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Having two positive definite matrices $A, B$, it holds that the product $ABA$ is positive definite.

I'm looking for a simple proof of this fact.

Rodrigo de Azevedo
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user1747134
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  • Welcome to MSE! It helps if you tell us what you have tried and where you are stuck. Also, if this is HW, it should be tagged as such. Regards – Amzoti Mar 09 '13 at 14:59
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    If $B$ is definite positive and $A$ is self-adjoint, then $ABA=C^*C$ with $C=\sqrt{B}A$. So it is definite positive. But the good/minimal answer is given by Dan Shved below. – Julien Mar 09 '13 at 15:11

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You don't even need positive definiteness of $A$, it is enough that $B$ is positive definite and $A$ is nonsingular and symmetric. Then for any nonzero column $x \in \mathbb{R}^n$ column $Ax$ is also nonzero, therefore $$ x^T(ABA)x = (Ax)^T B (Ax) > 0. $$

Dan Shved
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  • In question $B$ is not symmetric...So how you take $B$ as symmetric ? What happen for any two arbitrary positive definite matrices $A$ and $B$ ? – Empty Apr 24 '15 at 19:07
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    @S.Panja-1729 I'm using the definition of a positive-definite matrix as given on wikipedia. According to this definition, each positive-definite matrix is symmetric. If, however, you choose to define positive-definite matrices as any matrices that have $x^T A x > 0$ for each column $x \neq 0$, then the statement in the question simply becomes false. – Dan Shved Apr 25 '15 at 07:50