1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Amzoti Mar 9 '13 at 14:59
  • 1
    $\begingroup$ 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. $\endgroup$ – Julien Mar 9 '13 at 15:11
5
$\begingroup$

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 $$ x^T(ABA)x = (Ax)^T B (Ax) > 0. $$

$\endgroup$
  • $\begingroup$ In question $B$ is not symmetric...So how you take $B$ as symmetric ? What happen for any two arbitrary positive definite matrices $A$ and $B$ ? $\endgroup$ – Empty Apr 24 '15 at 19:07
  • $\begingroup$ @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. $\endgroup$ – Dan Shved Apr 25 '15 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.