Assume $A$ is symmetrical. Prove that $A$ is positive definite and $B$ is non-singular if, and only if, $BA B^T$ is positive definite.

Asked Sep 05 '15 at 23:56

Active Sep 05 '15 at 23:56

Viewed 53 times

I have no idea how to even approach this problem. I am not necessarily wanting the whole solution but can someone point me in the right direction?

asked Sep 05 '15 at 23:56

1

All you need is the definition, the rank property and the fact that for a non-singular matrix $B$ we have $y=Bx\ne 0$ iff $x\ne 0$. – A.Γ. Sep 06 '15 at 00:18
Brilliant! The fact that you reminded me that $Bx \neq 0$ if $x \neq 0 $ is a property of a non-singular matrix B, nudged me in the right direction. Additionally, if a matrix $C$ is non-singular then $C^T$ is also non-singular and vice-versa. These are the things needed to solve this proof. – Jan Izak Cornelius Vermaak Sep 07 '15 at 17:45

0 Answers0