most recent 30 from http://mathoverflow.net 2013-05-21T04:29:43Z http://mathoverflow.net/feeds/question/19942 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19942/rank-of-aba-where-b-is-positive-definite Frank Meulenaar 2010-03-31T12:18:18Z 2012-08-30T00:02:33Z

<p>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.</p>

http://mathoverflow.net/questions/19942/rank-of-aba-where-b-is-positive-definite/19954#19954 Johannes Hahn 2010-03-31T14:47:12Z 2010-03-31T14:47:12Z

<p>$A^T BAx = 0 \implies (Ax)^TB(Ax)=0 \implies Ax=0$ by positve definiteness of $B$. So $ker(M)=ker(A)$ and hence $rk(M)=rk(A)$.</p>

http://mathoverflow.net/questions/19942/rank-of-aba-where-b-is-positive-definite/105894#105894 S.A.A 2012-08-30T00:02:33Z 2012-08-30T00:02:33Z

<p>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: $Ker A^* \perp Im A$.</p>