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Suppose $A$ and $B$ are positive definite matrices (not necessarily symmetric).

Will $ABA$ , $ABA^*$ and $AB$ be positive definite matrices? $A^*$ is $A$ transpose.

I have no idea. Because I dealt with only symmetric matrices in this context. But $A$ , $B$ are not necessarily symmetric matrices here. Can somebody help me out?

  • how?@SahibaArora –  Jun 11 '17 at 20:09
  • By saying that you "deal with only symmetric matrices in this context" I take it that your definition is a matrix $A$ is positive definite iff it is symmetric and $\langle x, Ax \rangle \gt 0$ whenever $x$ is nonzero. – hardmath Jun 11 '17 at 20:13
  • please read the question once again..@hardmath –  Jun 11 '17 at 20:17
  • As far as I know ''positive definite'' is a property defined only for symmetric matrices: https://en.wikipedia.org/wiki/Positive-definite_matrix#Further_properties – Emilio Novati Jun 11 '17 at 20:18
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    The typical definition of positive definite requires the matrix be self-adjoint. It seems like this is not the case in your question, so I editted to make this assumption explicit. If I am wrong, rollback my edit. – Aweygan Jun 11 '17 at 20:20
  • I also knew it.. But https://math.stackexchange.com/questions/66520/if-a-and-b-are-positive-definite-matrices-is-ab-positive-definite confused me..@hardmath Emilio Novati –  Jun 11 '17 at 20:21
  • Thanks for clarification! – hardmath Jun 11 '17 at 20:24
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    "But $A,B$ are normal matrices here." Normal matrix is a technical term denoting a matrix satisfying $A^A=AA^$. Is that what you mean, or do just mean $A,B$ may not be symmetric? –  Jun 11 '17 at 20:25
  • U can answer supposing that $A$ and $B$ are symmetric. If they are not P.D then we will be done... –  Jun 11 '17 at 20:28
  • U can answer supposing that $A$ and $B$ are symmetric. If they are not P.D then we will be done...@hardmath –  Jun 11 '17 at 20:28
  • Note that $A^$ is ordinarily the conjugate transpose. For a real matrix this amounts to the same thing as transpose, but for complex matrices it is not generally the same. In the case of complex vector operations, positive definite* (defined as $\langle x,Ax \rangle \gt 0 $ for any nonzero $x$) implies self-adjoint (i.e. implies Hermitian). – hardmath Jun 12 '17 at 16:29

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