Stochastic Trace Estimation

January 26, 2023 by Ethan N. Epperly 6 Comments

I am delighted to share that me, Joel A. Tropp, and Robert J. Webber‘s paper XTrace: Making the Most of Every Sample in Stochastic Trace Estimation has recently been released as a preprint on arXiv. In it, we consider the implicit trace estimation problem:

Implicit trace estimation problem: Given access to a square matrix $A$ via the matrix–vector product operation $\omega \mapsto A\omega$ , estimate its trace $\tr A = \sum_{i=1}^n A_{ii}$ .

Algorithms for this task have many uses such as log-determinant computations in machine learning, partition function calculations in statistical physics, and generalized cross validation for smoothing splines. I described another application to counting triangles in a large network in a previous blog post.

Our paper presents new trace estimators XTrace and XNysTrace which are highly efficient, producing accurate trace approximations using a small budget of matrix–vector products. In addition, these algorithms are fast to run and are supported by theoretical results which explain their excellent performance. I really hope that you will check out the paper to learn more about these estimators!

For the rest of this post, I’m going to talk about the most basic stochastic trace estimation algorithm, the Girard–Hutchinson estimator. This seemingly simple algorithm exhibits a number of nuances and forms the backbone for more sophisticated trace estimates such as Hutch++, Nyström++, XTrace, and XNysTrace. Toward the end, this blog post will be fairly mathematical, but I hope that the beginning will be fairly accessible to all.

Girard–Hutchinson Estimator: The Basics

The Girard–Hutchinson estimator for the trace of a square matrix $A$ is

(1) $\hat{\tr} = \frac{1}{m} \sum_{i=1}^m \omega_i^* A \omega_i.$

Here, $\omega_1,\ldots,\omega_m$ are random vectors, usually chosen to be statistically independent, and ${}^*$ denotes the conjugate transpose of a vector or matrix. The Girard–Hutchinson estimator only depends on the matrix $A$ through the matrix–vector products $A\omega_1,\ldots,A\omega_m$ .

Unbiasedness

Provided the random vectors are isotropic

(2) $\mathbb{E} [\omega_i\omega_i^*] = I,$

the Girard–Hutchinson estimator is unbiased:

(3) $\mathbb{E} [\hat{\tr}] = \tr A.$

Let us confirm this claim in some detail. First, we use linearity of expectation to evaluate

(4) $\mathbb{E} [\hat{\tr}] = \mathbb{E} \left[ \frac{1}{m} \sum_{i=1}^m \omega_i^*A\omega_i \right] = \frac{1}{m} \sum_{i=1}^m \mathbb{E} \left[ \omega_i^* A \omega_i\right].$

Therefore, to prove that $\mathbb{E} [\hat{\tr}] = \tr A$ , it is sufficient to prove that $\mathbb{E} \left[\omega_i^*A\omega_i\right] = \tr A$ for each $i$ .

When working with traces, there are two tricks that solve 90% of derivations. The first trick is that, if we view a number as a $1\times 1$ matrix, then a number equals its trace, $x = \tr x$ . The second trick is the cyclic property: For a $k\times p$ matrix $B$ and a $p\times k$ matrix $C$ , we have $\tr (BC) = \tr (CB)$ . The cyclic property should be handled with care when one works with a product of three or more matrices. For instance, we have

$\tr[BCD] = \tr[(BC)D] = \tr[D(BC)] = \tr[DBC].$

However,

$\tr [BCD] \ne \tr[CBD] \quad \text{in general}.$

One should think of the matrix product $BCD$ as beads on a closed loop of string. One can move the last bead $D$ to the front of the other two, $\tr [BCD] = \tr[DBC]$ , but not interchange two beads, $\tr[BCD] \ne \tr[CBD]$ .

With this trick in hand, let’s return to proving that $\mathbb{E} \left[\omega_i^*A\omega_i\right] = \tr A$ for every $i$ . Apply our two tricks:

$\mathbb{E} \left[\omega_i^*A\omega_i\right] = \mathbb{E} \tr \left[\omega_i^*A\omega_i\right] = \mathbb{E} \tr \left[A\omega_i\omega_i^*\right].$

The expectation is a linear operation and the matrix $A$ is non-random, so we can bring the expectation into the trace as

$\mathbb{E} \left[\omega_i^*A\omega_i\right] = \mathbb{E} \tr \left[A\omega_i\omega_i^*\right] = \tr(A \mathbb{E}[\omega_i\omega_i^*] ).$

Invoke the isotropy condition (2) and conclude:

$\mathbb{E} \left[\omega_i^*A\omega_i\right] = \tr(A \mathbb{E}[\omega_i\omega_i^*] ) = \tr(A\cdot I) = \tr A.$

Plugging this into (4) confirms the unbiasedness claim (3).

Variance

Continue to assume that the $\omega_i$ ‘s are isotropic (3) and now assume that $\omega_1,\ldots,\omega_m$ are independent. By independence, the variance can be written as

$\Var(\hat{\tr}) = \frac{1}{m^2} \sum_{i=1}^m \Var(\omega_i^*A\omega_i).$

Assuming that $\omega_1,\ldots,\omega_m$ are identically distributed $\omega_1,\ldots,\omega_m \sim \omega$ , we then get

$\Var(\hat{\tr}) = \frac{1}{m} \Var(\omega^*A\omega).$

The variance decreases like $1/m$ , which is characteristic of Monte Carlo-type algorithms. Since $\hat{\tr}$ is unbiased (i.e, (3)), this means that the mean square error decays like $1/m$ so the average error (more precisely root-mean-square error) decays like

$\left| \hat{\tr} - \tr A \right| \lessapprox \frac{\mathrm{const}}{\sqrt{m}}.$

This type of convergence is very slow. If I want to decrease the error by a factor of $10$ , I must do $100\times$ the work!

Variance-reduced trace estimators like Hutch++ and our new trace estimator XTrace improve the rate of convergence substantially. Even in the worst case, Hutch++ and XTrace reduce the variance at a rate $1/m^2$ and (root-mean-square) error at rates $1/m$ :

$\Var(\hat{\tr}_{\text{H++ or X}}) \le \frac{\mathrm{const}}{m^2},\quad \left| \hat{\tr}_{\text{H++ or X}} - \tr A \right| \lessapprox \frac{\mathrm{const}}{m}.$

For matrices with rapidly decreasing singular values, the variance and error can decrease much faster than this.

Variance Formulas

As the rate of convergence for the Girard–Hutchinson estimator is so slow, it is imperative to pick a distribution on test vectors $\omega$ that makes the variance of the single–sample estimate $\omega^*A\omega$ as low as possible. In this section, we will provide several explicit formulas for the variance of the Girard–Hutchinson estimator. Derivations of these formulas will appear at the end of this post. These variance formulas help illuminate the benefits and drawbacks of different test vector distributions.

To express the formulas, we will need some notation. For a complex number $z = a + bi$ we use $\Re(z) = a$ and $\Im(z) = b$ to denote the real and imaginary parts. The variance of a random complex number $z$ is

$\Var(z) := \mathbb{E} |z - \mathbb{E} z|^2 = \Var(\Re z) + \Var(\Im z).$

The Frobenius norm of a matrix $A$ is

$\left\|A\right\|_{\rm F}^2 = \sum_{i,j} |A_{ij}|^2.$

If $A$ is real symmetric or complex Hermitian with (real) eigenvalues $\lambda_1,\ldots,\lambda_n$ , we have

(5) $\left\|A\right\|_{\rm F}^2 = \sum_{i=1}^n \lambda_i^2.$

$A^\top$ denotes the ordinary transpose of $A$ and $A^*$ denotes the conjugate transpose of $A$ .

Real-Valued Test Vectors

We first focus on real-valued test vectors $\omega$ . Since $\omega$ is real, we can use the ordinary transpose ${}^\top$ rather than the conjugate transpose ${}^*$ . Since $\omega^\top A\omega$ is a number, it is equal to its own transpose:

$\omega^\top A \omega = (\omega^\top A \omega)^\top = \omega^\top A^\top \omega.$

Therefore,

$\omega^\top A\omega = \frac{\omega^\top A \omega + \omega^\top A^\top \omega}{2} = \omega^\top \left( \frac{A + A^\top}{2} \right)\omega.$

The Girard–Hutchinson trace estimator applied to $A$ is the same as the Girard–Hutchinson estimator applied to the symmetric part of $A$ , $(A+A^\top)/2$ .

For the following results, assume $A$ is symmetric, $A = A^\top$ .

Real Gaussian: $\omega_1,\ldots,\omega_m$ are independent standard normal random vectors.
$\Var(\omega^\top A\omega) = 2 \left\|A\right\|_{\rm F}^2.$
Uniform signs (Rademachers): $\omega_1,\ldots,\omega_m$ are independent random vectors with uniform $\pm 1$ coordinates.
$\Var(\omega^\top A \omega) = 2\sum_{i\ne j} |A_{ij}|^2.$
Real sphere: Assume $\omega_1,\ldots,\omega_n$ are uniformly distributed on the real sphere of radius $\sqrt{n}$ : $\omega \sim \text{Uniform} \{x\in \mathbb{R}^n : x^\top x = n\}$ .
$\Var(\omega^\top A\omega) = \frac{2n}{n+2} \left( \left\|A\right\|_{\rm F}^2 - \frac{1}{n} |\tr A|^2 \right).$

These formulas continue to hold for nonsymmetric $A$ by replacing $A$ by its symmetric part $(A+A^\top)/2$ on the right-hand sides of these variance formulas.

Complex-Valued Test Vectors

We now move our focus to complex-valued test vectors $\omega$ . As a rule of thumb, one should typically expect that the variance for complex-valued test vectors applied to a real symmetric matrix $A$ is about half the natural real counterpart—e.g., for complex Gaussians, you get about half the variance than with real Gaussians.

A square complex matrix has a Cartesian decomposition

$A = A^{\rm H} + i A^{\rm SH}$

where

$A^{\rm H} = \frac{A+A^*}{2} ,\quad A^{\rm SH} = \frac{A - A^*}{2i}$

denote the Hermitian and skew-Hermitian parts of $A$ . Similar to how the imaginary part of a complex number is real, the skew-Hermitian part of a complex matrix is Hermitian (and $i A^{\rm SH}$ is skew-Hermitian). Since $A^{\rm H}$ and $A^{\rm SH}$ are both Hermitian, we have

$\Re(\omega^* A\omega) = \omega^* A^{\rm H} \omega, \quad \Im (\omega^* A \omega) = \omega^* A^{\rm SH} \omega.$

Consequently, the variance of $\omega^*A \omega$ can be broken into Hermitian and skew-Hermitian parts:

$\Var(\omega^* A\omega) = \Var(\omega^* A^{\rm H}\omega) + \Var(\omega^* A^{\rm SH}\omega).$

For this reason, we will state the variance formulas only for Hermitian $A$ , with the formula for general $A$ following from the Cartesian decomposition.

For the following results, assume $A$ is Hermitian, $A = A^*$ .

Complex Gaussian: $\omega_1,\ldots,\omega_m$ are independent standard complex random vectors, i.e., each $\omega_i$ has iid entries distributed as $(g_1+ig_2)/\sqrt{2}$ for $g_1,g_2$ standard normal random variables.
$\Var(\omega^* A\omega) = \left\|A\right\|_{\rm F}^2.$
Uniform phases (Steinhauses): $\omega_1,\ldots,\omega_m$ are independent random vectors whose entries are uniform on the complex unit circle $\{ z \in \complex : |z| \}$ .
$\Var(\omega^* A \omega) = \sum_{i\ne j} |A_{ij}|^2.$
Complex sphere: Assume $\omega_1,\ldots,\omega_n$ are uniformly distributed on the complex sphere of radius $\sqrt{n}$ : $\omega \sim \text{Uniform} \{x\in \complex^n : x^* x = n\}$ .
$\Var(\omega^* A\omega) = \frac{n}{n+1} \left( \left\|A\right\|_{\rm F}^2 - \frac{1}{n} |\tr A|^2 \right).$

Optimality Properties

Let us finally address the question of what the best choice of test vectors is for the Girard–Hutchinson estimator. We will state two results with different restrictions on $\omega_1,\ldots,\omega_m$ .

Our first result, due to Hutchinson, is valid for real symmetric matrices with real test vectors.

Optimality (independent test vectors with independent coordinates). If the test vectors $\omega_1,\ldots,\omega_m \in \mathbb{R}^n$ are isotropic (2), independent from each other, and have independent entries, then for any fixed real symmetric matrix $A$ , the minimum variance for $\hat{\tr}$ is obtained when $\omega_1,\ldots,\omega_m$ are populated with random signs $(\omega_i)_j \sim \textnormal{Uniform} \{\pm 1\}$ .

The next optimality results will have real and complex versions. To present the results for $\mathbb{R}$ -valued and an $\complex$ -valued test vectors on unified footing, let $\field$ denote either $\mathbb{R}$ or $\complex$ . We let a $\field$ -Hermitian matrix be either a real symmetric matrix (if $\field = \mathbb{R}$ ) or a complex Hermitian matrix (if $\field = \complex$ ). Let a $\field$ -unitary matrix be either a real orthogonal matrix (if $\field = \mathbb{R}$ ) or a complex unitary matrix (if $\field = \complex$ ).

The condition that the vectors $\omega_1,\ldots,\omega_m$ have independent entries is often too restrictive in practice. It rules out, for instance, the case of uniform vectors on the sphere. If we relax this condition, we get a different optimal distribution:

Optimality (independent test vectors). Consider any set $\mathscr{A}$ of $\field$ -Hermitian matrices which is invariant under $\field$ -unitary similary transformations:
$\text{If $A \in \mathscr{A}$ and $U$ is $\field$-unitary, then $U^*AU \in \mathscr{A}$.}$
Assume that the test vectors $\omega_1,\ldots,\omega_m$ are independent and isotropic (2). The worst-case variance $\sup_{A \in \mathscr{A}} \Var(\hat{\tr})$ is minimized by choosing $\omega_1,\ldots,\omega_m$ uniformly on the $\field$ -sphere: $\omega_1,\ldots,\omega_m \sim \text{Uniform} \{ x \in \field^n : x^*x =n \}$ .

More simply, if you wants your stochastic trace estimator to be effective for a class of inputs $\mathscr{A}$ (closed under $\field$ -unitary similarity transformations) rather than a single input matrix $A$ , then the best distribution are test vectors drawn uniformly from the sphere. Examples of classes of matrices $\mathscr{A}$ include:

Fixed eigenvalues. For fixed real eigenvalues $\lambda_1,\ldots,\lambda_n \in \mathbb{R}$ , the set of all $\field$ -Hermitian matrices with these eigenvalues.
Density matrices. The class of all trace-one psd matrices.
Frobenius norm ball. The class of all $\field$ -Hermitian matrices of Frobenius norm at most 1.

Derivation of Formulas

In this section, we provide derivations of the variance formulas. I have chosen to focus on derivations which are shorter but use more advanced techniques rather than derivations which are longer but use fewer tricks.

Real Gaussians

First assume $A$ is real. Since $A$ is real symmetric, $A$ has an eigenvalue decomposition $A = Q\Lambda Q^\top$ , where $Q$ is orthogonal and $\Lambda$ is a diagonal matrix reporting $A$ ‘s eigenvalues. Since the real Gaussian distribution is invariant under orthogonal transformations, $\omega^\top A\omega = (Q^\top \omega)^\top \Lambda (Q^\top\omega)$ has the same distribution as $\omega^\top \Lambda \omega$ . Therefore,

$\Var(\omega^\top A \omega) = \Var(\omega^\top \Lambda \omega) = \Var \left( \sum_{i=1}^n \lambda_i \omega_i^2 \right) = \sum_{i=1}^n \lambda_i^2 \Var(\omega_i^2) = 2\sum_{i=1}^n \lambda_i^2 = 2\left\|A\right\|_{\rm F}^2.$

Here, we used that the variance of a squared standard normal random variable is two.

For $A$ non-real matrix, we can break the matrix $A$ into its entrywise real and imaginary parts $A = \mathfrak{R}(A) + i \, \mathfrak{I}(A)$ . Thus,

$\Var(\omega^\top A \omega) = \Var(\omega^\top \mathfrak{R}(A) \omega) + \Var(\omega^\top \mathfrak{I}(A) \omega) = 2\left\|\mathfrak{R}(A)\right\|_{\rm F}^2 + 2\left\|\mathfrak{I}(A)\right\|_{\rm F}^2 = 2\left\|A\right\|_{\rm F}^2.$

Uniform Signs

First, compute

$\omega^\top A \omega - \mathbb{E}[\omega^\top A \omega] = \sum_{i,j=1}^n A_{ij} \omega_i\omega_j - \sum_{i=1}^n A_{ii} = \sum_{i\ne j} A_{ij} \omega_i\omega_j + \sum_{i=1}^n A_{ii}(\omega_i^2-1).$

For a vector $\omega$ of uniform random signs, we have $\omega_i^2 = 1$ for every $i$ , so the second sum vanishes. Note that we have assumed $A$ symmetric, so the sum over $i\ne j$ can be replaced by two times the sum over $i < j$ :

$\omega^\top A \omega - \mathbb{E}[\omega^\top A \omega] = 2\sum_{i< j} A_{ij} \omega_i\omega_j.$

Note that $\{ \omega_i \omega_j : i < j\}$ are pairwise independent. As a simple exercise, one can verify that the identity

$\Var(a_1 X_1+\cdots+a_kX_k) = |a_1|^2 \Var(X_1) + \cdots + |a_k|^2 \Var(X_k)$

holds for any pairwise independent family of random variances $X_1,\ldots,X_k$ and numbers $a_1,\ldots,a_k$ . Ergo,

$\begin{align*}\Var(\omega^\top A\omega) &= \Var(\omega^\top A \omega - \mathbb{E}[\omega^\top A \omega]) \\&= \Var\left(\sum_{i< j} 2A_{ij} \omega_i\omega_j\right) \\&= \sum_{i<j} 4 |A_{ij}|^2 \Var(\omega_i\omega_j) \\&= \sum_{i<j} 4 |A_{ij}|^2 \\&= 2 \sum_{i\ne j} |A_{ij}|^2.\end{align*}$

In the second-to-last line, we use the fact that $\omega_i\omega_j$ is a uniform random sign, which has variance $1$ . The final line is a consequence of the symmetry of $A$ .

Uniform on the Real Sphere

The simplest proof is I know is by the “camel principle”. Here’s the story (a lightly edited quotation from MathOverflow):

A father left 17 camels to his three sons and, according to the will, the eldest son was to be given a half of the camels, the middle son one-third, and the youngest son the one-ninth. The sons did not know what to do since 17 is not evenly divisible into either two, three, or nine parts, but a wise man helped the sons: he added his own camel, the oldest son took $18/2=9$ camels, the second son took $18/3=6$ camels, the third son $18/9=2$ camels and the wise man took his own camel and went away.

We are interested in a vector $\omega$ which is uniform on the sphere of radius $\sqrt{n}$ . Performing averages on the sphere is hard, so we add a camel to the problem by “upgrading” $\omega$ to a spherically symmetric vector $g$ which has a random length. We want to pick a distribution for which the computation $\Var(g^\top A g)$ is easy. Fortunately, we already know such a distribution, the Gaussian distribution, for which we already calculated $\Var(g^\top A g) = 2\left\|A\right\|_{\rm F}^2$ .

The Gaussian vector $g$ and the uniform vector $\omega$ on the sphere are related by

$g = \sqrt{\frac{a}{n}} \omega,$

where $a$ is the squared length of the Gaussian vector $g$ . In particular, $a$ has the distribution of the sum of $n$ squared Gaussian random variables, which is known as a $\chi^2$ random variable with $n$ degrees of freedom.

Now, we take the camel back. Compute the variance of $g^\top A g$ using the chain rule for variance:

$\Var(g^\top A g) = \mathbb{E}[\Var(g^\top A g \mid a)] + \Var(\mathbb{E}[g^\top A g \mid a]).$

Here, $\Var(\cdot \mid a)$ and $\mathbb{E}[ \cdot \mid a]$ denote the conditional variance and conditional expectation with respect to the random variable $a$ . The quick and dirty ways of working with these are to treat the random variable $a$ “like a constant” with respect to the conditional variance and expectation.

Plugging in the formula $g = \sqrt{a/n} \cdot \omega$ and treating $a$ “like a constant”, we obtain

$\begin{align*}\Var(g^\top A g) &= \mathbb{E}[\Var(a/n \cdot \omega^\top A \omega \mid a)] + \Var(\mathbb{E}[a/n \cdot \omega^\top A \omega \mid a]) \\&=\mathbb{E}[(a/n)^2\Var(\omega^\top A \omega)] + \Var(a/n \cdot \mathbb{E}[\omega^\top A \omega]) \\&= \frac{1}{n^2} \mathbb{E}[a^2] \cdot \Var(\omega^\top A \omega) + \frac{1}{n^2} \Var(a) |\mathbb{E} [\omega^\top A \omega]|^2.\end{align*}$

As we mentioned, $a$ is a $\chi^2$ random variable with $n$ degrees of freedom and $\mathbb{E}[a^2]$ and $\Var(a)$ are known quantities that can be looked up:

$\mathbb{E}[a^2] = n(n+2), \quad \Var(a) = 2n.$

We know $\Var(g^\top A g) = 2\left\|A\right\|_{\rm F}^2$ and $\mathbb{E} [\omega^\top A \omega] = \tr A$ . Plugging these all in, we get

$2\left\|A\right\|_{\rm F}^2 = \frac{n+2}{n} \Var(\omega^\top A\omega) + \frac{2}{n} |\tr A|^2.$

Rearranging, we obtain

$\Var(\omega^\top A\omega) = \frac{2n}{n+2} \left( \left\|A\right\|_{\rm F}^2 - \frac{1}{n}|\tr A|^2\right).$

Complex Gaussians

The trick is the same as for real Gaussians. By invariance of complex Gaussian random vectors under unitary transformations, we can reduce to the case where $A$ is a diagonal matrix populated with eigenvalues $\lambda_1,\ldots,\lambda_n$ . Then

$\Var(\omega^*A \omega) = \Var \left( \sum_{i=1}^n \lambda_i |\omega_i|^2 \right) = \sum_{i=1}^n \Var(|\omega_i|^2) \lambda_i^2 = \sum_{i=1}^n \lambda_i^2 = \left\|A\right\|_{\rm F}^2.$

Here, we use the fact that $2|\omega_i|^2$ is a $\chi^2$ random variable with two degrees of freedom, which has variance four.

Random Phases

The trick is the same as for uniform signs. A short calculation (remembering that $A$ is Hermitian and thus $\overline{A_{ij}} = A_{ji}$ ) reveals that

$\Var\left( \omega^* A \omega \right) = \Var \left( \sum_{i<j} 2 \Re(A_{ij} \overline{\omega_i} \omega_j) \right).$

The random variables $\{\overline{\omega_i} \omega_j : i < j\}$ are pairwise independent so we have

$\Var\left( \omega^* A \omega \right) = \Var \left( \sum_{i<j} 2 \Re(A_{ij} \overline{\omega_i} \omega_j) \right) = 4\sum_{i<j} \Var \left( \Re(A_{ij} \overline{\omega_i} \omega_j) \right).$

Since $\overline{\omega}_i \omega_j$ is uniformly distributed on the complex unit circle, we can assume without loss of generality that $A_{ij} = |A_{ij}|$ . Thus, letting $\phi$ be uniform on the complex unit circle,

$\Var\left( \omega^* A \omega \right) = 4\sum_{i<j} \Var \left( |A_{ij}|\Re(\phi)) \right) = 4\Var\left( \Re(\phi) \right)\sum_{i<j}|A_{ij}|^2.$

The real and imaginary parts of $\phi$ have the same distribution so

$1 = \Var(\phi) = \Var(\Re \phi) + \Var(\Im \phi) = 2 \Var(\Re \phi)$

so $\Var(\Re \phi) = 1/2$ . Thus

$\Var\left( \omega^* A \omega \right) = 2 \sum_{i<j}|A_{ij}|^2 = \sum_{i\ne j} |A_{ij}|^2.$

Uniform on the Complex Sphere: Derivation 1 by Reduction to Real Case

There are at least three simple ways of deriving this result: the camel trick, reduction to the real case, and Haar integration. Each of these techniques illustrates a trick that is useful in its own right beyond the context of trace estimation. Since we have already seen an example of the camel trick for the real sphere, I will present the other two derivations.

Let us begin with the reduction to the real case. Let $\mathfrak{R}(\cdot)$ and $\mathfrak{I}(\cdot)$ denote the real and imaginary parts of a vector or matrix, taken entrywise. The key insight is that if $\omega$ is a uniform random vector on the complex sphere of radius $\sqrt{n}$ , then

$\mathscr{R}(\omega) := \twobyone{\mathfrak{R}(\omega)}{\mathfrak{I}(\omega)}\in\real^{2n} \quad \text{is a uniform random vector on the real sphere of radius $\sqrt{n}$}.$

We’ve converted the complex vector $\omega$ into a real vector $\mathscr{R}(\omega)$ .

Now, we need to convert the complex matrix $A$ into a real matrix $\mathscr{R}(A)$ . To do this, recall that one way of representing complex numbers is by $2\times 2$ matrices:

$a + bi \iff \twobytwo{a}{-b}{b}{a}.$

Using this correspondence addition and multiplication of complex numbers can be carried by addition and multiplication of the corresponding matrices.

To convert complex matrices to real matrices, we use a matrix-version of the same representation:

$\mathscr{R}(A) = \twobytwo{\mathfrak{R}(A)}{-\mathfrak{I}(A)}{\mathfrak{I}(A)}{\mathfrak{R}(A)}.$

One can check that addition and multiplication of complex matrices can be carried out by addition and multiplication of the corresponding “realified” matrices, i.e.,

$\mathscr{R}(A + B) = \mathscr{R}(A) + \mathscr{R}(B), \quad \mathscr{R}(A\cdot B) = \mathscr{R}(A) \cdot \mathscr{R}(B)$

holds for all complex matrices $A$ and $B$ .

We’ve now converted complex matrix $A$ and vector $\omega$ into real matrix $\mathscr{R}(A)$ and vector $\mathscr{R}(\omega)$ . Let’s compare $\omega^*A\omega$ to $\mathscr{R}(\omega)^\top\mathscr{R}(A)\mathscr{R}(\omega)$ . A short calculation reveals

$\omega^*A\omega = \mathscr{R}(\omega)^\top \mathscr{R}(A)\mathscr{R}(\omega) .$

Since $\mathscr{R}(\omega)$ is a uniform random vector on the sphere of radius $\sqrt{n}$ , $\sqrt{2}\cdot \mathscr{R}(\omega)$ is a uniform random vector on the sphere of radius $\sqrt{2n}$ . Thus, by the variance formula for the real sphere, we get

$\Var(\omega^*A\omega) = \Var[(\sqrt{2}\mathscr{R}(\omega))^\top (\mathscr{R}(A)/2)(\sqrt{2}\mathscr{R}(\omega) )] = \frac{4n}{2n+2} \left[ \|\mathscr{R}(A)/2\|_{\rm F}^2 - \frac{1}{8n}(\tr\mathscr{R}(A))^2 \right].$

A short calculation verifies that $\tr \mathscr{R}(A) = 2\tr A$ and $\norm{\mathscr{R}(A)}_{\rm F}^2 = 2\|A\|_{\rm F}^2$ . Plugging this in, we obtain

$\Var(\omega^*A\omega)= \frac{n}{n+1} \left[ \|A\|_{\rm F}^2 - \frac{1}{n}(\tr A)^2 \right].$

Uniform on the Complex Sphere: Derivation 2 by Haar Integration

The proof by reduction to the real case requires some cumbersome calculations and requires that we have already computed the variance in the real case by some other means. The method of Haar integration is more slick, but it requires some pretty high-power machinery. Haar integration may be a little bit overkill for this problem, but this technique is worth learning as it can handle some truly nasty expected value computations that appear, for example, in quantum information.

We seek to compute

$\mathbb{E} [(\omega^*A \omega)^2].$

The first trick will be to write this expession using a single matrix trace using the tensor (Kronecker) product $\otimes$ . For those unfamiliar with the tensor product, the main properties we will be using are

(6) $(A\otimes B) (C\otimes D) = (AB) \otimes (CD), \quad \tr(A\otimes B) = \tr A \cdot \tr B.$

We saw in the proof of unbiasedness that

$\omega^* A \omega = \tr (\omega^*A\omega) = \tr (A \omega\omega^*).$

Therefore, by (6),

$(\omega^*A\omega)^2 = (\tr [A \omega\omega^*])^2 = \tr [A\omega\omega^* \otimes A\omega\omega^*] = \tr [(A\otimes A) (\omega\omega^* \otimes \omega\omega^*)].$

Thus, to evaluate $\mathbb{E}[(\omega^*A\omega)^2]$ , it will be sufficient to evaluate $\mathbb{E}[\omega\omega^* \otimes \omega\omega^*]$ . Forunately, there is a useful formula for these expectation provided by a field of mathematics known as representation theory (see Lemma 1 in this paper):

$\mathbb{E}[ \omega\omega^* \otimes \omega\omega^*] = \frac{2n}{n+1} \operatorname{Proj}_{\operatorname{Sym}^2(\complex^n)}.$

Here, $\operatorname{Proj}_{\operatorname{Sym}^2(\complex^n)}$ is the orthogonal projection onto the space of symmetric two-tensors $\operatorname{Sym}^2(\complex^n) = \operatorname{span} \{ v \otimes v : v \in \complex^n \}$ . Therefore, we have that

$\mathbb{E}[(\omega^*A\omega)^2] = \tr [(A\otimes A) \mathbb{E}(\omega\omega^* \otimes \omega\omega^*)] = \frac{2n}{n+1} \tr [(A\otimes A) \operatorname{Proj}_{\operatorname{Sym}^2(\complex^n)}].$

To evalute the trace on the right-hand side of this equation, there is another formula (see Lemma 6 in this paper):

$\tr \left[(A\otimes B) \operatorname{Proj}_{\operatorname{Sym}^2(\complex^n)}\right] = \frac{1}{2} \left( \tr(AB) + \tr A \cdot \tr B \right).$

Therefore, we conclude

$\begin{align*}\Var(\omega^* A \omega) &= \mathbb{E}[(\omega^*A\omega)^2] - (\mathbb{E}[\omega^*A\omega])^2 \\&= \frac{2n}{n+1}\tr [(A\otimes A) \operatorname{Proj}_{\operatorname{Sym}^2(\complex^n)}] - (\tr A)^2 \\&= \frac{n}{n+1}\left[ \tr A^2 + (\tr A)^2 \right] - (\tr A)^2 \\&= \frac{n}{n+1}\left[ \left\|A\right\|_{\rm F}^2 - \frac{1}{n} (\tr A)^2 \right].\end{align*}$

Proof of Optimality Properties

In this section, we provide proofs of the two optimality properties.

Optimality: Independent Vectors with Independent Coordinates

Assume $A$ is real and symmetric and suppose that $\omega$ is isotropic (2) with independent coordinates. The isotropy condition

$\mathbb{E}[\omega\omega^\top] = I$

implies that $\mathbb{E}[\omega_i\omega_j] = \delta_{ij}$ , where $\delta$ is the Kronecker symbol. Using this fact, we compute the second moment:

$\begin{align*}\mathbb{E}[ (\omega^*A \omega)^2] &= \mathbb{E}\left[ \left( \sum_{i=1}^n A_{ii} \omega_i^2 +2 \sum_{i<j} A_{ij}\omega_i\omega_j) \right)^2\right] \\&= \sum_{i=1}^n A_{ii}^2 \mathbb{E}[\omega_i^4] + \sum_{i<j} (2A_{ii}A_{jj}+4A_{ij}^2) \mathbb{E}[\omega_i^2]\mathbb{E}[\omega_j^2] \\&= \sum_{i=1}^n A_{ii}^2 \mathbb{E}[\omega_i^4] + \sum_{i<j} (2A_{ii}A_{jj}+4A_{ij}^2) .\end{align*}$

Thus

$\Var(\omega^*A\omega) = \mathbb{E}[ (\omega^*A \omega)^2] - (\mathbb{E}[\omega^* A \omega])^2 = \sum_{i=1}^n A_{ii}^2 (\mathbb{E}[|\omega_i|^4]-1) + 4\sum_{i<j} A_{ij}^2.$

The variance is minimized by choosing $\omega$ with $\mathbb{E} \omega_i^4$ as small as possible. Since $\mathbb{E} \omega_i^2 = 1$ , the smallest possible value for $\mathbb{E} \omega_i^4$ is $\mathbb{E} \omega_i^4 = 1$ , which is obtained by populating $\omega$ with random signs.

Optimality: Independent Vectors

This result appears to have first been proven by Richard Kueng in unpublished work. We use an argument suggested to me by Robert J. Webber.

Assume $\mathscr{A}$ is a class of $\field$ -Hermitian matrices closed under $\field$ -unitary similarity transformations and that $\omega$ is an isotropic random vector (2). Decompose the test vector as

$\omega = a \cdot s \quad \text{for} \quad a \in [0,+\infty), \: s \in\{x\in \field^n : x^*x = n \}.$

First, we shall show that the variance is reduced by replacing $s$ with a vector $t$ drawn uniformly from the sphere

(7) $\sup_{A\in\mathscr{A}} \Var(\tilde{\omega}^*A\tilde{\omega}) \le \sup_{A\in\mathscr{A}} \Var(\omega^*A\omega$

where

(8) $\tilde{\omega} = a\cdot t \quad \text{and}\quad t\sim \text{Uniform} \{ x \in \field^n :x^*x = n \} \quad \text{is independent of $a$}.$

Note that such a $t$ can be generated as $t = Qs$ for a uniformly random $\field$ -unitary matrix $Q$ . Therefore, we have

$\begin{align*}\sup_{A\in\mathscr{A}} \Var(\tilde{\omega}^*A\tilde{\omega})&= \sup_{A\in\mathscr{A}} \left[\mathbb{E}[(\tilde{\omega}^*A\tilde{\omega})^2] - (\tr A)^2\right]\\&= \sup_{A\in\mathscr{A}} \left[\mathbb{E}[a^2 \cdot s^*(Q^*AQ)s] - (\tr (Q^*AQ))^2\right].\end{align*}$

Now apply Jensen’s inequality only over the randomness in $Q$ to obtain

$\begin{align*}\sup_{A\in\mathscr{A}} \Var(\tilde{\omega}^*A\tilde{\omega})&= \sup_{A\in\mathscr{A}} \left[\mathbb{E}[a^2 \cdot s^*(Q^*AQ)s] - (\tr (Q^*AQ))^2\right] \\&\le \mathbb{E}_Q \sup_{A\in\mathscr{A}} \left[\mathbb{E}_{a,s}[a^2 \cdot s^*(Q^*AQ)s] - (\tr (Q^*AQ))^2\right].\end{align*}$

Finally, note that since $\mathscr{A}$ is closed under $\field$ -unitary similarity transformations, the supremum over $Q^*AQ$ for $A \in \mathscr{A}$ is the same as the supremum of $A \in \mathscr{A}$ , so we obtain

$\begin{align*}\sup_{A\in\mathscr{A}} \Var(\tilde{\omega}^*A\tilde{\omega})&\le \mathbb{E}_Q \sup_{A\in\mathscr{A}} \left[\mathbb{E}_{a,s}[a^2 \cdot s^*(Q^*AQ)s] - (\tr (Q^*AQ))^2\right] \\&= \mathbb{E}_Q \sup_{A\in\mathscr{A}} \left[\mathbb{E}_{a,s}[a^2 \cdot s^*As] - (\tr A)^2\right] \\&= \sup_{A\in\mathscr{A}} \Var(\omega^*A\omega).\end{align*}$

We have successfully proven (7). This argument is a specialized version of a far more general result which appears as Proposition 4.1 in this paper.

Next, we shall prove

(9) $\sup_{A\in\mathscr{A}} \Var(t^*At) \le \sup_{A\in\mathscr{A}} \Var(\tilde{\omega}^*A\tilde{\omega}),$

where $t$ is still defined as in (8). Indeed, using the chain rule for variance, we obtain

$\begin{align*}\Var(\tilde{\omega}^*A\tilde{\omega})&= \Var(a^2\cdot t^*At) \\&= \mathbb{E}[\Var(a^2\cdot t^* A t \mid a)] + \Var(\mathbb{E}[a^2\cdot t^* A t \mid a]) \\&= \mathbb{E}[a^4]\Var(t^* A t )+ (\tr A)^2\Var(a^2) \\&\ge \mathbb{E}[a^4]\Var(t^* A t ).\end{align*}$

Here, we have used that $t$ is uniform on the sphere and thus $\mathbb{E}[t^*At] = \tr A$ . By definition, $a$ is the length of $\omega$ divided by $\sqrt{n}$ . Therefore,

$\mathbb{E}[a^2] = \frac{1}{n}\mathbb{E}[\omega^*\omega] = \frac{1}{n} \mathbb{E}[\tr (\omega\omega^*)] = \frac{1}{n} \tr (\mathbb{E}[\omega\omega^*]) = \frac{\tr I}{n} = 1.$

Therefore, by Jensen’s inequality,

$\mathbb{E}[a^4] = \mathbb{E}[(a^2)^2] \ge (\mathbb{E}[a^2])^2 = 1.$

Thus

$\Var(\tilde{\omega}^*A\tilde{\omega}) \ge \mathbb{E}[a^4]\Var(t^* A t ) \ge \Var(t^*At) \quad \text{for every }A,$

which proves (9).

Chebyshev Polynomials

August 13, 2022 by Ethan N. Epperly 2 Comments

This post is co-written by my brother, Aidan Epperly, for the second Summer of Math Exposition (SoME2).

Let’s start with a classical problem: connect-the-dots. As we know from geometry, any two points in the plane are connected by one and only one straight line:

But what if we have more than two points? How should we connect them? One natural way is by parabola. Any three points (with distinct $x$ coordinates) are connected by one and only one parabola $ax^2+bx+c$ :

And we can keep extending this. Any $n+1$ points¹ (with distinct $x$ coordinates) are connected by a unique degree- $n$ polynomial $a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x+a_0$ :

This game of connect-the-dots with polynomials is known more formally as polynomial interpolation. We can use polynomial interpolation to approximate functions. For instance, we can approximate the function $\sin(x)$ on the interval $[-\pi,\pi]$ to visually near-perfect accuracy by connecting the dots between seven points $(-\pi,\sin(-\pi)), (-\tfrac{2}{3}\pi,\sin(-\tfrac{2}{3}\pi)),\ldots ,(\pi,\sin(\pi))$ :

But something very peculiar happens when we try and apply this trick to the specially chosen function $R(x) = 1/(1+25x^2)$ on the interval $[-1,1]$ :

Unlike $\sin(x)$ , the polynomial interpolant for $R(x)$ is terrible! What’s going on? Why doesn’t polynomial interpolation work here? Can we fix it? The answer to the last question is yes and the solution is Chebyshev polynomials.

Reverse-Engineering Chebyshev

The failure of polynomial interpolation for $R(x)$ is known as Runge’s phenomenon after Carl Runge who discovered this curious behavior in 1901. The function $R(x)$ is called the Runge function. Our goal is to find a fix for polynomial interpolation which crushes the Runge phenomenon, allowing us to reliably approximate every sensible² function with polynomial interpolation.

Carl Runge

Let’s put on our thinking caps and see if we can discover the fix for ourselves. In order to discover a fix, we must first identify the problem. Observe that the polynomial interpolant is fine near the center of the interval; it only fails near the boundary.

This leads us to a guess for what the problem might be; maybe we need more interpolation points near the boundaries of the interval. Indeed, tipping our hand a little bit, this turns out to be the case. For instance, connecting the dots for the following set of “mystery points” clustered at the endpoints works just fine:

Let’s experiment a little and see if we can discover a nice set of interpolation points, which we will call $x_0,x_1,\ldots,x_n$ , like this for ourselves. We’ll assume the interpolation points are given by a function $x_j = g(j/n)$ so we can form the polynomial interpolant for any desired polynomial degree $n$ .³ For instance, if we pick $g(t) = 2t^2-1$ , the points look like this:

Equally spaced points $j/n$ (shown on vertical axis) give rise to
non-equally spaced points $g(j/n)$ (shown on horizontal axis)

How should we pick the function $g(\cdot)$ ? First observe that, even for the Runge function, equally spaced interpolation points are fine near the center of the interval. We thus have at least two conditions for our desired $n+1$ interpolation points:

The interior points should maintain their spacing of roughly $2/n$ .
The points must cluster near both boundaries.

As a first attempt let’s divide the interval into thirds and halve the spacing of points except in the middle third. This leads to the function

$\begin{equation*}g(x) = \begin{cases} -1+x, & 0\le x < \tfrac{1}{3}, \\-\tfrac{2}{3} + 4\left( x-\tfrac{1}{3}\right), & \tfrac{1}{3} \le x < \tfrac{2}{3}, \\\tfrac{2}{3} + \left( x-\tfrac{2}{3}\right), & \tfrac{2}{3} \le x \le 1.\end{cases}\end{equation*}$

These interpolation points initially seem promising, even successfully approximating the Runge function itself.

Unfortunately, this set of points fails when we consider other functions. For instance, if we use the Runge-like function $S(x) = 1/(1+900x^2)$ , we see that these interpolation points now lead to a failure to approximate the function at the middle of the interval, even if we use a lot of interpolation points!

Maybe the reason this set of interpolation points didn’t work is that the points are too close at the endpoints. Or maybe we should have divided the interval as quarter–half–quarter rather than thirds. There are lots of variations of this strategy for choosing points to explore and all of them eventually lead to failure on some Runge-flavored example. We need a fundamentally different strategy then making the points $a$ times closer within distance $b$ of the endpoints.

Let’s try a different approach. The closeness of the points at the endpoints is determined by the slope of the function $g$ at $0$ and $1$ . The smaller that $|g'(0)|$ and $|g'(1)|$ are, the more clustered the points will be. For instance,

$\begin{equation*}g'(0) = g'(1) = 2 \quad \text{for equally spaced points}.\end{equation*}$

When we halved the distance between points, we instead had

$\begin{equation*}g'(0) = g'(1) = 1 \quad \text{when points at ends were twice as close together}.\end{equation*}$

So if we want the points to be much more clustered together, it is natural to require

$\begin{equation*}g'(0) = g'(1) = 0. \quad \text{(new requirement)}\end{equation*}$

It also makes sense for the function $g(\cdot)$ to cluster points equally near both endpoints, since we see no reason to preference one end over the other. Collecting together all the properties we want the function $g(\cdot)$ to have, we get the following list:

$g$ spans the whole range $[-1,1]$ ,
$g'(0) = g'(1) = 0$ , and
$g$ is symmetric about $1/2$ , $g(1/2+x) = -g(1/2-x)$ .

Mentally scrolling through our Rolodex of friendly functions, a natural one that might come to mind meeting these three criteria is the cosine function, specifically $g(t) = \cos(\pi t)$ . This function yields points which are more clustered at the endpoints:

The points

$\begin{equation*}x_j = \cos\left(\frac{j\pi}{n}\right)\end{equation*}$

we guessed our way into are known as the Chebyshev points.⁴ The Chebyshev points prove themselves perfectly fine for the Runge function:

As we saw earlier, success on the Runge function alone is not enough to declare victory for the polynomial interpolation problem. However, in this case, there are no other bad examples left to find. For any nice function with no jumps, polynomial interpolation through the Chebyshev points works excellently.⁵

Why the Chebyshev Points?

We’ve guessed our way into a solution to the polynomial interpolation problem, but we still really don’t know what’s going on here. Why are the Chebyshev points much better at polynomial interpolation than equally spaced ones?

Now that we know that the Chebyshev points are a right answer to the interpolation problem,⁶ let’s try and reverse engineer a principled reason for why we would expect them to be effective for this problem. To do this, we ask:

What is special about the cosine function?

From high school trigonometry, we know that $\cos \theta$ gives the $x$ coordinate of a point $\theta$ radians along the unit circle. This means that the Chebyshev points are the $x$ coordinates of equally spaced points on the unit circle (specifically the top half of the unit circle $0\le \theta\le \pi$ ).

Chebyshev points are the $x$ coordinates of equally spaced points on the unit circle.

This raises the question:

What does the interpolating polynomial $p(x)$ look like as a function of the angle $\theta$ ?

To convert between $x$ and $\theta$ we simply plug in $x = \cos \theta$ to $p(x)$ :

$\begin{equation*}p^\circ(\theta) = p(\cos \theta) = a_n \cos^n \theta + a_{n-1} \cos^{n-1} \theta + \cdots + a_0.\end{equation*}$

This new function depending on $\theta$ , which we can call $p^\circ(\theta)$ , is a polynomial in the variable $\cos \theta$ . Powers of cosines are not something we encounter every day, so it makes sense to try and simplify things using some trig identities. Here are the first couple powers of cosines:

$\begin{gather*}\cos^2 \theta = \frac{1}{2} + \frac{1}{2} \cos (2\theta), \\\cos^3 \theta = \frac{3}{4}\cos \theta + \frac{1}{4} \cos (3\theta), \\\cos^4 \theta = \frac{3}{8}+ \frac{1}{2} \cos (2\theta) + \frac{1}{8} \cos (4\theta),\\\vdots\end{gather*}$

A pattern has appeared! The powers $\cos^k \theta$ always take the form⁷

$\begin{equation*}\cos^k \theta = \textnormal{(some number)} \cdot \cos(k\theta) + \textnormal{(some number)} \cdot \cos((k-2)\theta) + \cdots .\end{equation*}$

The significance of this finding is that, by plugging in each of these formulas for $\cos^k \theta$ , we see that our polynomial $p(x)$ in the variable $x$ has morphed into a Fourier cosine series in the variable $\theta$ :

$\begin{equation*}p^\circ(\theta) = b_n \cos(n\theta) + b_{n-1} \cos((n-1)\theta) + \cdots + b_1 \cos \theta + b_0.\end{equation*}$

For anyone unfamiliar with Fourier series, we highly encourage the 3Blue1Brown video on the subject, which explains why Fourier series are both mathematically beautiful and practically useful. The basic idea is that almost any function can be expressed as a combination of waves (that is, sines and cosines) with different frequencies.⁸ In our case, this formula tells us that $p^\circ(\theta)$ is equal to $b_0$ units of frequency $0$ , plus $b_1$ units of frequency $1$ , all the way up to $b_n$ units of frequency $n$ . Different types of Fourier series are appropriate in different contexts. Since our Fourier series only possesses cosines, we call it a Fourier cosine series.

We’ve discovered something incredibly cool:

Polynomial interpolation through the Chebyshev points is equivalent to finding a Fourier cosine series for equally spaced angles $\theta$ .

We’ve arrived at an answer to why the Chebyshev points work well for polynomial interpolation.

Polynomial interpolation through the Chebyshev points is effective because Fourier cosine series through equally spaced angles $\theta$ is effective.

Of course, this explanation just raises the further question: Why do Fourier cosine series give effective interpolants through equally spaced angles $\theta$ ? This question has a natural answer as well, involving the convergence theory and aliasing formula (see Section 3 of this paper) for Fourier series. We’ll leave the details to the interested reader for investigation. The success of Fourier cosines series in interpolating equally spaced data is a fundamental observation that underlies the field of digital signal processing. Interpolation through the Chebyshev points effectively hijacks this useful fact and applies it to the seemingly unrelated problem of polynomial interpolation.

Another question this explanation raises is the precise meaning of “effective”. Just how good are polynomial interpolants through the Chebyshev points at approximating functions? As is discussed at length in another post on this blog, the degree to which a function can be effectively approximated is tied to how smooth or rough it is. Chebyshev interpolants approximate nice analytic functions like $\sin(x)$ or $1/(1+25x^2)$ with exponentially small errors in the number of interpolation points used. By contrast, functions with kinks like $|x|$ are approximated with errors which decay much more slowly. See theorems 2 and 3 on this webpage for more details.

Chebyshev Polynomials

We’ve now discovered a set of points, the Chebyshev points, through which polynomial interpolation works well. But how should we actually compute the interpolating polynomial

$\begin{equation*}p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_0?\end{equation*}$

Again, it will be helpful to draw on the connection to Fourier series. Computations with Fourier series are highly accurate and can be made lightning fast using the fast Fourier transform algorithm. By comparison, directly computing with a polynomial $p(x)$ through its coefficients $a_n,a_{n-1},\ldots,a_0$ is a computational nightmare.

In the variable $\theta$ , the interpolant takes the form

$\begin{equation*}p^\circ(\theta) = b_n \cos(n\theta) + b_{n-1} \cos((n-1)\theta) + \cdots + b_1 \cos \theta + b_0.\end{equation*}$

To convert back to $x = \cos \theta$ , we use the inverse function⁹ $\theta = \arccos x$ to obtain:

$\begin{equation*}p(x) = b_n \cos(n\arccos(x)) + \cdots + b_1 \cos(\arccos x) + b_0\end{equation*}$

This is a striking formula. Given all of the trigonometric functions, it’s not even obvious that $p(x)$ is a polynomial (it is)!

Despite its seeming peculiarity, this is a very powerful way of representing the polynomial $p(x)$ . Rather than expressing $p(x)$ using monomials $1,x,x^2,\ldots$ , we’ve instead written $p(x)$ as a combination of more exotic polynomials

$\begin{equation*}T_k(x) = \cos(k \arccos x) \quad \text{for $k=0,1,2,\ldots n$}.\end{equation*}$

The polynomials $T_0(x),T_1(x),T_2(x),\ldots$ are known as the Chebyshev polynomials,¹⁰ named after Pafnuty Chebyshev who studied the polynomials intensely.¹¹

Pafnuty Chebyshev

Writing out the first few Chebyshev polynomials shows they are indeed polynomials:

$\begin{gather*}T_0(x) = 1, \\T_1(x) = x, \\T_2(x) = 2x^2 - 1, \\ T_3(x) = 4x^3 - 3x, \\\vdots \end{gather*}$

The first four Chebyshev polynomials

To confirm that this pattern does continue, we can use trig identities to derive¹² the following recurrence relation for the Chebyshev polynomials:

$\begin{equation*}T_{k+1}(x) = 2x T_k(x) - T_{k-1}(x).\end{equation*}$

Since $T_0(x) = 1$ and $T_1(x) = x$ are both polynomials, every Chebyshev polynomial is as well.

We’ve arrived at the following amazing conclusion:

Under the change of variables $x = \cos \theta$ , the Fourier cosine series
$p^\circ(\theta) = b_n\cos(n\theta) + \cdots + b_1\cos\theta + b_0$
becomes the combination of Chebyshev polynomials
$p(x) = b_nT_n(x) + \cdots + b_1 T_1(x) + b_0.$

This simple and powerful observations allows us to apply the incredible speed and accuracy of Fourier series to polynomial interpolation.

Beyond being a neat idea with some nice mathematics, this connection between Fourier series and Chebyshev polynomials is a powerful tool for solving computational problems. Once we’ve accurately approximated a function by a polynomial interpolant, many quantities of interest (derivatives, integrals, zeros) become easy to compute—after all, we just have to compute them for a polynomial! We can also use Chebyshev polynomials to solve differential equations with much faster rates of convergence than other methods. Because of the connection to Fourier series, all of these computations can be done to high accuracy and blazingly fast via the fast Fourier transform, as is done in the software package Chebfun.

The Chebyshev polynomials have an array of amazing properties and they appear all over mathematics and its applications in other fields. Indeed, we have only scratched the surface of the surface. Many questions remain:

What is the connection between the Chebyshev points and the Chebyshev polynomials?
The cosine functions $1,\cos \theta,\cos(2\theta),\ldots$ are orthogonal to each other; are the Chebyshev polynomials?
Are the Chebyshev points the best points for polynomial interpolation? What does “best” even mean in this context?
Every “nice” even periodic function has an infinite Fourier cosine series which converges to it. Is there a Chebyshev analog? Is there a relation between the infinite Chebyshev series and the (finite) interpolating polynomial through the Chebyshev points?

All of these questions have beautiful and fairly simple answers. The book Approximation Theory and Approximation Practice is a wonderfully written book that answers all of these questions in its first six chapters, which are freely available on the author’s website. We recommend the book highly to the curious reader.

TL;DR: To get an accurate polynomial approximation, interpolate through the Chebyshev points.
To compute the resulting polynomial, change variables to $\theta = \arccos x$ , compute the Fourier cosine series interpolant, and obtain your polynomial interpolant as a combination of Chebyshev polynomials.

Don’t Solve the Normal Equations

July 26, 2022 by Ethan N. Epperly 2 Comments

The (ordinary) linear least squares problem is as follows: given an $m\times n$ matrix $A$ and a vector $b$ of length $m$ , find the vector $x$ such that $Ax$ is as close to $b$ as possible, when measured using the two-norm $\| \cdot \|$ . That is, we seek to

(1) $\begin{equation*} \mbox{find } x \in\mathbb{R}^n \mbox{ such that }\| b - Ax \|^2 = \sum_{i=1}^m \left(b_i - \sum_{j=1}^n A_{ij} x_j \right)^2 \mbox{ is minimized}. \end{equation*}$

From this equation, the name “least squares” is self-explanatory: we seek $x$ which minimizes the sum of the squared discrepancies between the entries of $b$ and $Ax$ .

The least squares problem is ubiquitous in science, engineering, mathematics, and statistics. If we think of each row $a_i$ of $A$ as an input and its corresponding entry $b_i$ of $b$ as an output, then the solution $x$ to the least squares model gives the coefficients of a linear model for the input–output relationship. Given a new previously unseen input $a_{\rm new}$ , our model predicts the output $b_{\rm new}$ is approximately $b_{\rm new} \approx a_{\rm new}^\top x = \sum_{i=1}^n x_i (a_{\rm new})_i$ . The vector $x$ consists of coefficients for this linear model. The least squares solution satisfies the property that the average squared difference between the output $b_i$ and the prediction $a_i^\top x$ is as small as it could possibly be for all choices of coefficient vectors $x$ .

How do we solve the least squares problem? A classical solution approach, ubiquitous in textbooks, is to solve a system of linear equations known as the normal equations. The normal equations associated with the least squares problem (1) are given by

(2) $\begin{equation*} A^\top A \,x = A^\top b. \end{equation*}$

This system of equations always has a solution. If $A$ has full column-rank, then $A^\top A$ is invertible and the unique least squares solution to (1) is given by $(A^\top A)^{-1} A^\top b$ . We assume that $A$ has full column-rankQ for the rest of this discussion. To solve the normal equations in software, we compute $A^\top A$ and $A^\top b$ and solve (2) using a linear solver like MATLAB’s “\”.¹ (As is generally true in matrix computations, it is almost never a good idea to explicitly form the inverse of the matrix $A^\top A$ , or indeed any matrix.) We also can solve the normal equations using an iterative method like (preconditioned) conjugate gradient.

The purpose of the article is to advocate against the use of the normal equations for solving the least squares problems, at least in most cases. So what’s wrong with the normal equations? The problem is not that the normal equations aren’t mathematically correct. Instead, the problem is that the normal equations often lead to poor accuracy for the least squares solution using computer arithmetic.

Most of the time when using computers, we store real numbers as floating point numbers.² At a coarse level, the right model to have in your head is that real numbers on a computer are stored in scientific notation with only 16 decimal digits after the decimal point.³ When two numbers are added, subtracted, multiplied, and divided, the answer is computed and then rounded to 16 decimal digits; any extra digits of information are thrown away. Thus, the result of our arithmetic on a computer is the true answer to the arithmetic problem plus a small rounding error. These rounding errors are small individually, but solving an even modestly sized linear algebra problem requires thousands of such operations. Making sure many small errors don’t pile up into a big error is part of the subtle art of numerical computation.

To make a gross simplification, if one solves a system of linear equations $Mx = c$ on a computer using a well-designed piece of software, one obtains an approximate solution $\hat{x}$ which is, after accounting for the accumulation of rounding errors, close to $x$ . But just how close the computed solution $\hat{x}$ and the true solution $x$ are depends on how “nice” the matrix $M$ is. The “niceness” of a matrix $M$ is quantified by a quantity known as the condition number of $M$ , which we denote $\kappa(M)$ .⁴ As a rough rule of thumb, the relative error between $x$ and $\hat{x}$ is roughly bounded as

(3) $\begin{equation*} \frac{\| \hat{x} - x \|}{\|x\|} \lessapprox \kappa(M)\times 10^{-16}. \end{equation*}$

The “ $10^{-16}$ corresponds to the fact we have roughly 16 decimal digits of accuracy in double precision floating point arithmetic. Thus, if the condition number of $M$ is roughly $10^{10}$ , then we should expect around $6$ digits of accuracy in our computed solution.

The accuracy of the least squares problem is governed by its own condition number $\kappa(A)$ . We would hope that we can solve the least squares problem with an accuracy like the rule-of-thumb error bound (3) we had for linear systems of equations, namely a bound like $\|\hat{x} - x\|/\|x\| \lessapprox \kappa(A)\times 10^{-16}$ . But this is not the kind of accuracy we get for the least squares problem when we solve it using the normal equations. Instead, we get accuracy like

(4) $\begin{equation*} \frac{\| \hat{x} - x \|}{\|x\|} \lessapprox \left(\kappa(A)\right)^2\times 10^{-16}. \end{equation*}$

By solving the normal equations we effectively square the condition number! Perhaps this is not surprising as the normal equations also more-or-less square the matrix $A$ by computing $A^\top A$ . This squared condition number drastically effects the accuracy of the computed solution. If the condition number of $A$ is $10^{8}$ , then the normal equations give us absolute nonsense for $\hat{x}$ ; we expect to get no digits of the answer $x$ correct. Contrast this to above, where we were able to get $6$ correct digits in the solution to $Mx = c$ despite the condition number of $M$ being $100$ times larger than $A$ !

All of this would be just a sad fact of life for the least squares problem if the normal equations and their poor accuracy properties were the best we could do for the least squares problem. But we can do better! One can solve linear least squares problems by computing a so-called QR factorization of the matrix $A$ .⁵ Without going into details, the upshot is that the least squares solution by a well-designed⁶ QR factorization requires a similar amount of time to solving the normal equations and has dramatically improved accuracy properties, achieving the desirable rule-of-thumb behavior⁷

(5) $\begin{equation*} \frac{\| \hat{x} - x \|}{\|x\|} \lessapprox \kappa(A)\times 10^{-16}. \end{equation*}$

I have not described how the QR factorization is accurately computed nor how to use the QR factorization to solve least squares problems nor even what the QR factorization is. All of these topics are explained excellently by the standard textbooks in this area, as well as by publicly available resources like Wikipedia. There’s much more that can be said about the many benefits of solving the least squares problem with the QR factorization,⁸ but in the interest of brevity let me just say this: TL;DR when presented in the wild with a least squares problem, the solution method one should default to is one based on a well-implemented QR factorization, not solving the normal equations.

Suppose for whatever reason we don’t have a high quality QR factorization algorithm at our disposal. Must we then resort to the normal equations? Even in this case, there is a way we can reduce the problem of solving a least squares problems to a linear system of equations without squaring the condition number! (For those interested, to do this, we recognize the normal equations as a Schur complement of a somewhat larger system of linear equations and then solve that. See Eq. (7) in this post for more discussion of this approach.)

The title of this post Don’t Solve the Normal Equations is deliberately overstated. There are times when solving the normal equations is appropriate. If $A$ is well-conditioned with a small condition number, squaring the condition number might not be that bad. If the matrix $A$ is too large to store in memory, one might want to solve the least squares problem using the normal equations and the conjugate gradient method.

However, the dramatically reduced accuracy of solving the normal equations should disqualify the approach from being the de-facto way of solving least squares problems. Unless you have good reason to think otherwise, when you see $A^\top A$ , solve a different way.

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.

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.

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.