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I have two positive-definite matrices $A$ and $B$ (not necessarily symmetric), and we have $AB=BA$, is there any theorem that ensures that the product of $A$ and $B$, $AB$ is positive definite? Or semi-positive definite?

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If $A$ and $B$ are positive definite then for nonzero $x$ we have $x^TAx >0$ and $x^T Bx>0$ so that $$x^TAxx^TBx >0$$ now $$x^TAxx^TBx = x^TA\|x\|^2Bx =\|x\|^2(x^TABx) >0$$ Since $\|x\|^2>0$ is the length of $x$ we have that also $x^TABx>0$ for all nonzero $x$. Thus $AB$ is also positive definite.

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    $\begingroup$ $xx^T \ne x^Tx = \|x\|^2$. $\endgroup$ – martini Jan 3 '14 at 13:16
  • $\begingroup$ @martini Ow yeah you're right, silly of me. $\endgroup$ – Slugger Jan 3 '14 at 13:17
  • $\begingroup$ Thanks for your answer, but this mistake in your derivation process but I don't know how to type in formula. But anyway thanks $\endgroup$ – Lencholucy Jan 3 '14 at 13:18
  • $\begingroup$ No problem, sorry for the mistake. I will probably delete my answer as soon as a correct one pops up :) $\endgroup$ – Slugger Jan 3 '14 at 13:21
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Since $A$ and B are commutable, $AB=BA$, hence they are joint diagonalizable with a unitary matrix, $U$, $U^\dagger U=I$:

$A=U^\dagger D_A U$

$B=U^\dagger D_B U$

hence

$AB=U^\dagger D_A U U^\dagger D_B U=U^\dagger D_A D_B U$.

Now if entries of $D_A$ and $D_B$ are positive (nonnegative), i.e., $A$ and $B$ are P.D. (P.S.D.) then $AB$ is P.D. (P.S.D).

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