Rank of ABA where B is positive definite - MathOverflow 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 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 Answer by Johannes Hahn for Rank of ABA where B is positive definite 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 Answer by S.A.A for Rank of ABA where B is positive definite 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>