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Given A, B two symmetric and positive-definite matrices, then is their product A B still positive-definite? If yes, how to prove this?

Jean Marie
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Francesco Ciardiello
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1 Answers1

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The product of two commuting symmetric positive definite matrices is also positive definite. However, in general $AB$ need not be symmetric anymore for symmetric $A$,$B$. So then there are counterexamples - see this MSE-question. Take $$A=\begin{pmatrix}2 & -1\\ -1 & 2\end{pmatrix},\qquad B = \begin{pmatrix}10 & 3\\ 3 & 1\end{pmatrix}. $$ Then $AB$ is not symmetric, and hence not positive definite. Also we have $x^t(AB)x=-1<0$ for $x=(0,1)^t$. So $AB$ is definitly not positive definite.

Dietrich Burde
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  • that counterexample in the link you have suggested is wrong simply because the product matrix is still positive definite. Clearly if matrices commute then their product remains positive definite. However. In case we have symmetry for matrices and not commutativity, can we still say that the product remains positive definite? I think that my question is different from what you are saying and the counterexample is wrong. – Francesco Ciardiello May 12 '17 at 11:59
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    @FrancescoCiardiello If the product does not commute, it is not a symmetric matrix, so it doesn't make sense to ask if it's positive definite. – Arnaud D. May 12 '17 at 12:04
  • @FrancescoCiardiello It is useful to recall that "positive definite" does not mean "has only strictly positive eigenvalues". It means $x^T A x>0$ for all nonzero vectors $x$, which automatically implies symmetry. – Ian May 12 '17 at 12:34
  • @ArnaudD. To your first remark. It is not said that matrices positive definite are necessarily symmetric. So, it makes sense to ask. – Francesco Ciardiello May 12 '17 at 12:49
  • @ArnaudD. To your second remark. Your are right. However, in the example suggested above trough the link, the product is positive definite anyway. So the counterexample does not work. – Francesco Ciardiello May 12 '17 at 12:52
  • @FrancescoCiardiello I've only made one remark (perhaps you wanted to ping Ian ). And in general, positive definiteness (and all the variants) is only defined for symmetric matrices, so it wouldn't make sense to ask wether a non-symmetric matrix is positive definite, which is what I meant in my comment. But if you want to extend the definition to any matrix as in Ian's comment, then the counterexample above works. – Arnaud D. May 12 '17 at 13:10
  • Please excuse me, my comment was mistaken. I am too accustomed to working on $\mathbb{C}^n$, where $x^* A x > 0$ for all $x$ implies $A$ is Hermitian. On $\mathbb{R}^n$ this does not hold. – Ian May 12 '17 at 13:29