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jay
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Bounding the eigenvalues of $B A B^T$ with the eigenvalues of $A$
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Bounding the eigenvalues of $B A B^T$ with the eigenvalues of $A$
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Bounding the eigenvalues of $B A B^T$ with the eigenvalues of $A$
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Bounding the eigenvalues of $B A B^T$ with the eigenvalues of $A$
@AlexM. I fixed the title. PSD means positive semi definite.
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Bounding the eigenvalues of $B A B^T$ with the eigenvalues of $A$
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Bounding the eigenvalues of $B A B^T$ with the eigenvalues of $A$