# Rank of ABA where B is positive definite

I have a n-by-k matrix A and a n-by-n matrix B, where B is positive definite. I can form the matrix $M = A^t B A$. Playing around, I always found $rk(M) = rk(A)$ but I can't prove this.

-
Hint: M and A have the same null space. –  Mark Meckes Mar 31 '10 at 12:26
Thanks, I actually found that, but missed that A and M have the same number of columns to finish the proof. –  Frank Meulenaar Mar 31 '10 at 12:28
$A^T BAx = 0 \implies (Ax)^TB(Ax)=0 \implies Ax=0$ by positve definiteness of $B$. So $ker(M)=ker(A)$ and hence $rk(M)=rk(A)$.
This is rather tangential, but I can't help mentioning that this is related to the fact that in a hilbert space, for a (compact) linear operator one has: $Ker A^* \perp Im A$.