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Assume that we have two real symmetric matrices A and B, where A is a positive diagonal matrix, and B is a symmetric matrix with one eigenvalue λ = 0. Assume that H= AB; is it possible to proof that the matrix H has the same properties than B (semi-positive, negative or indefinite matrix)? That is means the eigenvalues of H have the same signe than the eigenvalues of B.

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closed as off-topic by Stefan Kohl, Will Jagy, Steven Sam, José Figueroa-O'Farrill, Chris Godsil Sep 17 '14 at 23:39

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  • $\begingroup$ First, the product $AB$ is not symmetric in general, so its eigenvalues could be complex. $\endgroup$ – Chris Godsil Sep 17 '14 at 23:39
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The eigenvalues of $H$ are the same as the eigenvalues of $A^{1/2} B A^{1/2}$. See Sylvester's Law of Inertia.

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