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Let $A$, $B$ square matrices over $\mathbb{R}$ with the same dimension. If $A$ is positive definite and $B$ is positive semidefinite, is $AB$ positive semidefinite? If yes, prove it. If no, counterexample it.

What if $A$ and $B$ are symmetric?

artful_dodger
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  • I'm not sure if your question is well stated. Definition of positive (semi)definite matrix assumes symmetry, but the product of two symmetric matrices need not be symmetric. – Przemek Jul 18 '17 at 18:00
  • Positive definite = eigenvalues are $>0$. Positive semidefinite = eigenvalues are $\ge 0$. No symmetry required. – artful_dodger Jul 18 '17 at 18:42
  • @user459312 That is not what positive definite usually means, although there is a definition that allows for non-symmetric positive definite matrices. – Ben Grossmann Jul 18 '17 at 18:48
  • @Przemek in some areas, a real matrix $A$ is called positive definite if and only if $A + A^T$ is positive definite (in the usual sense). – Ben Grossmann Jul 18 '17 at 18:49
  • @Omnomnomnom in every area I encountered it, that was the definition. But no matter, I got the counterexample I wanted. – artful_dodger Jul 18 '17 at 18:53
  • @user459312 right. Note that this is not the same as saying that the eigenvalues are positive. It is, however, an established fact that positive definite matrices have eigenvalues with positive real part. – Ben Grossmann Jul 18 '17 at 18:57

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The matrices should be symmetric by definition of positive definiteness.

If $A$ and $B$ commute, the result is positive semidefinite again. For the case in which they don't commute, there is a counterexpample.

Proof for the case $AB = BA$: Since $A>0$, there exists a unique symmetric matrix $A^{1/2}>0$ such that $A^{1/2}A^{1/2} = A$. We can write

$$\langle AB x,x\rangle = \langle BA x,x\rangle = \langle A^{-1/2}A^{1/2} B A^{1/2}A^{1/2}x,x\rangle = \langle A^{1/2} B A^{1/2}x,x\rangle = \langle B(A^{1/2}x), A^{1/2}x\rangle =\langle Bz,z\rangle \ge 0,$$ with $z = A^{1/2}x$.

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DominikS
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