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Given positive definite matices $A \in \mathbb{C}^{n\times n}$ and $B \in \mathbb{C}^{n\times n}$ what can we say about the positive definiteness of $A^{-1} B A$?

Mike
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    If $A$ and $B$ do not share the same set of eigenvectors, the result will not even be symmetric. – Zeekless Jul 29 '24 at 16:43
  • Here are a couple of other MSE posts to check out. https://math.stackexchange.com/questions/113842/is-the-product-of-symmetric-positive-semidefinite-matrices-positive-definite https://math.stackexchange.com/questions/211453/the-inverse-of-a-positive-definite-matrix-is-also-positive-definite – Cameron L. Williams Jul 29 '24 at 16:46
  • To talk about positive definiteness with respect to Eigenvalues, you need an ordering of the Eigenvalues, that is you need them to be real. You could use (even without this, that complex matrices are similiar complex symmetric matrices and use factorizations that come with them look at Horn‘s book about Matrix Analysis). – dForga Jul 29 '24 at 18:01
  • @dForga The eigenvalues of $A^{-1} B A$ are guaranteed to be real (and strictly positive) as eigenvalues are invariant under cyclic permutation of square matrices. I.e., it has the same spectrum as $BAA^{-1} = B$ which has real and positive eigenvalues. – Rammus Aug 22 '24 at 15:03

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