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Suppose I have a matrix of size $(n \times p) $, denoted $B$, and a symmetric matrix of size $(n \times n) $, denoted $A$ with $n>p$.

If I know that the matrix $BB'A$ is positive definite, can I now tell if the matrix $B'AB$ is positive definite? If so why/why not?

JDoe2
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  • Should the assumption not be $BB'A$ is positive definite? – Rammus May 28 '20 at 11:05
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    Assuming we're considering operators on a real vector space (as you specified symmetric and not Hermitian). In the case $A$ commutes with $BB'$ we have $B B' A$ is symmetric and so it is positive-definite iff all of its eigenvalues are positive. Then we can use the fact that eigenvalues are invariant under cyclic permutations (see this SE post) to conclude that $B'AB$ has only positive eigenvalues. As $B'AB$ is symmetric and has positive eigenvalues it is PSD. – Rammus May 28 '20 at 11:18
  • $B'BA$ is not a legitimate product of matrices. – user1551 May 28 '20 at 13:08
  • Ah apologies yes I should have put $BB'$. Thank you @Rammus though, that answer was what I was looking for! – JDoe2 May 28 '20 at 17:46
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    @JDoe2 Happy to help :) – Rammus May 29 '20 at 07:58

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