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Take two (real) matrices $A$ and $B$, where both have (real) eigenvalues within the unit circle, $B$ is also a diagonal matrix.

Can I say something about the eigenvalues of the product $A \cdot B$ ?

I suspect that the eigenvalues of $A \cdot B$ are also within the unit circle. Any help/reference for a proof of that?

Thanks

TheoMarsy
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ivan
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  • Just an observation, which is weaker than what you need: The spectral radius is bounded from above by $|AB|$, and in turn $|AB|\le |A||B|$. If both $A$ and $B$ had a norm bounded by $1$, this would imply that the product's spectral radius is bounded by $1$ as well. However, from your assumptions it is likely, but not necessary, that $|A|\le 1$ (since $B$ is diagonal, you do know that $|B|\le 1$). I don't know if there is a counter-example for the case $|A|>1$. – DominikS Jan 25 '24 at 11:47
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    So, you are saying that if I can prove that ||A||<1, then I know that the eigenvalues of AB are within the unit circle. Right? – ivan Jan 25 '24 at 12:36
  • The spectral radius of $AB$ can be arbitrarily large. See my answer to another question for an illustration. – user1551 Jan 25 '24 at 14:46

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