^{2(n-1)} \\ \omega_n^0 & \omega_n^3 & \omega_n^6 & \cdots & \omega_n^{3(n-1)}\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\omega_n^0 & \omega_n^{n-1} & \omega_n^{2(n-1)} & \cdots & \omega_n^{(n-1)(n-1)}\end{bmatrix} \end{equation*}$

This is a highly structured matrix. The patterns in this matrix are more easily seen for a particular value of $n$ . We shall focus on $n = 8$ in this discussion, but what will follow will generalize in a straightforward way to $n$ any power of two (and in less straightforward ways to arbitrary $n$ —we will return to this point at the end).

Instantiating Eq. (3) with $n = 8$ (and writing $\omega = \omega_8$ ), we have

(4) $\begin{equation*} F_8 = \frac{1}{\sqrt{8}} \begin{bmatrix}\omega^{0} &\omega^{0} &\omega^{0} &\omega^{0} &\omega^{0} &\omega^{0} &\omega^{0} &\omega^{0} \\ \omega^{0} &\omega^{1} &\omega^{2} &\omega^{3} &\omega^{4} &\omega^{5} &\omega^{6} &\omega^{7} \\ \omega^{0} &\omega^{2} &\omega^{4} &\omega^{6} &\omega^{8} &\omega^{10} &\omega^{12} &\omega^{14} \\ \omega^{0} &\omega^{3} &\omega^{6} &\omega^{9} &\omega^{12} &\omega^{15} &\omega^{18} &\omega^{21} \\ \omega^{0} &\omega^{4} &\omega^{8} &\omega^{12} &\omega^{16} &\omega^{20} &\omega^{24} &\omega^{28} \\ \omega^{0} &\omega^{5} &\omega^{10} &\omega^{15} &\omega^{20} &\omega^{25} &\omega^{30} &\omega^{35} \\ \omega^{0} &\omega^{6} &\omega^{12} &\omega^{18} &\omega^{24} &\omega^{30} &\omega^{36} &\omega^{42} \\ \omega^{0} &\omega^{7} &\omega^{14} &\omega^{21} &\omega^{28} &\omega^{35} &\omega^{42} &\omega^{49} \\ \end{bmatrix} \end{equation*}$

To fully exploit the patterns in this matrix, we note that $\omega$ represents a clockwise rotation of the complex plane by an eighth of the way around the circle. So, for example $\omega^{21}$ is twenty-one eighths of a turn or simply just $21-16 = 5$ turns. Thus $\omega^{21} = \omega^5$ and more generally $\omega^m = \omega^{m \operatorname{mod} 8}$ . This allows us to simplify as follows:

(5) $\begin{equation*} F_8 = \frac{1}{\sqrt{8}}\begin{bmatrix}1 &1 &1 &1 &1 &1 &1 &1 \\ 1 &\omega^{1} &\omega^{2} &\omega^{3} &\omega^{4} &\omega^{5} &\omega^{6} &\omega^{7} \\ 1 &\omega^{2} &\omega^{4} &\omega^{6} &1 &\omega^{2} &\omega^{4} &\omega^{6} \\ 1 &\omega^{3} &\omega^{6} &\omega^{1} &\omega^{4} &\omega^{7} &\omega^{2} &\omega^{5} \\ 1 &\omega^{4} &1 &\omega^{4} &1 &\omega^{4} &1 &\omega^{4} \\ 1 &\omega^{5} &\omega^{2} &\omega^{7} &\omega^{4} &\omega^{1} &\omega^{6} &\omega^{3} \\ 1 &\omega^{6} &\omega^{4} &\omega^{2} &1 &\omega^{6} &\omega^{4} &\omega^{2} \\ 1 &\omega^{7} &\omega^{6} &\omega^{5} &\omega^{4} &\omega^{3} &\omega^{2} &\omega^{1} \\ \end{bmatrix} \end{equation*}$

Now notice that, since $\omega$ represents a clockwise rotation of an eighth of the way around the circle, $\omega^2$ represents a quarter turn of the circle. This fact leads to the surprising observation we can actually find the DFT matrix $F_4$ for $n = 4$ hidden inside the DFT matrix $F_8$ for $n = 8$ !

To see this, rearrange the columns of $F_8$ to interleave every other column. In matrix language this is represented by right-multiplying with an appropriate⁴ permutation matrix $\Pi$ :

(6) $\begin{equation*} F_8 \Pi = \frac{1}{\sqrt{8}}\begin{bmatrix} 1 &1 &1 &1 &1 &1 &1 &1 \\ 1 &\omega^{2} &\omega^{4} &\omega^{6} &\omega^{1} &\omega^{3} &\omega^{5} &\omega^{7} \\ 1 &\omega^{4} &1 &\omega^{4} &\omega^2 &\omega^{6} &\omega^{2} &\omega^{6} \\ 1 &\omega^{6} &\omega^{4} &\omega^{2} &\omega^{3} &\omega^{1} &\omega^{7} &\omega^{5} \\ 1 &1 &1 &1 &\omega^4 &\omega^{4} &\omega^4 &\omega^{4} \\ 1 &\omega^{2} &\omega^{4} &\omega^{6} &\omega^{5} &\omega^{7} &\omega^{1} &\omega^{3} \\ 1 &\omega^{4} &1 &\omega^{4} &\omega^{6} &\omega^{2} & \omega^6 &\omega^{2} \\ 1 &\omega^{6} &\omega^{4} &\omega^2 & \omega^7 & \omega^5 & \omega^3 & \omega^1 \end{bmatrix} \end{equation*}$

The top-left $4\times 4$ sub-block is precisely $F_4$ (up to scaling). In fact, defining the diagonal matrix $\Omega_4 = \operatorname{diag}(1,\omega,\omega^2,\omega^3)$ (called the twiddle factor) and noting that $\omega^4 = -1$ , we have

(7) $\begin{equation*} F_8 \Pi = \frac{1}{\sqrt{2}} \begin{bmatrix} F_4 & \Omega_4 F_4 \\ F_4 & -\Omega_4 F_4 \end{bmatrix}. \end{equation*}$

The matrix $F_8$ is entirely built up of simple scalings of the smaller DFT matrix $F_4$ ! This suggests the following decomposition to compute $F_8x$ :

(8) $\begin{equation*} F_8 x = (F_8 \Pi)(\Pi^\top x) = \frac{1}{\sqrt{2}} \begin{bmatrix} F_4 & \Omega_4 F_4 \\ F_4 & -\Omega_4 F_4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \frac{F_4 x_1 + \Omega_4 (F_4 x_2)}{\sqrt{2}} \\ \frac{F_4 x_1 - \Omega_4 (F_4 x_2)}{\sqrt{2}}\end{bmatrix}. \end{equation*}$

Here $x_1$ represent the even-indexed entries of $x$ and $x_2$ the odd-indexed entries. Thus, we see that we can evaluate $F_8 x$ by evaluating the two expressions $F_4x_1$ and $F_4x_2$ . We have broken our problem into two smaller problems, which we then recombine into a solution of our original problem.

How then, do we compute the smaller DFTs $F_4x_1$ and $F_4x_2$ ? We just use the same trick again, breaking, for example, the product $F_4x_1$ into further subcomputations $F_2x_{11}$ and $F_2x_{12}$ . Performing this process one more time, we need to evaluate expressions of the form $F_1x_{111}$ , which are simply given by $F_1 x_{111}= x_{111}$ since the matrix $F_1$ is just a $1\times 1$ matrix whose single entry is $1$ .

This procedure is an example of a recursive algorithm: we designed an algorithm which solves a problem by breaking it down into one or more smaller problems, solve each of the smaller problems by using this same algorithm, and then reassemble the solutions of the smaller problems to solve our original problem. Eventually, we will break our problems into such small pieces that they can be solved directly, which is referred to as the base case of our recursion. (In our case, the base case is multiplication by $F_1$ ). Algorithms using this recursion in this way are referred to as divide-and-conquer algorithms.

Let us summarize this recursive procedure we’ve developed. We want to compute the DFT $y = F_n x$ where $n$ is a power of two. First, we use the DFT to recursively compute $F_{n/2}x_1$ and $F_{n/2}x_2$ . Next, we combine these computations to evaluate $y = F_n x$ by the formula

(9) $\begin{equation*} F_n x = \begin{bmatrix} \frac{F_{n/2} x_1 + \Omega_4 (F_{n/2} x_2)}{\sqrt{2}} \\ \frac{F_{n/2} x_1 - \Omega_4 (F_{n/2} x_2)}{\sqrt{2}}\end{bmatrix}. \end{equation*}$

This procedure is the famous fast Fourier transform (FFT), whose modern incarnation was presented by Cooley and Tukey in 1965 with lineage that can be traced back to work by Gauss in the early 1800s. There are many variants of the FFT using similar ideas.

Let us see why the FFT is considered “fast” by analyzing its operation count. As is common for divide-and-conquer algorithms, the number of operations for computing $F_n x$ using the FFT can be determined by solving a certain recurrence relation. Let $T(n)$ be the number of operations required by the FFT. Then the cost of computing $F_n x$ consists of

proportional-to- operations (or operations, in computer science language⁵ $\mathcal{O}(\cdot)$ refers to big-O notation. Saying an algorithm takes $\mathcal{O}(n\log n)$ operations is stating that, more or less, the algorithm takes less than some multiple of $n\log n$ operations to complete.) to:
- add, subtract, and scale vectors and
- multiply by the diagonal matrix $\Omega_{n/2} = \operatorname{diag}(1,\omega_{2n},\ldots,\omega_{2n}^{n-1})$ and
two recursive computations of $F_{n/2}x_j$ for $j= 1,2$ , each of which requires $T(n/2)$ operations.

This gives us the recurrence relation

(10) $\begin{equation*} T(n) = 2T(n/2) + \mathcal{O}(n). \end{equation*}$

Solving recurrences is a delicate art in general, but a wide class of recurrences are immediately solved by the flexible master theorem for recurrences. Appealing to this result, we deduce that the FFT requires $T(n) = \mathcal{O}(n\log n)$ operations. This is a dramatic improvement of the $\mathcal{O}(n^2)$ operations to compute $F_n x$ directly using Eq. (1). This dramatic improvement in speed is what makes the FFT “fast”.

Extending the FFT Idea

The FFT is a brilliant algorithm. It exploits the structure of the discrete Fourier transform problem Eq. (1) for dramatically lower operation counts. And as we shall see a taste of, the FFT is useful in a surprisingly broad range of applications. Given the success of the FFT, we are naturally led to the question: can we learn from our success with the FFT to develop fast algorithms for other problems?

I think the FFT speaks to the power of a simple problem-solving strategy for numerical algorithm design⁶: whenever you have a linear transformation, write it as a matrix-vector product; whenever you have a matrix, write it down and see if there are any patterns.⁷ We often like to present mathematics with each step of a derivation follows almost effortlessly from the last from a firm basis of elegant mathematical intuition. Often, however, noticing patterns by staring at symbols on a page can be more effective than reasoning grounded in intuition. Once the pattern has been discovered, intuition and elegance sometimes will follow quickly behind.

The most natural generalization of the FFT is the fast inverse discrete Fourier transform, providing a fast algorithm to compute the inverse discrete Fourier transform Eq. (2). The inverse FFT is quite an easy generalization of the FFT presented in the past section; it is a good exercise to see if you can mimic the development in the previous section to come up with this generalization yourself. The FFT can also be generalized to other discrete trigonometric transforms and 2D and 3D discrete Fourier transforms.

I want to consider a problem more tangentially related to the FFT, the evaluation of expressions of the form $y = (A \otimes B)x$ , where $A$ is an $m_1\times n_1$ matrix, $B$ is an $m_2\times n_2$ matrix, $x$ is a vector of length $n_1n_2$ , and $\otimes$ denotes the Kronecker product. For the unitiated, the Kronecker product of $A$ and $B$ is a $m_1m_2\times n_1n_2$ matrix defined as the block matrix

(11) $\begin{equation*} A\otimes B = \begin{bmatrix} A_{11} B & A_{12} B & \cdots & A_{1n_1} B\\ A_{21}B & A_{22} B & \cdots & A_{2n_1} B \\ \vdots & \vdots & \ddots & \vdots \\ A_{m_1 1} B & A_{m_1 2} B & \cdots & A_{m_1 n_1} B\end{bmatrix}. \end{equation*}$

We could just form this $m_1 m_2\times n_1n_2$ matrix and compute the matrix-vector product $y = (A\otimes B)x$ directly, but this takes a hefty $\mathcal{O}(m_1m_2n_1n_2)$ operations.⁸ We can do better.

The insight is much the same as with the FFT: scaled copies of the matrix $B$ are embedded in $A\otimes B$ . In the FFT, we needed to rearrange the columns of the DFT matrix $F_n$ to see this; for the Kronecker product, this pattern is evident in the natural ordering. To exploit this fact, chunk the vectors $x$ and $y$ into pieces $x_1,x_2,\ldots,x_{n_1}$ and $y_1,y_2,\ldots,y_{m_1}$ of length $n_2$ and $m_2$ respectively so that our matrix vector product can be written as⁹

(12) $\begin{equation*} \underbrace{\begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_{m_1} \end{bmatrix}}_{=y} = \underbrace{\begin{bmatrix} A_{11} B & A_{12} B & \cdots & A_{1n_1} B\\ A_{21}B & A_{22} B & \cdots & A_{2n_1} B \\ \vdots & \vdots & \ddots & \vdots \\ A_{m_1 1} B & A_{m_1 2} B & \cdots & A_{m_1 n_1} B\end{bmatrix}}_{=A\otimes B} \underbrace{\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_{n_1} \end{bmatrix}}_{=x}. \end{equation*}$

To compute this product efficiently, we proceed in two steps. First, we compute the products $Bx_1,Bx_2,\ldots,Bx_{n_1}$ which takes time $\mathcal{O}(m_2n_2\cdot n_1)$ in total. Next, we compute each component $y_1,\ldots,y_{m_1}$ by using the formula

(13) $\begin{equation*} y_j = A_{j1} (Bx_1) + A_{j2} (Bx_2) + \cdots + A_{jn_1} (Bx_{n_1}), \end{equation*}$

which takes a total of $\mathcal{O}(m_1 n_1 \cdot m_2)$ operations to compute all the $y_j$ ‘s. This leads to a total operation count of $\mathcal{O}(m_2n_1(m_1+n_2))$ for computing the matrix-vector product $y = (A\otimes B)x$ , much better than our earlier operation count of $\mathcal{O}(m_1m_2n_1n_2)$ .¹⁰

While this idea might seem quite far from the FFT, if one applies this idea iteratively, one can use this approach to rapidly evaluate a close cousin of the DFT called the Hadamard-Walsh transform. Using the Kronecker product, the Hadamard-Walsh transform $\hat{f}$ of a vector $f$ is defined to be

(14) $\begin{equation*} \hat{f} = \left( \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \otimes \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \otimes \cdots \otimes \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \right) f. \end{equation*}$

If one applies the Kronecker product trick we developed repeatedly, this gives an algorithm to evaluate the Hadamard-Walsh transform of a vector $f$ of length $n$ in $\mathcal{O}(n \log n)$ operations, just like the FFT.

The Hadamard-Walsh transform can be thought of as a generalization of the discrete Fourier transform to Boolean functions, which play an integral role in computer science. The applications of the Hadamard-Walsh transform are numerous and varied, from everything to voting systems to quantum computing. This is really just the tip of the iceberg. The ideas behind the FFT (and related ideas from the fast multipole method) allow for the rapid evaluation of a large number of transformations, some of which are connected by deep and general theories.

Resisting the temptation to delve into these interesting subjects in any more depth, we return to our main idea: when presented with a linear transformation, write it as a matrix-vector product; whenever you have a matrix, write it down and see if there are any patterns. The FFT exploits one such pattern, noticing that (after a reordering) a matrix contains many scaled copies of the same matrix. Rapidly evaluation expressions of the form $y = (A\otimes B)x$ involves an even simpler application of the same idea. But there are many other patterns that can be exploited: sparsity, (approximate) low rank, off-diagonal blocks approximately of low rank, and displacement structure are other examples. Very often in applied math, our problems have additional structure that can be exploited to solve problems much faster, and sometimes finding that structure is as easy as just trying to look for it.

An Application of the FFT

A discussion of the FFT would be incomplete without exploring at least one reason why you’d want to compute the discrete Fourier transform. To focus our attention, let us consider another linear algebraic calculation which appears to have no relation to the FFT on its face: computing a matrix-vector product with a Toeplitz matrix. A matrix $T$ is said to be Toeplitz if it has the following structure:

(15) $\begin{equation*} T = \begin{bmatrix} t_0 & t_1 & t_2 & \cdots & t_{n-1} \\ t_{-1} & t_0 & t_1 & \cdots & t_{n-2} \\ t_{-2} & t_{-1} & t_0 & \cdots & t_{n-3} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ t_{-(n-1)} & t_{-(n-2)} & t_{-(n-3)} & \cdots & t_0 \end{bmatrix}. \end{equation*}$

Toeplitz matrices and their relatives appear widely across applications of applied mathematics including control and systems theory, time series, numerical partial differential equations, and signal processing.

We seek to compute the matrix-vector product $y = Tx$ . Let us by considering a special case of a Toeplitz matrix, a circulant matrix. A circulant matrix $C$ has the form

(16) $\begin{equation*} \begin{bmatrix} c_0 & c_{n-1} & c_{n-2} & \cdots & c_1 \\ c_1 & c_0 & c_{n-1} & \cdots & c_2\\ c_2 & c_1 & c_0 & \cdots & c_3 \\ \vdots & \vdots & \vdots &\ddots & \vdots \\ c_{n-1} & c_{n-2} & c_{n-3} & \cdots & c_0 \end{bmatrix}. \end{equation*}$

By direct computation, the matrix-vector product $y = Cx$ is given by

(17) $\begin{equation*} y_k = \sum_{j=0}^{n-1}x_j c_{\operatorname{mod}(k-j,n)}. \end{equation*}$

A surprising and non-obvious fact is that the circulant matrix $C$ is diagonalized by the discrete Fourier transform. Specifically, we have $C = F_n^* \operatorname{diag}(\sqrt{n} F_n c) F_n$ where $c=(c_0,\ldots,c_{n-1})$ . This gives a fast algorithm to compute $y = Cx$ in time $\mathcal{O}(n\log n)$ : compute the DFTs of $x$ and $c$ and multiply them together entrywise, take the inverse Fourier transform, and scale by $\sqrt{n}$ .

There is a connection with signal processing and differential equations that may help to shed light on why technique works for those familiar with those areas. In the signal processing context, the matrix-vector product $y = Cx$ can be interpreted as the discrete convolution of $x$ with $c$ (see Eq. (17)) which is a natural extension of the convolution $f* g$ of two functions $f$ and $g$ on the real line. It is an important fact that the Fourier transform of a convolution is the same as multiplication of the Fourier transforms: $\widehat{f * g} = \hat{f} \cdot \hat{g}$ (up to a possible normalizing constant).¹¹ The fact that the DFT diagonalizes a circulant matrix is just the analog of this fact for the discrete Fourier transform and the discrete convolution.

This fast algorithm for circulant matrix-vector products is already extremely useful. One can naturally reframe the problems of multiplying integers and polynomials as discrete convolutions, which can then be computed rapidly by applying the $\mathcal{O}(n\log n)$ algorithm for fast circulant matrix-vector products. This video gives a great introduction to the FFT with this as its motivating application.

Let’s summarize where we’re at. We are interested in computing the Toeplitz matrix-vector product $y = Tx$ . We don’t know how to do this for a general Toeplitz matrix yet, but we can do it for a special Toeplitz matrix called a circulant matrix $C$ . By use of the FFT, we can compute the circulant matrix-vector product $y = Cx$ in $\mathcal{O}(n\log n)$ operations.

We can now leverage what we’ve done with circulant matrices to accelerate Toeplitz matrix-vector product. The trick is very simple: embedding. We construct a big circulant matrix which contains the Toeplitz matrix as a sub-matrix and then use multiplications by the bigger matrix to compute multiplications by the smaller matrix.

Consider the following circulant matrix, which contains as $T$ as defined in Eq. (15) a sub-matrix in its top-left corner:

(18) $\begin{equation*} C_T = \begin{bmatrix} t_0 & t_1 & \cdots & t_{n-1} & 0 & 0 & \cdots \\ t_{-1} & t_0 & \cdots & t_{n-2} & t_{n-1} & 0 & \cdots\\ \vdots & \vdots & \ddots & \vdots & \vdots &\vdots &\ddots\\ t_{-(n-1)} & t_{-(n-2)} & \cdots & t_0 & t_1 & t_2&\cdots\\ 0 & t_{-(n-1)}&\cdots&t_{-1} & t_0 & t_1& \cdots\\ 0 & 0 & \cdots & t_{-2} & t_{-1} & t_0& \cdots\\ \vdots& \vdots & \ddots & \vdots & \vdots & \vdots &\ddots\\ 0 & 0 &\cdots&&&& \cdots\\ t_{n-1} & 0 & \cdots&&&&\cdots\\ t_{n-2} & t_{n-1} & \cdots &&&&\cdots\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots &\ddots\\ t_1 & t_2 & \cdots & 0 & 0 & 0 & \cdots \end{bmatrix} \end{equation*}$

This matrix is hard to write out, but essentially we pad the Toeplitz matrix with extra zeros to embed it into a circulant matrix. The “ $c$ ” vector for this larger circulant matrix is obtained from the parameters $t_0,t_{\pm 1},\ldots,t_{\pm (n-1)}$ of the Toeplitz matrix Eq. (15) by $c = (t_0,t_{-1},\ldots,t_{-(n-1)},0,\ldots,0,t_{n-1},t_{n-2},\ldots,t_1)$ .

Here comes another clever observation: we can choose the number of padding zeros used cleverly to make the size of $C_T$ exactly equal to a power of two. This is useful because it allows us to compute matrix-vector products $w=C_Tz$ with the power-of-two FFT described above, which we know is fast.

Finally, let’s close the loop and use fast multiplications with $C_T$ to compute fast multiplications with $T$ . We wish to compute the product $y = Tx$ fast. To do this, vector $x$ into a larger vector $z$ by padding with zeros to get

(19) $\begin{equation*} C_T z = \begin{bmatrix} T & \star \\ \star & \star \end{bmatrix} \begin{bmatrix} x \\ 0 \end{bmatrix} = \begin{bmatrix} y \\ \star \end{bmatrix}, \end{equation*}$

where we use $\star$ to denote matrix or vector entries which are immaterial to us. We compute $Tx$ by using our fast algorithm to compute $C_T z = w$ and then discarding everything but the first entries of $w$ to obtain $y$ . If you’re careful to analyze how much padding we need to make this work, we see that this algorithm also takes only $\mathcal{O}(n \log n)$ operations. Thus, we’ve completed our goal: we can compute Toeplitz matrix-vector products in a fast $\mathcal{O}(n\log n)$ operations.

Finally, let us bring this full circle and see a delightfully self-referential use of this algorithm: we can use the FFT-accelerated fast Toeplitz matrix-vector multiply to compute DFT itself. Recall that the FFT algorithm we presented above was particularized to $n$ which were powers of $2$ . There are natural generalizations of the along the lines of what we did above to more general $n$ which are highly composite and possess many small prime factors. But what if we want to evaluate the DFT for $n$ which is a large prime?

Recall that the DFT matrix $F_n$ has $jk$ th entry $(F_n)_{jk} = \tfrac{1}{\sqrt{n}} \omega_n^{jk}$ . We now employ a clever trick. Let $D$ be a diagonal matrix with the $jj$ th entry equal to $\omega_n^{-\tfrac{1}{2} j^2}$ . Then, defining $T = D F_n D$ , we have that $T_{jk} = \tfrac{1}{\sqrt{n}} \omega_n^{-\tfrac{1}{2} j^2 + jk - \tfrac{1}{2} k^2} = \tfrac{1}{\sqrt{n}} \omega_n^{-\tfrac{1}{2}(j-k)^2}$ , which means $T$ is a Toeplitz matrix! (Writing out the matrix $T$ entrywise may be helpful to see this.)

Thus, we can compute the DFT $\hat{f} = F_n f$ for any size $n$ by evaluating the DFT as $\hat{f} = D^{-1}(T(D^{-1} f))$ , where the product $Tx$ is computed using the fast Toeplitz matrix-vector product. Since our fast Toeplitz matrix-vector product only requires us to evaluate power-of-two DFTs, this technique allows us to evaluate DFTs of arbitrary size $n$ in only $\mathcal{O}(n\log n)$ operations.

Upshot: The discrete Fourier transform (DFT) is an important computation which occurs all across applied mathematics. The fast Fourier transform (FFT) reduces the operation count of evaluating the DFT of a vector of length $n$ to proportional to $n \log n$ , down from proportional to $n^2$ for direct evaluation. The FFT is an example of a broader matrix algorithm design strategy of looking for patterns in the numbers in a matrix and exploiting these patterns to reduce computation. The FFT can often have surprising applications, such as allowing for rapid computations with Toeplitz matrices.

The Better Way to Convert an SVD into a Symmetric Eigenvalue Problem

March 18, 2021 by Ethan N. Epperly 7 Comments

A singular value decomposition of an $m\times n$ matrix $B$ is a factorization of the form $B = U\Sigma V^\top$ , where $U$ and $V$ are square, orthogonal matrices and $\Sigma$ is a diagonal matrix with $(i,i)$ th entry $\sigma_i \ge 0$ .¹ The diagonal entries of $\Sigma$ are referred to as the singular values of $B$ and are conventionally ordered $\sigma_{\rm max} = \sigma_1 \ge \sigma_2 \ge \cdots \ge \sigma_{\min(m,n)} = \sigma_{\rm min}$ . The columns of the matrices $U$ and $V$ are referred to as the right- and left- singular vectors of $B$ and satisfy the relations $Bv_i = \sigma_i u_i$ and $B^\top u_i = \sigma_i v_i$ .

One can obtain the singular values and right and left singular vectors of $B$ from the eigenvalues and eigenvectors of $B^\top B$ and $BB^\top$ . This follows from the calculations $B^\top B = V\Sigma^2 V^\top$ and $B^\top B = U\Sigma^2 U^\top$ . In other words, the nonzero singular values of $B$ are the square roots of the nonzero eigenvalues of $B^\top B$ and $BB^\top$ . If one merely solves one of these problems, computing $\Sigma$ along with $U$ or $V$ , one can obtain the other matrix $V$ or $U$ by computing $U = BV \Sigma^{-1}$ or $V = B^\top U \Sigma^{-1}$ . (These formulas are valid for invertible square matrices $B$ , but similar formulas hold for singular or rectangular $B$ to compute the singular vectors with nonzero singular values.)

This approach is often unundesirable for several reasons. Here are a few I’m aware of:

Accuracy: Roughly speaking, in double-precision arithmetic, accurate stable numerical methods can resolve differences on the order of 16 orders of magnitude. This means an accurately computed SVD of $B$ can resolve the roughly 16 orders of magnitude of decaying singular values, with singular values smaller than that difficult to compute accurately. By computing $B^\top B$ , we square all of our singular values, so resolving 16 orders of magnitude of the eigenvalues of $B^\top B$ means we only resolve 8 orders of magnitude of the singular values of $B$ .² The dynamic range of our numerical computations has been cut in half!
Loss of orthogonality: While $U = BV \Sigma^{-1}$ and $V = B^\top U \Sigma^{-1}$ are valid formulas in exact arithmetic, they fair poorly when implemented numerically. Specifically, the numerically computed values $U_{\rm numerical}$ and $V_{\rm numerical}$ may not be orthogonal matrices with, for example, $U_{\rm numerical}^\top U_{\rm numerical}$ not even close to the identity matrix. One can, of course, orthogonalize the computed $U$ or $V$ , but this doesn’t fix the underlying problem that $U$ or $V$ have not been computed accurately.
Loss of structure: If $B$ possesses additional structure (e.g. sparsity), this structure may be lost or reduced by computing the product $B^\top B$ .
Nonlinearity: Even if we’re not actually computing the SVD numerically but doing analysis with pencil and paper, finding the SVD of $B$ from $B^\top B$ has the disadvantage of performing a nonlinear transformation on $B$ . This prevents us from utilizing additive perturbation theorems for sums of symmetric matrices in our analysis.³

There are times where these problems are insignificant and this approach is sensible: we shall return to this point in a bit. However, these problems should disqualify this approach from being the de facto way we reduce SVD computation to a symmetric eigenvalue problem. This is especially true since we have a better way.

The better way is by constructing the so-called Hermitian dilation⁴ of $B$ , which is defined to be the matrix

(1) $\begin{equation*} \mathcal{H}(B) = \begin{bmatrix} 0 & B \\ B^\top & 0 \end{bmatrix}. \end{equation*}$

One can show that the nonzero eigenvalues of $\mathcal{H}(B)$ are precisely plus-or-minus the singular values of $B$ . More specifically, we have

(2) $\begin{equation*} \mathcal{H}(B) \begin{bmatrix} u_i \\ \pm v_i \end{bmatrix} = \pm \sigma_i \begin{bmatrix} u_i \\ \pm v_i \end{bmatrix}. \end{equation*}$

All of the remaining eigenvalues of $\mathcal{H}(B)$ not of this form are zero.⁵ Thus, the singular value decomposition of $B$ is entirely encoded in the eigenvalue decomposition of $\mathcal{H}(B)$ .

This approach of using the Hermitian dilation $\mathcal{H}(B)$ to compute the SVD of $B$ fixes all the issues identified with the “ $B^\top B$ ” approach. We are able to accurately resolve a full 16 orders of magnitude of singular values. The computed singular vectors are accurate and numerically orthogonal provided we use an accurate method for the symmetric eigenvalue problem. The Hermitian dilation $\mathcal{H}(B)$ preserves important structural characteristics in $B$ like sparsity. For purposes of theoretical analysis, the mapping $B \mapsto \mathcal{H}(B)$ is linear.⁶

Often one can work with the Hermitian dilation only implicitly: the matrix $\mathcal{H}(B)$ need not actually be stored in memory with all its extra zeros. The programmer designs and implements an algorithm with $\mathcal{H}(B)$ in mind, but deals with the matrix $B$ directly for their computations. In a pinch, however, forming $\mathcal{H}(B)$ directly in software and utilizing symmetric eigenvalue routines directly is often not too much less efficient than a dedicated SVD routine and can cut down on programmer effort significantly.

As with all things in life, there’s no free lunch here. There are a couple of downsides to the Hermitian dilation approach. First, $\mathcal{H}(B)$ is, except for the trivial case $B = 0$ , an indefinite symmetric matrix. By constast, $B^\top B$ and $BB^\top$ are positive semidefinite, which can be helpful in some contexts.⁷ Further, if $n\ll m$ (respectively, $m \ll n$ ), then $B^\top B$ (respectively, $BB^\top$ ) is tiny compared to $\mathcal{H}(B)$ , so it might be considerably cheaper to compute an eigenvalue decomposition of $B^\top B$ (or $BB^\top$ ) than $\mathcal{H}(B)$ .

Despite the somewhat salacious title of this article, the $B^\top B$ and Hermitian dilation approaches both have their role, and the purpose of this article is not to say the $B^\top B$ approach should be thrown in the dustbin. However, in my experience, I frequently hear the $B^\top B$ approach stated as the definitive way of converting an SVD into an eigenvalue problem, with the Hermitian dilation approach not even mentioned. This, in my opinion, is backwards. For accuracy reasons alone, the Hermitian dilation should be the go-to tool for turning SVDs into symmetric eigenvalue problems, with the $B^\top B$ approach only used when the problem is known to have singular values which don’t span many orders of magnitude or $B$ is tall and skinny and the computational cost savings of the $B^\top B$ approach are vital.

Big Ideas in Applied Math: Galerkin Approximation

August 9, 2020 by Ethan N. Epperly 6 Comments

My first experience with the numerical solution of partial differential equations (PDEs) was with finite difference methods. I found finite difference methods to be somewhat fiddly: it is quite an exercise in patience to, for example, work out the appropriate fifth-order finite difference approximation to a second order differential operator on an irregularly spaced grid and even more of a pain to prove that the scheme is convergent. I found that I liked the finite element method a lot better¹ as there was a unifying underlying functional analytic theory, Galerkin approximation, which showed how, in a sense, the finite element method computed the best possible approximate solution to the PDE among a family of potential solutions. However, I came to feel later that Galerkin approximation was, in a sense, the more fundamental concept, with the finite element method being one particular instantiation (with spectral methods, boundary element methods, and the conjugate gradient method being others). In this post, I hope to give a general introduction to Galerkin approximation as computing the best possible approximate solution to a problem within a certain finite-dimensional space of possibilities.

Systems of Linear Equations

Let us begin with a linear algebraic example, which is unburdened by some of the technicalities of partial differential equations. Suppose we want to solve a very large system of linear equations $Ax = b$ , where the matrix $A$ is symmetric and positive definite (SPD). Suppose that $A$ is $N\times N$ where $N$ is so large that we don’t even want to store all $N$ components of the solution $x$ on our computer. What can we possibly do?

One solution is to consider only solutions $x$ lying in a subspace $\mathcal{X}$ of the set of all possible solutions $\mathbb{R}^N$ . If this subspace has a basis $x_1,x_2,\ldots,x_M \in \mathcal{X}$ , then the solution $x \in \mathcal{X}$ can be represented as $x = a_1x_1 + \cdots + a_Mx_M$ and one only has to store the $M < N$ numbers $a_1,\ldots,a_M$ . In general, $x$ will not belong to the subspace $\mathcal{X}$ and we must settle for an approximate solution $\hat{x} \in \mathcal{X}$ .

The next step is to convert the system of linear equations $Ax = b$ into a form which is more amenable to approximate solution on a subspace $\mathcal{X}$ . Note that the equation $Ax = b$ encodes $n$ different linear equations $a_i^\top x = b_i$ where $a_i^\top$ is the $i$ th row of $A$ and $b_i$ is the $i$ th element of $b$ . Note that the $i$ th equation is equivalent to the condition $e_i^\top A x = e_i^\top b$ , where $e_i$ is the vector with zeros in all entries except for the $i$ th entry which is a one. More generally, by multiplying the equation $Ax = b$ by an arbitrary test row vector $y^\top$ , we get $y^\top Ax = y^\top b$ for all $y \in \mathbb{R}^N$ . We refer to this as a variational formulation of the linear system of equations $Ax = b$ . In fact, one can easily show that the variational problem is equivalent to the system of linear equations:

(1) $\begin{equation*} Ax = b \:\mbox{ if, and only if, }\: y^\top Ax = y^\top b \mbox{ for every } y \in \mathbb{R}^N. \end{equation*}$

Since we are seeking an approximate solution from the subspace $\mathcal{X}$ , it is only natural that we also restrict our test vectors $y$ to lie in the subspace $\mathcal{X}$ . Thus, we seek an approximate solution $\hat{x}$ to the system of equations $Ax = b$ as the solution of the variational problem

(2) $\begin{equation*} y^\top A\hat{x} = y^\top b \mbox{ for every } y \in\mathcal{X}. \end{equation*}$

One can relatively easily show this problem possesses a unique solution $\hat{x}$ .² In what sense is $\hat{x}$ a good approximate solution for $Ax = b$ ? To answer this question, we need to introduce a special way of measuring the error to an approximate solution to $Ax = b$ . We define the $A$ -inner product of a vector $x$ and $y$ to be $\langle x, y \rangle_A := y^\top Ax$ and the associated $A$ -norm $\|x\|_A = \sqrt{\langle x, x\rangle_A} = \sqrt{x^\top A x}$ .³ All of the properties satisfied by the familiar Euclidean inner product and norm carry over to the new $A$ -inner product and norm (e.g., the Pythagorean theorem). Indeed, for those familiar, one can show $\langle \cdot, \cdot \rangle_A$ satisfies all the axioms for an inner product space.

We shall now show that the error $x - \hat{x}$ between $x$ and its Galerkin approximation $\hat{x}$ is $A$ -orthogonal to the space $\mathcal{X}$ in the sense that $\langle y, x - \hat{x}\rangle_A = 0$ for all $y \in \mathcal{X}$ . This follows from the straightforward calculation, for $y \in \mathcal{X}$ ,

(3) $\begin{equation*} \langle y, x - \hat{x} \rangle_A = y^\top A (x - \hat{x}) = y^\top A x - y^\top A \hat{x} = y^\top b - y^\top b = 0, \end{equation*}$

where $y^\top A x = y^\top b$ since $x$ solves the variational problem Eq. (1) and $y^\top A \hat{x} = y^\top b$ since $x$ solves the variational problem Eq. (2).

The fact that the error $x - \hat{x}$ is $A$ -orthogonal to $\mathcal{X}$ can be used to show that $\hat{x}$ is, in a sense, the best approximate solution to $Ax = b$ in the subspace $\mathcal{X}$ . First note that, for any approximate solution $z \in \mathcal{X}$ to $Ax = b$ , the vector $\hat{x} - z \in \mathcal{X}$ is $A$ -orthogonal to $x - \hat{x}$ . Thus, by the Pythagorean theorem,

(4) $\begin{equation*} \|x - z\|^2_A = \|(x - \hat{x}) + (\hat{x}-z)\|_A^2 = \|x-\hat{x}\|_A^2 + \|\hat{x} - z\|_A^2 \ge \|x - \hat{x} \|_A^2. \end{equation*}$

Thus, the Galerkin approximation $\hat{x}$ is the best approximate solution to $Ax = b$ in the subspace $\mathcal{X}$ with respect to the $A$ -norm, $\|x - z\|_A \ge \|x - \hat{x} \|_A$ for every $z \in \mathcal{X}$ . Thus, if one picks a subspace $\mathcal{X}$ for which the solution $x$ almost lies in $\mathcal{X},$ ⁴ then $\hat{x}$ will be a good approximate solution to $Ax = b$ , irrespective of the size of the subspace $\mathcal{X}$ .

Variational Formulations of Differential Equations

As I hope I’ve conveyed in the previous section, Galerkin approximation is not a technique that only works for finite element methods or even just PDEs. However, differential and integral equations are one of the most important applications of Galerkin approximation since the space of all possible solution to a differential or integral equation is infinite-dimensional: approximation in a finite-dimensional space is absolutely critical. In this section, I want to give a brief introduction to how one can develop variational formulations of differential equations amenable to Galerkin approximation. For simplicity of presentation, I shall focus on a one-dimensional problem which is described by an ordinary differential equation (ODE) boundary value problem. All of this generalized wholesale to partial differential equations in multiple dimensions, though there are some additional technical and notational difficulties (some of which I will address in footnotes). Variational formulation of differential equations is a topic with important technical subtleties which I will end up brushing past. Rigorous references are Chapters 5 and 6 from Evans’ Partial Differential Equations or Chapters 0-2 from Brenner and Scott’s The Mathematical Theory of Finite Element Methods.

As our model problem for which we seek a variational formulation, we will focus on the one-dimensional Poisson equation, which appears in the study of electrostatics, gravitation, diffusion, heat flow, and fluid mechanics. The unknown $u$ is a real-valued function on an interval which take to be $[0,1]$ .⁵ We assume Dirichlet boundary conditions that $u$ is equal to zero on the boundary $u(0) = u(1) = 0$ .⁶ Poisson’s equations then reads⁷

(5) $\begin{equation*} -u''(x) = f(x) \mbox{ for every } x \in (0,1), \quad u(0) = u(1) = 0. \end{equation*}$

We wish to develop a variational formulation of this differential equation, similar to how we develop a variational formulation of the linear system of equations in the previous section. To develop our variational formulation, we take inspiration from physics. If $u(x)$ represents, say, the temperature at a point $x$ , we are never able to measure $u(x)$ exactly. Rather, we can measure the temperature in a region around $x$ with a thermometer. No matter how carefully we engineer our thermometer, our thermometer tip will have some volume occupying a region $R$ in space. The temperature $u_{\rm meas}$ measured by our thermometer will be the average temperature in the region $R$ or, more generally, a weighted average $u_{\rm meas} = \int_0^1 u(x) v(x) \, dx$ where $v(\cdot)$ is a weighting function which is zero outside the region $R$ . Now let’s use our thermometer to “measure” our differential equation:

(6) $\begin{equation*} \int_0^1-v(x) u''(x) \, dx = \int_0^1 v(x)f(x) \, dx. \end{equation*}$

This integral expression is some kind of variational formulation of our differential equation, as it is an equation involving the solution to our differential equation $u$ which must hold for every averaging function $v$ . (The precise meaning of every will be forthcoming.) It will benefit us greatly to make this expression more “symmetric” with respect to $u$ and $v$ . To do this, we shall integrate by parts:⁸

(7) $\begin{equation*} \int_0^1-v(x)u''(x) \, dx = \int_0^1 v'(x)u'(x) \, dx - v(0) u'(0) - v(1)u'(1). \end{equation*}$

In particular, if $v$ is zero on the boundary $v(0) = v(1) = 0$ , then the second two terms vanish and we’re left with the variational equation

(8) $\begin{equation*} \int_0^1 v'(x)u'(x) \, dx = \int_0^1 v(x) f(x) \, dx \mbox{ for all \textit{nice} functions $v$ on $[0,1]$ with } v(0) = v(1) = 0. \end{equation*}$

Compare the variational formulation of the Poisson equation Eq. (8) to the variational formulation of the system of linear equations $Ax = b$ in Eq. (1). The solution vector $x$ in the differential equation context is a function $u$ satisfying the boundary condition of $u$ being zero on the boundary $u(0) = u(1) = 0$ . The right-hand side $b$ is replaced by a function $f$ on the interval $[0,1]$ . The test vector $y$ is replaced by a test function $v$ on the interval $[0,1]$ . The matrix product expression $y^\top A x$ is replaced by the integral $\int_0^1 v'(x)u'(x) \, dx$ . The product $y^\top b$ is replaced by the integral $\int_0^1 v(x) f(x) \, dx$ . As we shall soon see, there is a unifying theory which treats both of these contexts simultaneously.

Before this unifying theory, we must address the question of which functions $v$ we will consider in our variational formulation. One can show that all of the calculations we did in this section hold if $v$ is a continuously differentiable function on $[0,1]$ which is zero away from the endpoints $0$ and $1$ and $u$ is a twice continuously differentiable function on $[0,1]$ . Because of technical functional analytic considerations, we shall actually want to expand the class of functions in our variational formulation to even more functions $v$ . Specifically, we shall consider all functions $v$ which are (A) square-integrable ( $\int_0^1|v(x)|^2 \,dx$ is finite), (B) possess a square integrable derivative⁹ $v'$ ( $\int_0^1|v'(x)|^2 \,dx$ is finite), and (C) are zero on the boundary. We refer to this class of functions as the Sobolev space $H_0^1((0,1))$ .¹⁰

Now this is where things get really strange. Note that it is possible for a function $u$ to satisfy the variational formulation Eq. (8) but for $u$ not to satisfy the Poisson equation Eq. (5). A simple example is when $f$ possesses a discontinuity (say, for example, a step discontinuity where $f$ is $0$ and then jumps to $1$ ). Then no continuously differentiable $u$ will satisfy Eq. (5) at every point in $\Omega$ and yet a solution $u$ to the variational problem Eq. (8) exists! The variational formulation actually allows us to give a reasonable definition of “solving the differential equation” when a classical solution to $-u'' = f$ does not exist. Our only requirement for the variational problem is that $u$ , itself, belongs to the space $H_0^1((0,1))$ . A solution to the variational problem Eq. (8) is called a weak solution to the differential equation Eq. (5) because, as we have argued, a weak solution to Eq. (8) need not always solve Eq. (5).¹¹

The Lax-Milgram Theorem

Let us now build up an abstract language which allows us to use Galerkin approximation both for linear systems of equations and PDEs (as well as other contexts). If one compares the expressions $y^\top A x$ from the linear systems context and $\int_0^1 v'(x)u'(x) \, dx$ from the differential equation context, one recognizes that both these expressions are so-called bilinear forms: they depend on two arguments ( $x$ and $y$ or $u$ and $v$ ) and are a linear transformation in each argument independently if the other one is fixed. For example, if one defines $a(x,y) = y^\top A x$ one has $a(x,\alpha_1 y_1 + \alpha_2 y_2) = \alpha_1 a(x,y_1) + \alpha_2 a(x,y_2)$ . Similarly, if one defines $a(u,v) = \int_0^1 v'(x)u'(x) \, dx$ , then $a(\alpha_1 u_1 + \alpha_2 u_2,v) = \alpha_1 a(u_1,v) + \alpha_2 a(u_2, v)$ .

Implicitly swimming in the background is some space of vectors or function which this bilinear form $a(\cdot,\cdot)$ is defined upon. In the linear system of equations context, this space $\mathbb{R}^N$ of $N$ -dimensional vectors and in the differential context, this space is $H_0^1((0,1))$ as defined in the previous section.¹² Call this space $\mathcal{V}$ . We shall assume that $\mathcal{V}$ is a special type of linear space called a Hilbert space, an inner product space (with inner product $\langle \cdot, \cdot \rangle_\mathcal{V}$ ) where every Cauchy sequence converges to an element in $\mathcal{V}$ (in the inner product-induced norm).¹³ The Cauchy sequence convergence property, also known as metric completeness, is important because we shall often deal with a sequence of entries $u_1,u_2,\ldots \in \mathcal{V}$ which we will need to establish convergence to a vector $u \in \mathcal{V}$ . (Think of $u_1,u_2,\ldots$ as a sequence of Galerkin approximations to a solution $u$ .)

With these formalities, an abstract variational problem takes the form

(9) $\begin{equation*} \mbox{Find $u \in \mathcal{V}$ such that }a(u,v) = \ell(v) \mbox{ for all } v \in \mathcal{V}, \end{equation*}$

where $a(\cdot,\cdot)$ is a bilinear form on $\mathcal{V}$ and $\ell(\cdot)$ is a linear form on $\mathcal{V}$ (a linear map $\ell: \mathcal{V} \to \mathbb{R}$ ). There is a beautiful and general theorem called the Lax-Milgram theorem which establishes existence and uniqueness of solutions to a problem like Eq. (9).

Theorem (Lax-Milgram): Let $a(\cdot,\cdot)$ and $f$ satisfy the following properties:

(Boundedness of $a$ ) There exists a constant $C \ge 0$ such that every $u,v \in \mathcal{V}$ , $|a(u,v)| \le C \|u\|_{\mathcal{V}}\|v\|_{\mathcal{V}}$ .
(Coercivity) There exists a positive constant $c > 0$ such that $a(u,u) \ge c \|u\|_{\mathcal{V}}^2$ for every $u \in \mathcal{V}$ .
(Boundedness of $\ell$ ) There exists a constant $K$ such that $|\ell(v)| \le K \|v\|_{\mathcal{V}}$ for every $v\in \mathcal{V}$ .

Then the variational problem Eq. (9) possesses a unique solution.

For our cases, $a$ will also be symmetric $a(u,v) = a(v,u)$ for all $u,v \in \mathcal{V}$ . While the Lax-Milgram theorem holds without symmetry, let us continue our discussion with this additional symmetry assumption. Note that, taken together, properties (1-2) say that the $a$ -inner product, defined as $\langle u, v \rangle_a = a(u,v)$ , is no more than so much bigger or smaller than the standard inner product $\langle u, v\rangle_{\mathcal{V}}$ of $u$ and $v$ .¹⁴

Let us now see how the Lax-Milgram theorem can apply to our two examples. For a reader who wants a more “big picture” perspective, they can comfortably skip to the next section. For those who want to see Lax-Milgram in action, see the discussion below.

Applying the Lax-Milgram Theorem

Begin with the linear system of equations with $\mathcal{V} = \mathbb{R}^N$ with inner product $\langle x, y \rangle_{\mathbb{R}^N} = y^\top x$ , $a(x,y) = y^\top Ax$ , and $\ell(y) = y^\top b$ . Note that we have the inequality $\lambda_{\rm min} x^\top x \le x^\top A x \le \lambda_{\rm max} x^\top x$ .¹⁵ In particular, we have that $\|x\|_A = \sqrt{x^\top A x} \le \sqrt{\lambda_{\rm max}} \|x\|$ . Property (1) then follows from the Cauchy-Schwarz inequality applied to the $A$ -inner product: $|a(x,y)| = |\langle x, y\rangle_A| \le \|x\|_A \|y\|_A \le \lambda_{\rm max} \|x\|\|y\|$ . Property (2) is simply the established inequality $a(x,x) = x^\top Ax \ge \lambda_{\min} x^\top x$ . Property (3) also follows from the Cauchy-Schwarz inequality: $|\ell(y)| = |y^\top b| \le \|b\|\|y\|$ . Thus, by Lax-Milgram, the variational problem $y^\top Ax = y^\top b$ for $y \in \mathbb{R}^N$ has a unique solution $x$ . Note that the linear systems example shows why the coercivity property (2) is necessary. If $A$ is positive semi-definite but not positive-definite, then there exists an eigenvector $v$ of $A$ with eigenvalue $0$ . Then $v^\top Av = 0 \not\ge c v^\top v$ for any positive constant $c$ and $A$ is singular, so the variational formulation of $Ax = b$ has no solution for some choices of the vector $b$ .

Applying the Lax-Milgram theorem to differential equations can require powerful inequalities. In this case, the $\mathcal{V} = H_0^1((0,1))$ -inner product is given by $\langle u, v \rangle_{H_0^1((0,1))} = \int_0^1 v(x)u(x) + v'(x)u'(x) \, dx$ , $a(u,v) = \int_0^1 v'(x)u'(x) \, dx$ , and $\ell(v) = \int_0^1 v(x)f(x) \, dx$ . Condition (1) is follows from a application of the Cauchy-Schwarz inequality for integrals:¹⁶

(10) $\begin{equation*} \begin{split} |a(u,v)| &\le \int_0^1 |v'(x)||u'(x)| \, dx \\ &\le \left(\int_0^1 |v'(x)|^2 \, dx\right)^{1/2} \left(\int_0^1 |u'(x)|^2 \, dx\right)^{1/2} \\ &\le \|v\|_{H_0^1((0,1))}\|u\|_{H_0^1((0,1))}. \end{split} \end{equation*}$

Let’s go line-by-line. First, we note that the absolute value of integral is less than the integral of absolute value. Next, we apply the Cauchy-Schwarz inequality for integrals. Finally, we note that $\int_0^1|v'(x)|^2 \, dx \le \int_0^1|v(x)|^2 + |v'(x)|^2 \, dx = \langle v, v\rangle_{H_0^1((0,1))} = \|v\|_{H_0^1((0,1))}^2$ . This establishes Property (1) with constant $C = 1$ . As we already see one third of the way into verifying the hypotheses of Lax-Milgram, establishing these inequalities can require several steps. Ultimately, however, strong knowledge of just a core few inequalities (e.g. Cauchy-Schwarz) may be all that’s needed.

Proving coercivity (Property (2)) actually requires a very special inequality, Poincaré’s inequality.¹⁷ In it’s simplest incarnation, the inequality states that there exists a constant $k$ such that, for all functions $u \in H_0^1((0,1))$ ,¹⁸

(11) $\begin{equation*} \int_0^1|u'(x)|^2 \, dx \ge k\int_0^1|u(x)|^2 \, dx. \end{equation*}$

With this inequality in tow, property (2) follows after another lengthy string of inequalities:¹⁹

(12) $\begin{equation*} \begin{split} a(u,u) &= \int_0^1 |u'(x)|^2 \, dx \\ &= \frac{1}{2} \int_0^1 |u'(x)|^2 \, dx + \frac{1}{2} \int_0^1 |u'(x)|^2 \, dx \\ &\ge \frac{1}{2} \int_0^1 |u'(x)|^2 \, dx + \frac{k}{2} \int_0^1 |u(x)|^2 \, dx \\ &\ge \min \left(\frac{1}{2},\frac{k}{2}\right) \left( \int_0^1 |u'(x)|^2 \, dx + \int_0^1|u(x)|^2 \, dx\right) \\ &= \min \left(\frac{1}{2},\frac{k}{2}\right) \|u\|_{H_0^1((0,1))}^2. \end{split} \end{equation*}$

For Property (3) to hold, the function $f$ must be square-integrable. With this hypothesis, Property (3) is much easier than Properties (1-2) and we leave it as an exercise for the interested reader (or to a footnote²⁰ for the uninterested reader).

This may seem like a lot of work, but the result we have achieved is stunning. We have proven (modulo a lot of omitted details) that the Poisson equation $-u'' = f$ has a unique weak solution as long as $f$ is square-integrable!²¹ What is remarkable about this proof is that it uses the Lax-Milgram theorem and some inequalities alone: no specialized knowledge about the physics underlying the Poisson equation were necessary. Going through the details of Lax-Milgram has been a somewhat lengthy affair for an introductory post, but hopefully this discussion has illuminated the power of functional analytic tools (like Lax-Milgram) in studying differential equations. Now, with a healthy dose of theory in hand, let us return to Galerkin approximation.

General Galerkin Approximation

With our general theory set up, Galerkin approximation for general variational problem is the same as it was for a system of linear equations. First, we pick an approximation space $\mathcal{X}$ which is a subspace of $\mathcal{V}$ . We then have the Galerkin variational problem

(13) $\begin{equation*} \mbox{Find $\hat{u} \in \mathcal{X}$ such that } a(\hat{u},v) = \ell(v) \mbox{ for every } v \in \mathcal{X}. \end{equation*}$

Provided $a$ and $\ell$ satisfy the conditions of the Lax-Milgram theorem, there is a unique solution $\hat{u}$ to the problem Eq. (13). Moreover, the special property of Galerkin approximation holds: the error $u-\hat{u}$ is $a$ -orthogonal to the subspace $\mathcal{X}$ . Consequently, $\hat{u}$ is te best approximate solution to the variational problem Eq. (9) in the $a$ -norm. To see the $a$ -orthogonality, we have that, for any $v \in \mathcal{X}$ ,

(14) $\begin{equation*} \langle u-\hat{u}, v\rangle_a = a(u-\hat{u},v) = a(u,v) - a(\hat{u},v) = \ell(v) - \ell(v) = 0, \end{equation*}$

where we use the variational equation Eq. (9) for $a(u,v) = \ell(v)$ and Eq. (13) for $a(\hat{u},v) = \ell(v)$ . Note the similarities with Eq. (3). Thus, using the Pythagorean theorem for the $a$ -norm, for any other approximation solution $w \in \mathcal{X}$ , we have²²

(15) $\begin{equation*} \|u - w\|^2_a = \|(u - \hat{u}) + (\hat{u}-w)\|_a^2 = \|u-\hat{u}\|_a^2 + \|\hat{u} - w\|_a^2 \ge \|u - \hat{u} \|_a^2. \end{equation*}$

Put simply, $\hat{u}$ is the best approximation to $u$ in the $a$ -norm.²³

Galerkin approximation is powerful because it allows us to approximate an infinite-dimensional problem by a finite-dimensional one. If we let $\phi_1,\ldots,\phi_M$ be a basis for the space $\mathcal{X}$ , then the approximate solution $\hat{u}$ can be represented as $\hat{u} = x_1 \phi_1 + \cdots + x_M \phi_M$ . Since $\phi_1,\ldots,\phi_N$ form a basis of $\mathcal{X}$ , to check that the Galerkin variational problem Eq. (13) holds for all $v \in \mathcal{X}$ it is sufficient to check that it holds for $v = \phi_1, v=\phi_2,\ldots,v=\phi_M$ .²⁴ Thus, plugging in $\hat{u} = \sum_{j=1}^M x_j \phi_j$ and $v = \phi_i$ into Eq. (13), we get (using bilinearity of $a$ )

(16) $\begin{equation*} a(\hat{u},v) = a\left(\sum_{j=1}^M x_j \phi_j, \phi_i\right) = \sum_{j=1}^M a(\phi_j,\phi_i) x_j = \ell(v) = \ell(\phi_i), \quad i =1,2,\ldots,M. \end{equation*}$

If we define $a_{ij} = a(\phi_j,\phi_i)$ and $b_i = \ell(\phi_i)$ , then this gives us a matrix equation $Ax = b$ for the unknowns $x_1,\ldots,x_M$ parametrizing $\hat{u}$ . Thus, we can compute our Galerkin approximation by solving a linear system of equations.

We’ve covered a lot of ground so let’s summarize. Galerkin approximation is a technique which allows us to approximately solve a large- or infinite-dimensional problem by searching for an approximate solution in a smaller finite-dimensional space $\mathcal{X}$ of our choosing. This Galerkin approximation is the best approximate solution to our original problem in the $a$ -norm. By choosing a basis $\phi_1,\ldots,\phi_M$ for our approximation space $\mathcal{X}$ , we reduce the problem of computing a Galerkin approximation to a linear system of equations.

Design of a Galerkin approximation scheme for a variational problem thus boils down to choosing the approximation space $\mathcal{X}$ and a basis $\phi_1,\ldots,\phi_M$ . Picking $\mathcal{X}$ to be a space of piecewise polynomial functions (splines) gives the finite element method. Picking $\mathcal{X}$ to be a space spanned by a collection of trigonometric functions gives a Fourier spectral method. One can use a space spanned by wavelets as well. The Galerkin framework is extremely general: give it a subspace $\mathcal{X}$ and it will give you a linear system of equations to solve to give you the best approximate solution in $\mathcal{X}$ .

Two design considerations factor into the choice of space $\mathcal{X}$ and basis $\phi_1,\ldots,\phi_M$ . First, one wants to pick a space $\mathcal{X}$ , where the solution $u$ almost lies in. This is the rationale behind spectral methods. Smooth functions are very well-approximated by short truncated Fourier expansions, so, if the solution $u$ is smooth, spectral methods will converge very quickly. Finite element methods, which often use low-order piecewise polynomial functions, converge much more slowly to a smooth $u$ . The second design consideration one wants to consider is the ease of solving the system $Ax = b$ resulting from the Galerkin approximation. If the basis function $\phi_1,\ldots,\phi_M$ are local in the sense that most pairs of basis functions $\phi_i$ and $\phi_j$ aren’t nonzero at the same point $x$ (more formally, $\phi_i$ and $\phi_j$ have disjoint supports for most $i$ and $j$ ), the system $Ax = b$ will be sparse and thus usually much easier to solve. Traditional spectral methods usually result in a harder-to-solve dense linear systems of equations.²⁵ It should be noted that both spectral and finite element methods lead to ill-conditioned matrices $A$ , making integral equation-based approaches preferable if one needs high-accuracy.²⁶ Integral equations, themselves, are often solved using Galerkin approximation, leading to so-called boundary element methods.

Upshot: Galerkin approximation is a powerful and extremely flexible methodology for approximately solving large- or infinite-dimensional problems by finding the best approximate solution in a smaller finite-dimensional subspace. To use a Galerkin approximation, one must convert their problem to a variational formulation and pick a basis for the approximation space. After doing this, computing the Galerkin approximation reduces down to solving a system of linear equations with dimension equal to the dimension of the approximation space.

Big Ideas in Applied Math: Sparse Matrices

July 18, 2020 by Ethan N. Epperly 4 Comments

Sparse matrices are an indispensable tool for anyone in computational science. I expect there are a very large number of simulation programs written in scientific research across the country which could be faster by ten to a hundred fold at least just by using sparse matrices! In this post, we’ll give a brief overview what a sparse matrix is and how we can use them to solve problems fast.

A matrix is sparse if most of its entries are zero. There is no precise threshold for what “most” means; Kolda suggests that a matrix have at least 90% of its entries be zero for it to be considered sparse. The number of nonzero entries in a sparse matrix $A$ is denoted by $\operatorname{nnz}(A)$ . A matrix that is not sparse is said to be dense.

Sparse matrices are truly everywhere. They occur in finite difference, finite element, and finite volume discretizations of partial differential equations. They occur in power systems. They occur in signal processing. They occur in social networks. They occur in intermediate stages in computations with dense rank-structured matrices. They occur in data analysis (along with their higher-order tensor cousins).

Why are sparse matrices so common? In a word, locality. If the $ij$ th entry $a_{ij}$ of a matrix $A$ is nonzero, then this means that row $i$ and column $j$ are related in some way to each other according to the the matrix $A$ . In many situations, a “thing” is only related to a handful of other “things”; in heat diffusion, for example, the temperature at a point may only depend on the temperatures of nearby points. Thus, if such a locality assumption holds, every row will only have a small number of nonzero entries and the matrix overall will be sparse.

Storing and Multiplying Sparse Matrices

A sparse matrix can be stored efficiently by only storing its nonzero entries, along with the row and column in which these entries occur. By doing this, a sparse matrix can be stored in $\mathcal{O}(\operatorname{nnz}(A))$ space rather than the standard $\mathcal{O}(N^2)$ for an $N\times N$ matrix $A$ .¹ For the efficiency of many algorithms, it will be very beneficial to store the entries row-by-row or column-by-column using compressed sparse row and column (CSR and CSC) formats; most established scientific programming software environments support sparse matrices stored in one or both of these formats. For efficiency, it is best to enumerate all of the nonzero entries for the entire sparse matrix and then form the sparse matrix using a compressed format all at once. Adding additional entries one at a time to a sparse matrix in a compressed format requires reshuffling the entire data structure for each new nonzero entry.

There exist straightforward algorithms to multiply a sparse matrix $A$ stored in a compressed format with a vector $x$ to compute the product $b = Ax$ . Initialize the vector $b$ to zero and iterate over the nonzero entries $a_{ij}$ of $A$ , each time adding $a_{ij}x_j$ to $b_i$ . It is easy to see this algorithm runs in $\mathcal{O}(\operatorname{nnz}(A))$ time.² The fact that sparse matrix-vector products can be computed quickly makes so-called Krylov subspace iterative methods popular for solving linear algebraic problems involving sparse matrices, as these techniques only interact with the matrix $A$ by computing matrix-vector products $x \mapsto Ax$ (or matrix-tranpose-vector products $y \mapsto A^\top y$ ).

Lest the reader think that every operation with a sparse matrix is necessarily fast, the product of two sparse matrices $A$ and $B$ need not be sparse and the time complexity need not be $\mathcal{O}(\operatorname{nnz}(A) + \operatorname{nnz}(B))$ . A counterexample is

(1) $\begin{equation*} A = B^\top = \begin{bmatrix} a_1 & 0 & \cdots & 0 \\ a_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & 0 \\ a_n & 0 & \cdots & 0 \end{bmatrix} \in \mathbb{R}^n \end{equation*}$

for $a_1,\ldots,a_n \ne 0$ . We have that $\operatorname{nnz}(A) = \operatorname{nnz}(B) = N$ but

(2) $\begin{equation*} AB = \begin{bmatrix} a_1^2 & a_1 a_2 & \cdots & a_1a_N \\ a_2a_1 & a_2^2 & \cdots & a_2a_N \\ \vdots & \vdots & \ddots & \vdots \\ a_Na_1 & a_Na_2 & \cdots & a_N^2 \end{bmatrix} \end{equation*}$

which has $\operatorname{nnz}(AB) = N^2$ nonzero elements and requires $\mathcal{O}(N^2)$ operations to compute. However, if one does the multiplication in the other order, one has $\operatorname{nnz}(BA) = 1$ and the multiplication can be done in $\mathcal{O}(N)$ operations. Thus, some sparse matrices can be multiplied fast and others can’t. This phenomena of different speeds for different sparse matrices is very much also true for solving sparse linear systems of equations.

Solving Sparse Linear Systems

The question of how to solve a sparse system of linear equations $Ax = b$ where $A$ is sparse is a very deep problems with fascinating connections to graph theory. For this article, we shall concern ourselves with so-called sparse direct methods, which solve $Ax = b$ by means of computing a factorization of the sparse matrix $A$ . These methods produce an exact solution to the system $Ax = b$ if all computations are performed exactly and are generally considered more robust than inexact and iterative methods. As we shall see, there are fundamental limits on the speed of certain sparse direct methods, which make iterative methods very appealing for some problems.

Note from the outset that our presentation will be on illustrating the big ideas rather than presenting the careful step-by-step details needed to actually code a sparse direct method yourself. An excellent reference for the latter is Tim Davis’ wonderful book Direct Methods for Sparse Linear Systems.

Let us begin by reviewing how $LU$ factorization works for general matrices. Suppose that the $(1,1)$ entry of $A$ is nonzero. Then, $LU$ factorization proceeds by subtracting scaled multiples of the first row from the other rows to zero out the first column. If one keeps track of these scaling, then one can write this process as a matrix factorization, which we may demonstrate pictorially as

(3) $\begin{equation*} \underbrace{\begin{bmatrix} * & * & \cdots & * \\ * & * & \cdots & * \\ \vdots & \vdots & \ddots & \vdots \\ * & * & \cdots & *\end{bmatrix}}_{=A} = \begin{bmatrix} 1 & & &\\ * & 1 & & \\ \vdots & & \ddots & \\ * & & & 1\end{bmatrix} \begin{bmatrix} * & * & \cdots & * \\ & * & \cdots & * \\ & \vdots & \ddots & \vdots \\ & * & \cdots & *\end{bmatrix}. \end{equation*}$

Here, $*$ ‘s denote nonzero entries and blanks denote zero entries. We then repeat the process on the $(N-1)\times (N-1)$ submatrix in the bottom right (the so-called Schur complement). Continuing in this way, we eventually end up with a complete $LU$ factorization

(4) $\begin{equation*} \underbrace{\begin{bmatrix} * & * & \cdots & * \\ * & * & \cdots & * \\ \vdots & \vdots & \ddots & \vdots \\ * & * & \cdots & *\end{bmatrix}}_{=A} = \underbrace{\begin{bmatrix} 1 & & &\\ * & 1 & & \\ \vdots & \vdots & \ddots & \\ * &* &\cdots & 1\end{bmatrix}}_{=L} \underbrace{\begin{bmatrix} * & * & \cdots & * \\ & * & \cdots & * \\ & & \ddots & \vdots \\ & & & *\end{bmatrix}}_{=U}. \end{equation*}$

In the case that $A$ is symmetric positive definite (SPD), one has that $U = DL^\top$ for $D$ a diagonal matrix consisting of the entries on $U$ . This factorization $A = LDL^\top$ is a Cholesky factorization of $A$ .³ For general non-SPD matrices, one needs to incorporate partial pivoting for Gaussian elimination to produce accurate results.⁴

Let’s try the same procedure for a sparse matrix. Consider a sparse matrix with the following sparsity pattern:

(5) $\begin{equation*} A = \begin{bmatrix} * & * & * & & * \\ * & * & & * & \\ * & & * & & \\ & * & & * & \\ * & & & & * \end{bmatrix}. \end{equation*}$

When we eliminate the $(1,1)$ entry, we get the following factorization:

(6) $\begin{equation*} \underbrace{\begin{bmatrix} * & * & * & & * \\ * & * & & * & \\ * & & * & & \\ & * & & * & \\ * & & & & * \end{bmatrix}}_{=A} = \begin{bmatrix} 1 & & & & \\ * & 1& & & \\ * & & 1 & & \\ & & & 1 & \\ * & & & & 1 \end{bmatrix} \begin{bmatrix} * & * & * & & * \\ & * & \bullet & * & \bullet \\ & \bullet & * & & \bullet \\ & * & & * & \\ & \bullet & \bullet & & * \end{bmatrix} \end{equation*}$

Note that the Schur complement has new additional nonzero entries (marked with a $\bullet$ ) not in the original sparse matrix $A$ . The Schur complement of $A$ is denser than $A$ was; there are new fill-in entries. The worst-case scenario for fill-in is the arrowhead matrix:

(7) $\begin{equation*} \underbrace{\begin{bmatrix} * & * & * & \cdots & * \\ * & * & & & \\ * & & * & & \\ \vdots & & & \ddots & \\ * & & & & * \end{bmatrix}}_{=A} = \begin{bmatrix} 1 & & & & \\ * & 1& & & \\ * & & 1 & & \\ \vdots & & & \ddots & \\ * & & & & 1 \end{bmatrix} \begin{bmatrix} * & * & * & * & * \\ & * & \bullet & \cdots & \bullet \\ & \bullet & * & \cdots & \bullet \\ & \vdots & \vdots & \ddots & \vdots \\ & \bullet & \bullet & \cdots & * \end{bmatrix} \end{equation*}$

After one step of Gaussian elimination, we went from a matrix with $\mathcal{O}(N)$ nonzeros to a fully dense Schur complement! However, the arrowhead matrix also demonstrates a promising strategy. Simply construct a permutation matrix which reorders the first entry to be the last⁵ and then perform Gaussian elimination on the symmetrically permuted matrix $PAP^\top$ instead. In fact, the entire $LU$ factorization can be computed without fill-in:

(8) $\begin{equation*} \underbrace{\begin{bmatrix} * & & & & * \\ & * & & & * \\ & & * & & * \\ & & & \ddots & \vdots \\ * & * & * & \cdots & * \end{bmatrix}}_{=PAP^\top} = \begin{bmatrix} 1 & & & & \\ & 1& & & \\ & & 1 & & \\ & & & \ddots & \\ * & * & * & \cdots & 1 \end{bmatrix} \begin{bmatrix} * & & & & * \\ & * & & & * \\ & & * & & * \\ & & & \ddots & \vdots \\ & & & & * \end{bmatrix}. \end{equation*}$

This example shows the tremendous importance of reordering of the rows and columns when computing a sparse $LU$ factorization.

The Best Reordering

As mentioned above, when computing an $LU$ factorization of a dense matrix, one generally has to reorder the rows (and/or columns) of the matrix to compute the solution accurately. Thus, when computing the $LU$ factorization of a sparse matrix, one has to balance the need to reorder for accuracy and to reorder to reduce fill-in. For these reasons, for the remainder of this post, we shall focus on computing Cholesky factorizations of SPD sparse matrices, where reordering for accuracy is not necessary.⁶ Since we want the matrix to remain SPD, we must restrict ourselves to symmetric reordering strategies where $A$ is reordered to $PAP^\top$ where $P$ is a permutation matrix.

Our question is deceptively simple: what reordering produces the least fill-in? In matrix language, what permutation $P$ minimizes $\operatorname{nnz}(L)$ where $LDL^\top = PAP^\top$ is the Cholesky factorization of $PAP^\top$ ?

Note that, assuming no entries in the Gaussian elimination process exactly cancel, then the Cholesky factorization depends only on the sparsity pattern of $A$ (the locations of the zeros and nonzeros) and not on the actual numeric values of $A$ ‘s entries. This sparsity structure is naturally represented by a graph $\mathcal{G}(A)$ whose nodes are the indices $\{1,\ldots,N\}$ with an edge between $i \ne j$ if, and only if, $a_{ij} \ne 0$ .

Now let’s see what happens when we do Gaussian elimination from a graph point-of-view. When we eliminate the $(1,1)$ entry from matrix, this results in all nodes of the graph adjacent to $1$ becoming connected to each other.⁷

This shows why the arrowhead example is so bad. By eliminating the a vertex connected to every node in the graph, the eliminated graph becomes a complete graph.

Reordering the matrix corresponds to choosing in what order the vertices of the graph are eliminated. Choosing the elimination order is then a puzzle game; eliminate all the vertices of the graph in the order that produces the fewest fill-in edges (shown red).⁸

Finding the best elimination ordering for a sparse matrix (graph) is a good news/bad news situation. For the good news, many graphs possess a perfect elimination ordering, in which no fill-in is produced at all. There is a simple algorithm to determine whether a graph (sparse matrix) possesses a perfect elimination ordering and if so, what it is.⁹ Some important classes of graphs can be eliminated perfectly (for instance, trees). More generally, the class of all graphs which can be eliminated perfectly is precisely the set of chordal graphs, which are well-studied in graph theory.

Now for the bad news. The problem of finding the best elimination ordering (with the least fill-in) for a graph is NP-Hard. This means, assuming the widely conjectured result that ${\rm P} \ne {\rm NP}$ , that finding the best elimination ordering would be a hard computational problem than the worst-case $\mathcal{O}(N^3)$ complexity for doing Gaussian elimination in any ordering! One should not be too pessimistic about this result, however, since (assuming ${\rm P} \ne {\rm NP}$ ) all it says is that there exists no polynomial time algorithm guaranteed to produce the absolutely best possible elimination ordering when presented with any graph (sparse matrix). If one is willing to give up on any one of the bolded statements, further progress may be possible. For instance, there exists several good heuristics, which find reasonably good elimination orderings for graphs (sparse matrices) in linear $\mathcal{O}(\operatorname{nnz}(A))$ (or nearly linear) time.

Can Sparse Matrices be Eliminated in Linear Time?

Let us think about the best reordering question in a different way. So far, we have asked the question “Can we find the best ordering for a sparse matrix?” But another question is equally important: “How efficiently can we solve a sparse matrix, even with the best possible ordering?”

One might optimistically hope that every sparse matrix possesses an elimination ordering such that its Cholesky factorization can be computed in linear time (in the number of nonzeros), meaning that the amount of time needed to solve $Ax = b$ is proportional to the amount of data needed to store the sparse matrix $A$ .

When one tests a proposition like this, one should consider the extreme cases. If the matrix $A$ is dense, then it requires $\mathcal{O}(N^3)$ operations to do Gaussian elimination,¹⁰ but $A$ only has $\operatorname{nnz}(A) = N^2$ nonzero entries. Thus, our proposition cannot hold in unmodified form.

An even more concerning counterexample is given by a matrix $A$ whose graph $\mathcal{G}(A)$ is a $\mathcal{O}(\sqrt{N}) \times \mathcal{O}(\sqrt{N})$ 2D grid graph.

Sparse matrices with this sparsity pattern (or related ones) appear all the time in discretized partial differential equations in two dimensions. Moreover, they are truly sparse, only having $\operatorname{nnz}(A) = \mathcal{O}(N)$ nonzero entries. Unforunately, no linear time elimination ordering exists. We have the following theorem:

Theorem: For any elimination ordering for a sparse matrix $A$ with $\mathcal{G}(A)$ being a $\sqrt{N}\times \sqrt{N}$ 2D grid graph, in any elimination ordering, the Cholesky factorization $PAP^\top = LDL^\top$ requires $\Omega(N^{3/2})$ operations and satisfies $\operatorname{nnz}(L)= \Omega(N\log N)$ .¹¹

The proof is contained in Theorem 10 and 11 (and the ensuing paragraph) of classic paper by Lipton, Rose, and Tarjan. Natural generalizations to $d$ -dimensional grid graphs give bounds of $\Omega(N^{3(d-1)/d})$ time and $\operatorname{nnz}(L)=\Omega(N^{2(d-1)/d})$ for $d > 2$ . In particular, for 2D finite difference and finite element discretizations, sparse Cholesky factorization takes $\Omega(N^{3/2})$ operations and produces a Cholesky factor with $\operatorname{nnz}(L)= \Omega(N\log N)$ in the best possible ordering. In 3D, sparse Cholesky factorization takes $\Omega(N^{2})$ operations and produces a Cholesky factor with $\operatorname{nnz}(L)= \Omega(N^{4/3})$ in the best possible ordering.

Fortunately, at least these complexity bounds are attainable: there is an ordering which produces a sparse Cholesky factorization with $PAP^\top = LDL^\top$ requiring $\Theta(N^{3/2})$ operations and with $\operatorname{nnz}(L)= \Theta(N\log N)$ nonzero entries in the Cholesky factor.¹² One such asymptotically optimal ordering is the nested dissection ordering, one of the heuristics alluded to in the previous section. The nested dissection ordering proceeds as follows:

Find a separator $S$ consisting of a small number of vertices in the graph $\mathcal{G}(A)$ such that when $S$ is removed from $\mathcal{G}(A)$ , $\mathcal{G}(A)$ is broken into a small number of edge-disjoint and roughly evenly sized pieces $\mathcal{G}_1,\ldots,\mathcal{G}_k$ .¹³
Recursively use nested dissection to eliminate each component $\mathcal{G}_1,\ldots, \mathcal{G}_k$ individually.
Eliminate $S$ in any order.

For example, for the 2D grid graph, if we choose $S$ to be a cross through the center of the 2D grid graph, we have a separator of size $|S| = \Theta(\sqrt{N})$ dividing the graph into $4$ roughly $\sqrt{N}/2\times \sqrt{N}/2$ pieces.

Let us give a brief analysis of this nested dissection ordering. First, consider the sparsity of the Cholesky factor $\operatorname{nnz}(L)$ . Let $S(N)$ denote the number of nonzeros in $L$ for an elimination of the $\sqrt{N}\times \sqrt{N}$ 2D grid graph using the nested dissection ordering. Then step 2 of nested dissection requires us to recursively eliminate four $\sqrt{N/4} \times \sqrt{N/4}$ 2D grid graphs. After doing this, for step 3, all of the vertices of the separator might be connected to each other, so the separator graph will potentially have as many as $\mathcal{O}(|S|^2) = \mathcal{O}(N)$ edges, which result in nonzero entries in $L$ . Thus, combining the fill-in from both steps, we get

(9) $\begin{equation*} S(N) = 4S\left(\frac{N}{4}\right) + \mathcal{O}(N). \end{equation*}$

Solving this recurrence using the master theorem for recurrences gives $\operatorname{nnz}(L) = S(N) = \mathcal{O}(N\log N)$ . If one instead wants the time $T(N)$ required to compute the Cholesky factorization, note that for step 3, in the worst case, all of the vertices of the separator might be connected to each other, leading to a $\sqrt{N}\times \sqrt{N}$ dense matrix. Since a $\sqrt{N}\times \sqrt{N}$ matrix requires $\mathcal{O}((\sqrt{N})^3) = \mathcal{O}(N^{3/2})$ , we get the recurrence

(10) $\begin{equation*} T(N) = 4T\left(\frac{N}{4}\right) + \mathcal{O}(N^{3/2}), \end{equation*}$

which solves to $T(N) = \mathcal{O}(N^{3/2})$ .

Conclusions

As we’ve seen, sparse direct methods (as exemplified here by sparse Cholesky) possess fundamental scalability challenges for solving large problems. For the important class of 2D and 3D discretized partial differential equations, the time to solve $Ax = b$ scales like $\Theta(N^{3/2})$ and $\Theta(N^2)$ , respectively. For truly large-scale problems, these limitations may be prohibitive for using such methods.

This really is the beginning of the story, not the end for sparse matrices however. The scalability challenges for classical sparse direct methods has spawned many exciting different approaches, each of which combats the scalability challenges of sparse direct methods for a different class of sparse matrices in a different way.

Upshot: Sparse matrices occur everywhere in applied mathematics, and many operations on them can be done very fast. However, the speed of computing an $LU$ factorization of a sparse matrix depends significantly on the arrangement of its nonzero entries. Many sparse matrices can be factored quickly, but some require significant time to factor in any reordering.