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Suppose $A$ is real-valued negative-definite $d\times d$ matrix with norm < 1.

How could I estimate $A$, up to rotation, from values of $\operatorname{Tr}e^{A},\operatorname{Tr}e^{2A},\ldots,\operatorname{Tr}e^{kA}$ where $e^A$ is matrix exponential? How big does $k$ need to be for accurate result?

In my application $I+A\approx e^A$, so one could instead consider the problem of obtaining $A$ from values of $\operatorname{Tr}A,\operatorname{Tr}A^2,\ldots$ but the algebraic solution is not practical for numerical stability reasons.

In the continuous case, empirical density of $A$ eigenvalues and $g(k)=\operatorname{Tr}e^{Ak}$ are related through the Laplace transform. Can this connection be utilized to solve the problem at hand?

Yaroslav Bulatov
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1 Answers1

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If the minimal polynomial of $A$ has degree $d$ and if you know $Tr(A^k)$ up to $k\le 2d-1$ then it suffices to do the Gram-Schmidt orthonormalization of the innner product $\langle R,S\rangle = Tr(RS)$ on $\Bbb{R}[A]$, obtaining a basis $B_1,\ldots,B_d$ of $\Bbb{R}[A]$ such that $\langle B_n,B_m\rangle=1_{n=m}$, and to let $$C_{nm}=Tr(A B_nB_m)$$

Concretely we set $b_1=I,B_1=\frac{b_1}{\sqrt{Tr(b_1^2)}},b_2= A - Tr(A)B_1,B_2=\frac{b_2}{\sqrt{Tr(b_2^2)}}$, $b_3=A^2-Tr(A^2) b_1-Tr(b_2(A^2-Tr(A^2) b_1))b_2,\frac{b_3}{\sqrt{Tr(b_3^2)}}$ and so on, obtaining the coefficients of the $B_n$ in the $I,A,A^2,\ldots,A^{d-1}$ basis of $\Bbb{R}[A]$.

Then $C$ is the matrix of the multiplication by $A$ on the $B_n$ basis, so it has the same characteristic polynomial as $A$. And since $C$ is symmetric it means that $$C=P A P^\top$$ with $P$ orthonormal.

reuns
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