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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.
Rodrigo de Azevedo
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user3498429
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1 Answers1

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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.

A.Γ.
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    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... – user3498429 Jul 17 '15 at 23:05
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    $B^{1/2}AB^{1/2}$ has real eigenvalues as it is symmetric. – davcha Jul 17 '15 at 23:06
  • 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 $U\Sigma^{1/2}U^T$. – A.Γ. Jul 17 '15 at 23:10
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    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. – A.Γ. Jul 17 '15 at 23:12
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    @A.G. Ok I think I got it! Since $B$ is positive definite, its square root is positive definite as well (Wikipedia). Since the product $MNM$ is positive definite if both $M$ and $N$ are positive definite (Wikipedia), then $B^{1/2}AB^{1/2}$ is positive definite – user3498429 Jul 17 '15 at 23:23
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    @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). – A.Γ. Jul 17 '15 at 23:29
  • @A.G. your penultimate explanation was perfectly clear, I haven't seen before commenting. Thank you again! – user3498429 Jul 17 '15 at 23:37
  • @user3498429 You are welcome! – A.Γ. Jul 17 '15 at 23:39
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    Does the original problem statement still hold if the matrices are merely self-adjoint but not necessarily positive? In this case, we can't really take the square root, but is it fundamentally a different situation? – keej Jul 11 '17 at 17:58
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    @keej Positivity of eigenvalues for $AB$ is clearly out of question (for example, $A=I$, $B=-I$). To see that eigenvalues may be non-real: try $A=\begin{pmatrix}1 & 0\0 & -1\end{pmatrix}$, $B=\begin{pmatrix}0 & 1\1 & 0\end{pmatrix}$. – A.Γ. Jul 11 '17 at 18:45
  • @A.Γ. Did you assumed $A$ and $B$ are commutative in your argument? Because otherwise $AB$ is not symmetric. – Shah Dec 03 '19 at 15:41
  • @YMA No, I didn't. OP says that $AB$ does not need to be symmetric. – A.Γ. Dec 03 '19 at 16:25