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Assuming we know the eigenvalues of both $AB$ and $B$. Is there any way of determining the eigenvalues of the matrix $A$?

More precise formulation of my problem: I have two non-hermitian matrices $A$ and $B$ such that:

  1. $AB$ is stable (i.e. all eigenvalues have negative real part)
  2. All eigenvalues of $B$ are equal to 1 (i.e. $\lambda(B)=1$ and $\det(B)=1$)

Intuitively, I would think that this implies that A is stable too (since $\det(AB)=\det(A)$). However I have not been able to prove this.

A-D
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1 Answers1

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In general, $\det(AB)=\det(A)\det(B)$ is the only relationship between the eigenvalues of $AB$ and those of $A$ and $B$. See my previous answer for instance.

Your intuition about the stability of $A$ is also untrue. Consider e.g. $$ A=\pmatrix{-1&-3\\ 1&1+t},\ B=\pmatrix{1&-2\\ 0&1},\ AB=\pmatrix{-1&-1\\ 1&-1+t}. $$ When $t>0$ is small, $AB$ is stable because it is a small perturbation of $\pmatrix{-1&-1\\ 1&-1}$, but $A$ isn't stable because it has a positive trace.

user1551
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