Consider the system of linear equations:

$$ Ax=b \tag{1} \label{eq1} $$


  • $A\in\mathbb F^{n\times n}$, diagonalizable dense matrix, over the field $\mathbb F$ of real or complex numbers,
  • $x\in \mathbb F^{n\times 1}$ is a vector of unknowns
  • $b\in \mathbb F^{n\times 1}$ is a known right-hand side vector
  • $n$ is in the order of 1000–10000

Unlike the usual system of linear equations, I do not have access to $A$ itself; however, I have access to the matrix exponential $e^A$. The matrix exponential is accessible both as an explicit matrix and, consequently, as a function acting on a vector.

What are my options for finding the solution of $\eqref{eq1}$ knowing only the matrix exponential? Computing matrix logarithm does not seem the best option as I doubt being able to get anything numerically stable and reasonably efficient.

Am I missing something simple?

  • $\begingroup$ 1. Have you tried some ready-to-use methods for the matrix logarithm such as Matlab's logm(E) \ b, before writing it off as unstable / inefficient? 2. Are $A$ and $E=\exp(A)$ sparse or dense? $\endgroup$ Commented Apr 11, 2020 at 7:32
  • $\begingroup$ Is $B=e^A$ close to unity? Then you can use Taylor expansion $A = log(B) = (B-I) + (B-I)^2/2 - (B-I)^3/3 + ...$ $\endgroup$ Commented Apr 11, 2020 at 14:55
  • 1
    $\begingroup$ @FedericoPoloni $A$ and $E$ are dense. I would have to export the matrix to Matlab and try it there, which I should do. That is a "bruteforce" method in my initial assessment – and I certainly have to try it at least on a couple of examples. $\endgroup$
    – Anton Menshov
    Commented Apr 11, 2020 at 19:02
  • $\begingroup$ @MaximUmansky nope, $e^A$ is not close to unity. $\endgroup$
    – Anton Menshov
    Commented Apr 11, 2020 at 19:02
  • $\begingroup$ (What is $\|A\|$, just to have an idea?) $\endgroup$ Commented Apr 11, 2020 at 19:05

3 Answers 3


You are effectively asking how to compute $y=(\log M )^{-1}b$, where $M=e^A$ is the given matrix. There are several methods for computing $f(M)b$ without forming $f(M)$, and they are reviewed here. One general method is to use Cauchy’s theorem, $$y=\dfrac{1}{2\pi i}\int_\Gamma f(z)(zI - M)^{-1}b\,dz,$$ with $f(x) = 1/\log(x)$. $\Gamma$ is a contour that encloses all the eigenvalues of $M$, so you need to first estimate the magnitude of the largest eigenvalue, say, with a power method. Then you approximate the integral with the trapezoidal rule. You need to solve several shifted systems of the form $(zI-M)x =b$, for which a preliminary reduction to Hessenberg form is useful.

  • $\begingroup$ great reference. It seems like exactly what I hoped for. I will have to analyze it against computing $\log M$, but now I have an alternative approach. $\endgroup$
    – Anton Menshov
    Commented Apr 11, 2020 at 19:05

I am expanding my comment into an answer. I don't think that the method is efficient, but I think that it can be used to obtain the matrix $A$ from $e^{A}$.

We know that

$$\frac{d e^{tA}}{dt} = e^{tA} A\, ,$$

so, we could use

$$\left.\frac{d e^{tA}}{dt}\right|_{t=0} = A\, ,$$

if we can approximate the derivative

$$\frac{d e^{tA}}{dt} \approx D(A)\, .$$

For example, we could use a forward finite difference

$$\left.\frac{d e^{tA}}{dt}\right|_{t=0} \approx \frac{e^{hA} - I}{h}\, ,$$

but the problem is that we need to compute the fractional power of the matrix $e^{A}$. Maybe we could take advantage of higher-order approximations and just use integer powers of the matrix, but a couple I tried didn't work properly.

It seems to work, but I doubt it is efficient.

import numpy as np
from scipy.linalg import logm, fractional_matrix_power as powm
import matplotlib.pyplot as plt

eA = np.array([
    [1, -1],
    [1, 2]])
A = logm(eA)
rel_error = []
steps = [1, 1e-1, 1e-2, 1e-3, 1e-4, 1e-5]
for h in steps:
    A1 = np.real((powm(eA, h) - np.eye(2))/h)
    rel_error.append(np.linalg.norm(A - A1)/np.linalg.norm(A))

plt.loglog(steps, rel_error)
plt.xlabel("Relative error")
plt.savefig("matexp.png", dpi=300, bbox_inches="tight")

enter image description here

  • $\begingroup$ I think you used a pretty bad example to test your method here. Based on OP's question, his matrix is in the order of $1000 \times 1000$ to $10000 \times 10000$, and I believe your method won't converge at all for a $1000 \times 1000$ matrix. Replace your $2 \times 2$ matrix by this n = 1000 eA = np.random.rand(n,n) and you would see by decreasing $h$ you won't get a convergence. $\endgroup$ Commented Apr 12, 2020 at 3:07
  • $\begingroup$ @AloneProgrammer, I tested it with the example provided by Federico Poloni. $\endgroup$
    – nicoguaro
    Commented Apr 12, 2020 at 3:31
  • 1
    $\begingroup$ I guess your finite-difference expression can be also interpreted as a consequence of the property (or definition) of the exponential function, $exp(x) = lim(1+x/n)^n$ for large n, or $x = lim (n(e^{x/n}-1))$. If we had a good way to find the $n^{th}$ root of our given matrix $e^A$ then this would be indeed a possible way to approximate matrix A. $\endgroup$ Commented Apr 12, 2020 at 6:14
  • 1
    $\begingroup$ Here is a good survey of calculation methods for finding a nth root of a matrix, cis.upenn.edu/~cis610/bini_higham_pth_root_na_2005.pdf. $\endgroup$ Commented Apr 12, 2020 at 7:05
  • $\begingroup$ Thanks for the explanation, now I understand what you mean! $\endgroup$ Commented Apr 12, 2020 at 7:50

If we diagonalize matrix $A$ by finding the transformation S such that $A = S D S^{-1}$ where D is a diagonal matrix and the diagonal elements of $D$ are the eigenvalues $\lambda_k$ then the same transformation makes $e^A$ diagonal, and the eigenvalues are $e^{\lambda_k}$. So diagonalizing $e^A$ and taking log of eigenvalues we find matrix $D$, which is sufficient to solve the linear system; and using the transformation $S$ we can find the original matrix $A.$

Alternatively, if $B=e^A$ is close enough to unity, you can use Taylor expansion to find $A$, $A = log(B) = (B-I) + (B-I)^2/2 - (B-I)^3/3 + ...$

  • 1
    $\begingroup$ I think that's just how matrix logarithms work $\endgroup$ Commented Apr 11, 2020 at 5:53
  • 2
    $\begingroup$ That’s how matrix logarithm’s work in exact arithmetic and for normal matrices in floating point. For general diagonalizable matrices, $S$ mat be ill-conditioned so that significant errors may result from this approach. $\endgroup$ Commented Apr 11, 2020 at 8:21
  • 1
    $\begingroup$ You are slowly re-inventing the standard methods to compute logarithms of dense matrices. :) The next step is using the identity $\log(A) = 2^s \log(A^{1/2^s})$ to get a matrix closer to the identity, and replace Taylor with Padé which has a better convergence radius. $\endgroup$ Commented Apr 11, 2020 at 20:33
  • $\begingroup$ @Federico Poloni How about other expansions for $ln(x)$, in particular the second one in math.com/tables/expansion/log.htm? I guess it can be applied for matrices too, if we first calculate the inverse of our given $e^A$; and convergence of this series is not restricted to norm $|| e^A-I || $ < 1. $\endgroup$ Commented Apr 12, 2020 at 5:51
  • $\begingroup$ It's possible that it could be made to work, but I have no direct experience. Convergence of that series is restricted to matrices with no eigenvalues $\lambda < 1/2$, if I understand correctly (not clear how this generalizes to complex numbers). The problem is, it is easy to ensure that a matrix has no large eigenvalues (small norm implies that), but it is more difficult to ensure that it has no small eigenvalues). $\endgroup$ Commented Apr 12, 2020 at 7:48

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