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Suppose that you've got an $n \times n$ irreducible matrix $A$ with strictly positive real entries and eigenvalues $\lambda_i$, $i=1,...,m$, arranged so that $|\lambda_1| > \cdots > |\lambda_m|$. Suppose that $D$ is a diagonal matrix with strictly positive real entries, and define $$ \tilde{A} := D^{-1} A' D. $$ What can be said about the eigenvalues of the matrix $A \tilde{A}$? Note that $\tilde{A}$ is self adjoint with respect to the inner product $\langle \cdot, \cdot \rangle_D$ given by $$ \langle f, g \rangle_D = f' \overline{g} D $$ and that one could build a linear isometry between this inner-product space and the one with the usual Euclidean inner product, under which $A\tilde{A}$ would be Hermitian (real symmetric in fact).

My sense is that the eigenvalues of $A\tilde{A}$ would be the squares of the absolute values of the eigenvalues of $A$. Am I right? And does it share the eigenvalues of $\tilde{A} A$?

Thanks!

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The eigenvalues of $AB$ and $BA$ are always the same.

As for the eigenvalues of $A \tilde{A}$ being the squares of the absolute values of the eigenvalues of $A$, have you tried some examples? Try e.g. $A = \pmatrix{1 & 2\cr 1 & 1\cr}$ and $D = \pmatrix{2 & 0\cr 0 & 1\cr}$.

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  • $\begingroup$ Thanks Robert. I tried to remove the specifics of the problem I was working on, hoping something more general than I needed was true, but now I realize that I said something silly. This was helpful for knocking me back into reality though. $\endgroup$ – user139388 Apr 12 '14 at 1:37

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