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Given two real positive definite (and therefore, symmetric) matrices $A$ and $B$, are all the eigenvalues of $AB$ real and positive?

  • Wikipedia says $AB$ is positive definite if $A$ and $B$ are positive definite and commute, but I don't need $AB$ to be symmetric.
  • Between the lines of this question the asking user somehow prove that yes, "the eigenvalues of $AB$ are hence real and strictly positive" but I couldn't understand if that is confirmed in the answer.
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If we call $B^{1/2}$ the symmetric matrix such that $B^{1/2}B^{1/2}=B$ (i.e. the standard square root of a positive definite matrix) then $$ AB=AB^{1/2}B^{1/2}=B^{-1/2}(B^{1/2}AB^{1/2})B^{1/2}, $$ that is $AB$ is similar to the positive definite matrix $B^{1/2}AB^{1/2}$, sharing all eigenvalues. It makes the eigenvalues of $AB$ be positive.

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  • $\begingroup$ Thank you @A.G. ! I suppose we can also say that they're real since $B^{1/2}AB^{1/2}$ is positive definite. Correct? Anyway I can't see why $B^{1/2}AB^{1/2}$ is positive definite... $\endgroup$ – user3498429 Jul 17 '15 at 23:05
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    $\begingroup$ $B^{1/2}AB^{1/2}$ has real eigenvalues as it is symmetric. $\endgroup$ – davcha Jul 17 '15 at 23:06
  • $\begingroup$ Sure. $B^{1/2}$ is real. Look, the eigenvalues of a symmetric matrix are real, thus the eigenvectors too. So the Jordan form is real $B=U\Sigma U^T$ with $\Sigma$ being diagonal positive and $U$ orthogonal. Then construct $B^{1/2}$ as $U\Sigma^{1/2}U^T$. $\endgroup$ – A.Γ. Jul 17 '15 at 23:10
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    $\begingroup$ By definition $x^TB^{1/2}AB^{1/2}x=z^TAz>0$ for $z=B^{1/2}x\ne 0$ $\Leftrightarrow$ $x\ne 0$. Thus positive definite. $\endgroup$ – A.Γ. Jul 17 '15 at 23:12
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    $\begingroup$ @user3498429 Right. One can say even stronger that $MNM$ is positive definite for all symmetric non-singular $M$ as long as $N$ is positive definite. Just by definition (in my previous post). $\endgroup$ – A.Γ. Jul 17 '15 at 23:29

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