# Positive definite matrix multiplication $B^T\cdot (M\cdot M^T)\cdot B$

I'm trying to determine if $B^T(MM^T)B$ can be said to be positive definite or semi-positive definite. Both B and M are assumed to be made up of real numbers and are invertible.

I've read that $B^TB$ is positive definite for a real invertible matrix. (Though this may only be the case for real, invertible symmetric matrices?) And $MM^T$ (for a real invertible matrix) can be said to be semi-positive definite.

I think I saw a property that said if $MM^T$ is positive definite than $B^T(MM^T)B$ is positive definite but I cannot find where I read that or how $MM^T$ being semi-positive definite would change things.

Thanks!

• $B^TMM^TB=A^TA$ for $A=B^T M$. For real $A$, $A^TA$ is positive semidefinite. For real nonsingular $A$, $A^TA$ is positive definite. – Lord Shark the Unknown Sep 14 '17 at 15:06
• Thanks! That's so simple! For some reason I thought it was way more complicated.... Now to try and determine if A is real in my case... :) – Katzchen Sep 14 '17 at 15:13

A square matrix $A$ is positive semi-definite if the quadratic form $x^TAx \ge 0$ for any conformable $x \ne 0$. Consider: $$(x^TB^T)MM^T(Bx) = (y^TM)(M^Ty) = z^Tz \ge 0$$ because the last expression is a sum of squares. So $B^TMM^TB$ is indeed positive semi-definite.