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Randomized matrix-free trace and log-determinant estimators

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Abstract

We present randomized algorithms for estimating the trace and determinant of Hermitian positive semi-definite matrices. The algorithms are based on subspace iteration, and access the matrix only through matrix vector products. We analyse the error due to randomization, for starting guesses whose elements are Gaussian or Rademacher random variables. The analysis is cleanly separated into a structural (deterministic) part followed by a probabilistic part. Our absolute bounds for the expectation and concentration of the estimators are non-asymptotic and informative even for matrices of low dimension. For the trace estimators, we also present asymptotic bounds on the number of samples (columns of the starting guess) required to achieve a user-specified relative error. Numerical experiments illustrate the performance of the estimators and the tightness of the bounds on low-dimensional matrices, and on a challenging application in uncertainty quantification arising from Bayesian optimal experimental design.

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Notes

  1. The square matrix \(\mathbf {I}\) denotes the identity, with ones on the diagonal and zeros everywhere else.

  2. The superscript \(\dagger \) denotes the Moore–Penrose inverse.

  3. independent and identically distributed.

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Authors and Affiliations

Authors

Corresponding author

Correspondence to Arvind K. Saibaba.

Additional information

The third author acknowledges the support from the XDATA Program of the Defense Advanced Research Projects Agency (DARPA), administered through Air Force Research Laboratory contract FA8750-12-C-0323 FA8750-12-C-0323.

Appendices

Appendix 1: Gaussian random matrices

In this section, we state a lemma on the pseudo-inverse of a rectangular Gaussian random matrix, and use this result to prove both parts of Lemma 4.

1.1 Pseudo-inverse of a Gaussian random matrix

We state a result on the large deviation bound of the pseudo-inverse of a Gaussian random matrix [19, Proposition 10.4].

Lemma 6

Let \(\mathbf {G}\in \mathbb {R}^{k\times (k+p)}\) be a random Gaussian matrix and let \(p\ge 2\). For all \(t \ge 1\),

$$\begin{aligned} \mathbb {P}\left[ \, \Vert \mathbf {G}^\dagger \Vert _2 \ge \frac{e\sqrt{k+p}}{p+1}\cdot t \right] \le t^{-(p+1)}. \end{aligned}$$
(23)

1.2 Proof of Lemma 4

Proof

From [45, Corollary 5.35] we have

$$\begin{aligned} \mathbb {P}\left[ \, \Vert \mathbf {G}_2\Vert _2 > \sqrt{n-k} + \sqrt{k+p} + t \right] \le \exp (-t^2/2). \end{aligned}$$

Recall from (3) \(\mu = \sqrt{n-k} + \sqrt{k+p}\). From the law of the unconscious statistician [16, Proposition S4.2],

$$\begin{aligned} \mathbb {E}\left[ \, \Vert \mathbf {G}_2\Vert _2^2 \right] =&\> \int _0^\infty 2t \mathbb {P}\left[ \, \Vert \mathbf {G}_2\Vert _2> t \right] dt\\ \le&\> \int _0^{\mu } 2t dt + \int _{\mu }^\infty 2t \mathbb {P}\left[ \, \Vert \mathbf {G}_2\Vert _2> t \right] dt \\ \le&\> \mu ^2 + \int _{0}^\infty 2(u+\mu ) \exp (-u^2/2)du = \> \mu ^2 + 2\left( 1+ \mu \sqrt{\frac{\pi }{2}}\right) . \end{aligned}$$

This concludes the proof for (16).

Next consider (17). Using Lemma 6, we have for \(t > 0\)

$$\begin{aligned} \mathbb {P}\left[ \, \Vert \mathbf {G}_1^\dagger \Vert _2 \ge t \right] \le \> D t^{-(p+1)} \qquad D \equiv \> \frac{1}{\sqrt{2\pi (p+1)}}\left( \frac{e\sqrt{k+p}}{p+1}\right) . \end{aligned}$$
(24)

As before, we have

$$\begin{aligned} \mathbb {E}\left[ \, \Vert \mathbf {G}_1^{\dagger }\Vert _2^2 \right]&= \int _0^\infty 2t \mathbb {P}\left[ \, \Vert \mathbf {G}_1^\dagger \Vert _2> t \right] dt\\&\le \int _0^{\beta } 2t dt + \int _{\beta }^\infty 2t \mathbb {P}\left[ \, \Vert \mathbf {G}_1^\dagger \Vert _2 > t \right] dt\\ & \le \beta ^2 + \int _{\beta }^\infty 2t D t^{-(p+1)} dt = \beta ^2 + 2D\frac{\beta ^{1-p}}{p-1}. \end{aligned}$$

Minimizing w.r.t. \(\beta \), we get \(\beta = (D)^{1/(p+1)}\). Substitute this value for \(\beta \) and simplify.

\(\square \)

Appendix 2: Rademacher random matrices

In this section, we state the matrix Chernoff inequalities [43] and other useful concentration inequalities and use these results to prove Theorem 10.

1.1 Useful concentration inequalities

The proof of Theorem 10 relies on the matrix concentration inequalities developed in [43]. We will need the following result [43, Theorem 5.1.1] in what follows.

Theorem 11

(Matrix Chernoff) Let \(\{\mathbf {X}_k\}\) be finite sequence of independent, random, \(d\times d\) Hermitian matrices. Assume that \(0 \le \lambda _{\min }(\mathbf {X}_k)\) and \(\lambda _{\max }(\mathbf {X}_k) \le L\) for each index k. Let us define

$$\begin{aligned} \mu _{\min } \equiv \lambda _{\min }\left( \sum _k\mathbb {E}\left[ \, \mathbf {X}_k \right] \right) \qquad \mu _{\max } \equiv \lambda _{\max }\left( \sum _k\mathbb {E}\left[ \, \mathbf {X}_k \right] \right) , \end{aligned}$$

and let \(g(x) \equiv e^x(1+x)^{-(1+x)}.\) Then for any \(\epsilon > 0\)

$$\begin{aligned} \mathbb {P}\left[ \, \lambda _{\max }\left( \sum _k \mathbf {X}_k \right) \ge (1+\epsilon )\mu _{\max } \right] \le d g(\epsilon )^{\mu _{\max }/L}, \end{aligned}$$

and for any \(0 \le \epsilon <1 \)

$$\begin{aligned} \mathbb {P}\left[ \, \lambda _{\min }\left( \sum _k \mathbf {X}_k \right) {\le } (1-\epsilon )\mu _{\min } \right] \le d g(-\epsilon )^{\mu _{\min }/L}. \end{aligned}$$

The following result was first proved by Ledoux [25] but we reproduce the statement from [42, Proposition 2.1].

Lemma 7

Suppose \(f:\mathbb {R}^n \rightarrow \mathbb {R}\) is a convex function that satisfies the following Lipschitz bound

$$\begin{aligned} |f(\mathbf {x})-f(\mathbf {y})| \le L \Vert \mathbf {x}-\mathbf {y}\Vert _2 \qquad \text {for all }\>\mathbf {x},\mathbf {y}\in \mathbb {R}^n. \end{aligned}$$

Let \(\mathbf {z}\in \mathbb {R}^n\) be a random vector with entries drawn from an i.i.d. Rademacher distribution. Then, for all \(t \ge 0\),

$$\begin{aligned} \mathbb {P}\left[ \, f(\mathbf {z}) \ge \mathbb {E}\left[ \, f(\mathbf {z}) \right] + Lt \right] \le e^{-t^2/8}. \end{aligned}$$

Lemma 8

Let \(\mathbf {V}\) be a \(n\times r\) matrix with orthonormal columns and let \(n\ge r\). Let \(\mathbf {z}\) be an \(n\times 1\) vector with entries drawn from an i.i.d. Rademacher distribution. Then, for \(0< \delta < 1\),

$$\begin{aligned} \mathbb {P}\left[ \, \Vert \mathbf {V}^*\mathbf {z}\Vert _2 \ge \sqrt{r} + \sqrt{8\log \left( \frac{1}{\delta }\right) } \right] \le \delta . \end{aligned}$$

Proof

Our proof follows the strategy in [42, Lemma 3.3]. Define the function \(f(\mathbf {x}) = \Vert \mathbf {V}^*\mathbf {x}\Vert _2\). We observe that f satisfies the assumptions of Lemma 7, with Lipschitz constant \(L=1\); the latter follows from

$$\begin{aligned} |\Vert \mathbf {V}^*\mathbf {x}\Vert _2-\Vert \mathbf {V}^*\mathbf {y}\Vert _2| \le \Vert \mathbf {V}^*(\mathbf {x}-\mathbf {y})\Vert _2 \le \Vert \mathbf {x}-\mathbf {y}\Vert _2. \end{aligned}$$

Furthermore, using Hölder’s inequality

$$\begin{aligned} \mathbb {E}\left[ \, f(\mathbf {z}) \right] \le [\mathbb {E}\left[ \, f(\mathbf {z})^2 \right] ]^{1/2} = \Vert \mathbf {V}\Vert _F = \sqrt{r}. \end{aligned}$$

Using Lemma 7 with \(t_\delta = \sqrt{8\log \left( 1/\delta \right) }\) we have

$$\begin{aligned} \mathbb {P}\left[ \, f(\mathbf {z}) \ge \sqrt{r} + t_\delta \right] \le \mathbb {P}\left[ \, f(\mathbf {z}) \ge \mathbb {E}\left[ \, f(\mathbf {z}) \right] + t_\delta \right] \le e^{-t_\delta ^2/8} = \delta . \end{aligned}$$

\(\square \)

Lemma 9

Let \(X_i\) for \(i=1,\ldots ,n\) be a sequence of i.i.d. random variables. If for each \(i=1,\ldots ,n\), \(\mathbb {P}\left[ \, X_i \ge a \right] \le \xi \) holds, where \(\xi \in (0,1]\), then

$$\begin{aligned} \mathbb {P}\left[ \, \max _{i=1,\cdots ,n} X_i \ge a \right] \le n\xi . \end{aligned}$$

Proof

Since \( \mathbb {P}\left[ \, X_i \ge a \right] \le \xi \) then \(\mathbb {P}\left[ \, X_i < a \right] \ge 1 - \xi \). We can bound

$$\begin{aligned} \mathbb {P}\left[ \, \max _{i=1,\cdots ,n} X_i \ge a \right] =&\left( 1-\mathbb {P}\left[ \, \max _{i=1,\cdots ,n} X_i< a \right] \right) \\ =&\left( 1- \prod _{i=1}^n\mathbb {P}\left[ \, X_i <a \right] \right) \le 1 - (1-\xi )^n. \end{aligned}$$

The proof follows from Bernoulli’s inequality [40, Theorem 5.1] which states \((1-\xi )^n \ge 1 - n\xi \) for \(\xi \in [0,1]\) and \(n\ge 1\). \(\square \)

1.2 Proof of Theorem 10

Proof

Recall that \(\varvec{\Omega }_1 = \mathbf {U}^*_1 \varvec{\Omega }\) and \(\varvec{\Omega }_2 = \mathbf {U}_2^*\varvec{\Omega }\) where \(\varvec{\Omega }\) is random matrix with entries chosen from an i.i.d. Rademacher distribution. The proof proceeds in three steps.

1. Bound for \(\Vert \varvec{\Omega }_2\Vert _2^2\) The proof uses the matrix Chernoff concentration inequality. Let \(\varvec{\omega }_i \in \mathbb {R}^{n\times 1}\) be the i-th column of \(\varvec{\Omega }\). Note \( \varvec{\Omega }_2\varvec{\Omega }_2^* \in \mathbb {C}^{(n-k)\times (n-k)}\) and

$$\begin{aligned} \mathbb {E}\left[ \, \varvec{\Omega }_2\varvec{\Omega }_2^* \right] = \sum _{i=1}^\ell \mathbf {U}_2^*\mathbb {E}\left[ \, \varvec{\omega }_i\varvec{\omega }_i^{*} \right] \mathbf {U}_2 = \ell \mathbf {I}_{n-k}. \end{aligned}$$

Furthermore, define \(\mu _\text {min} ( \varvec{\Omega }_2\varvec{\Omega }_2^*) \equiv \lambda _\text {min} (\mathbb {E}\left[ \, \varvec{\Omega }_2\varvec{\Omega }_2^* \right] )\) and \(\mu _\text {max} (\varvec{\Omega }_2\varvec{\Omega }_2^*) \equiv \lambda _\text {max} (\mathbb {E}\left[ \, \varvec{\Omega }_2\varvec{\Omega }_2^* \right] )\). Clearly \(\mu _\text {min} = \mu _\text {max} = \ell \). Note that here we have expressed \( \varvec{\Omega }_2\varvec{\Omega }_2^*\) as a finite sum of \(\ell \) rank-1 matrices, each with a single nonzero eigenvalue \(\varvec{\omega }_i^* \mathbf {U}_2\mathbf {U}_2^*\varvec{\omega }_i \). We want to obtain a probabilistic bound for the maximum eigenvalue i.e., \(L_2 = \max _{i=1,\cdots ,\ell } \Vert \mathbf {U}_2^*\varvec{\omega }_i\Vert _2^2\). Using Lemma 8 we can write with probability at most \(e^{-t^2/8}\)

$$\begin{aligned} \left( \sqrt{n-k} + t\right) ^2 \le \Vert \mathbf {U}_2^*\varvec{\omega }_i\Vert _2^2 = \varvec{\omega }_i^* \mathbf {U}_2\mathbf {U}_2^*\varvec{\omega }_i . \end{aligned}$$

Since \(\Vert \mathbf {U}_2^*\varvec{\omega }_i\Vert _2^2\) are i.i.d., applying Lemma 9 gives

$$\begin{aligned} \mathbb {P}\left[ \, \max _{i=1,\cdots ,\ell } \Vert \mathbf {U}_2^*\varvec{\omega }_i\Vert _2 \ge \sqrt{n-k} + t \right] \le \ell e^{-t^2/8}. \end{aligned}$$

Take \(t = \sqrt{ 8\log (4\ell /\delta )}\) to obtain

$$\begin{aligned} \mathbb {P}\left[ \, L_2 \ge C_u^2 \right] \le \delta /4, \qquad C_u \equiv \sqrt{n-k} + \sqrt{ 8\log \left( \frac{4\ell }{\delta }\right) }. \end{aligned}$$
(25)

The matrix \(\varvec{\Omega }_2\) satisfies the conditions of the matrix Chernoff theorem 11; for \(\eta \ge 0\) we have

$$\begin{aligned} \mathbb {P}\left[ \, \lambda _\text {max}(\varvec{\Omega }_2\varvec{\Omega }_2^*) \ge (1+\eta ) \ell \right] \le (n-k) g(\eta )^\frac{\ell }{L_2}, \end{aligned}$$

where the function \(g(\eta )\) is defined in Theorem 11. For \(\eta > 1\) the Chernoff bounds can be simplified [30, Section 4.3] since \(g(\eta ) \le e^{-\eta /3}\), to obtain

$$\begin{aligned} \mathbb {P}\left[ \, \lambda _\text {max}(\varvec{\Omega }_2\varvec{\Omega }_2^*) \ge (1+\eta ) \ell \right] \le (n-k) \exp \left( -\frac{\eta \ell }{3L_2}\right) . \end{aligned}$$

Choose the parameter

$$\begin{aligned} \eta _\delta = C_{\ell ,\delta }C_u^2 = 3\ell ^{-1}C_u^2\log \left( \frac{4(n-k)}{\delta }\right) , \end{aligned}$$

so that

$$\begin{aligned} \mathbb {P}\left[ \, \Vert \varvec{\Omega }_2\Vert _2^2 \ge (1+\eta _\delta )\ell \right] \le&\> (n-k)\exp \left( -\frac{C_u^2}{L_2}\log \frac{4(n-k)}{\delta }\right) \\ =&\> (n-k)\left( \frac{\delta }{4(n-k)}\right) ^{C_u^2/L_2}. \end{aligned}$$

Finally, we want to find a lower bound for \(\Vert \varvec{\Omega }_2\Vert _2^2\). Define the events

$$\begin{aligned} A = \>\left\{ \varvec{\Omega }_2 \mid L_2 < C_u^2 \right\} , \qquad B = \>\left\{ \varvec{\Omega }_2 \mid \Vert \varvec{\Omega }_2\Vert _2^2 \ge (1+\eta _\delta )\ell \right\} . \end{aligned}$$

Note that \(\mathbb {P}\left[ \, A^c \right] \le \delta /4\) and under event A we have \(C_u^2 >L_2\) so that

$$\begin{aligned} \mathbb {P}\left[ \, B\mid A \right] \le (n-k)\left( \frac{\delta }{4(n-k)}\right) ^{C_u^2/L_2} \le \delta /4. \end{aligned}$$

Using the law of total probability

$$\begin{aligned} \mathbb {P}\left[ \, B \right] =&\> \mathbb {P}\left[ \, B\mid A \right] \mathbb {P}\left[ \, A \right] + \mathbb {P}\left[ \, B\mid A^c \right] \mathbb {P}\left[ \, A^c \right] \\ \le&\> \mathbb {P}\left[ \, B\mid A \right] + \mathbb {P}\left[ \, A^c \right] , \end{aligned}$$

we can obtain a bound for \(\mathbb {P}\left[ \, B \right] \) as

$$\begin{aligned} \mathbb {P}\left[ \, \Vert \varvec{\Omega }_2\Vert _2^2 \ge \ell \left( 1 + C_u^2 C_{\ell ,\delta }\right) \right] \le \delta /2. \end{aligned}$$

2. Bound for \(\Vert \varvec{\Omega }_1^\dagger \Vert _2^2\) The steps are similar and we again use the matrix Chernoff concentration inequality. Consider \(\varvec{\Omega }_1\varvec{\Omega }_1^*\in \mathbb {C}^{k\times k}\), and as before, write this matrix as the sum of rank-1 matrices to obtain

$$\begin{aligned} \mathbb {E}\left[ \, \varvec{\Omega }_1\varvec{\Omega }_1^* \right] = \sum _{i=1}^\ell \mathbf {U}_1^*\mathbb {E}\left[ \, \varvec{\omega }_i\varvec{\omega }_i^{*} \right] \mathbf {U}_1 = \ell \mathbf {I}_{k}, \end{aligned}$$

and \(\mu _\text {min} (\varvec{\Omega }_1\varvec{\Omega }_1^*) = \ell \). Each summand in the above decomposition of \(\varvec{\Omega }_1\varvec{\Omega }_1^*\) has one nonzero eigenvalue \(\varvec{\omega }_i^* \mathbf {U}_1\mathbf {U}_1^*\varvec{\omega }_i \). Following the same strategy as in Step 1, we define \(L_1 \equiv \max _{i=1,\ldots ,\ell }\Vert \mathbf {U}_1^*\varvec{\omega }_i\Vert _2^2 \) and apply Lemma 8 to obtain

$$\begin{aligned} \mathbb {P}\left[ \, \max _{i=1,\cdots ,\ell } \Vert \mathbf {U}_1^*\varvec{\omega }_i\Vert _2 \ge \sqrt{k} + t \right] \le \ell e^{-t^2/8} \le ne^{-t^2/8}. \end{aligned}$$

Take \(t = \sqrt{ 8\log (4n/\delta )}\) to obtain

$$\begin{aligned} \mathbb {P}\left[ \, L_1 \ge C_l^2 \right] \le \delta /4, \qquad C_l \equiv \sqrt{k} + \sqrt{ 8\log \left( \frac{4n}{\delta }\right) }. \end{aligned}$$
(26)

A straightforward application of the Chernoff bound in Theorem 11 gives us

$$\begin{aligned} {\mathbb {P}\left[ \, \lambda _\text {min}(\varvec{\Omega }_1\varvec{\Omega }_1^*) \le (1-\rho ) \ell \right] \le k g(-\rho )^\frac{\ell }{L_1}. } \end{aligned}$$

Next, observe that \(-\log g(-\rho )\) has the Taylor series expansion in the region \(0 {<} \rho {<} 1\)

$$\begin{aligned} -\log g(-\rho ) = {\rho } + (1-\rho )\log (1-\rho ) = \frac{\rho ^2}{2} + \frac{\rho ^3}{6} + \frac{\rho ^4}{12} + \cdots \end{aligned}$$

so that \(-\log g(-\rho ) \ge \rho ^2/2\) for \(0< \rho < 1\) or \(g(-\rho ) \le e^{-\rho ^2/2}\). This gives us

$$\begin{aligned} \mathbb {P}\left[ \, \Vert \varvec{\Omega }_1^\dagger \Vert _2^2 \ge \frac{1}{ (1-\rho ) \ell } \right] \> \le \> k \exp \left( -\frac{\rho ^2\ell }{2L_1}\right) , \end{aligned}$$
(27)

where we have used \(\lambda _\text {min}(\varvec{\Omega }_1\varvec{\Omega }_1^*) = 1/\Vert \varvec{\Omega }_1^\dagger \Vert _2^2\) assuming \(\mathsf {rank}(\varvec{\Omega }_1) = k\).

With the number of samples as defined in Theorem 3

$$\begin{aligned} \ell \ge 2\rho ^{-2}C_l^2 \log \left( \frac{4k}{\delta }\right) , \end{aligned}$$

the Chernoff bound (27) becomes

$$\begin{aligned} \mathbb {P}\left[ \, \Vert \varvec{\Omega }_1^\dagger \Vert _2^2 \ge \frac{1}{ (1-\rho ) \ell } \right] \> \le k\left( \frac{\delta }{4k}\right) ^{C_l^2/L_1}. \end{aligned}$$

Define the events

$$\begin{aligned} C = \left\{ \varvec{\Omega }_1 \mid \Vert \varvec{\Omega }_1^\dagger \Vert _2^2 \ge \frac{1}{ (1-\rho ) \ell }\right\} , \qquad D = \{ \varvec{\Omega }_1 \mid L_1 < C_\ell ^2\}. \end{aligned}$$

Note that \(\mathbb {P}\left[ \, D^c \right] \le \delta /4\) from (26). Then since the exponent is strictly greater than 1, we have

$$\begin{aligned} \mathbb {P}\left[ \, C \mid D \right] \> \le k\left( \frac{\delta }{4k}\right) ^{C_l^2/L_1} \le \delta /4. \end{aligned}$$

Using the conditioning argument as before gives \(\mathbb {P}\left[ \, C \right] \le \delta /2\).

3. Combining bounds Define the event

$$\begin{aligned} E = \left\{ \varvec{\Omega }\mid \Vert \varvec{\Omega }_1^\dagger \Vert _2^2 \ge \frac{1}{(1-\rho )\ell }\right\} , \qquad F = \left\{ \varvec{\Omega }\mid \Vert \varvec{\Omega }_2\Vert _2^2 \ge (1 + C_{\ell ,\delta }C_u^2)\ell \right\} , \end{aligned}$$

where \(C_{\ell ,\delta } \) is defined in Step 1, \(\mathbb {P}\left[ \, E \right] \le \delta /2\) and from Step 2, \(\mathbb {P}\left[ \, F \right] \le \delta /2\). It can be verified that

$$\begin{aligned} \left\{ \varvec{\Omega }\mid \Vert \varvec{\Omega }_2\Vert _2^2 \Vert \varvec{\Omega }_1^\dagger \Vert _2^2\ge \frac{1}{1-\rho }(1 + C_{\ell ,\delta }C_u^2) \right\} \subseteq E \cup F, \end{aligned}$$

and therefore, we can use the union bound

$$\begin{aligned} \mathbb {P}\left[ \, \Vert \varvec{\Omega }_2\Vert _2^2 \Vert \varvec{\Omega }_1^\dagger \Vert _2^2 \ge \frac{1}{1-\rho }(1 + C_{\ell ,\delta }C_u^2) \right] \le \mathbb {P}\left[ \, E \right] + \mathbb {P}\left[ \, F \right] \le \delta . \end{aligned}$$

Plugging in the value of \(C_{\ell ,\delta }\) and \(C_u^2\) from Step 1 gives the desired result. \(\square \)

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Saibaba, A.K., Alexanderian, A. & Ipsen, I.C.F. Randomized matrix-free trace and log-determinant estimators. Numer. Math. 137, 353–395 (2017). https://doi.org/10.1007/s00211-017-0880-z

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