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.



Rank of ABA where B is positive definite
$A^T BAx = 0 \implies (Ax)^TB(Ax)=0 \implies Ax=0$ by positve definiteness of $B$. So $ker(M)=ker(A)$ and hence $rk(M)=rk(A)$.


