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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.

flag	asked Mar 31 2010 at 12:18 Frank Meulenaar 47●6

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Hint: M and A have the same null space. – Mark Meckes Mar 31 2010 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 2010 at 12:28

Johannes Hahn · Accepted Answer · 2010-03-31 14:47:12Z UTC

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

answered Mar 31 2010 at 14:47

Johannes Hahn
1,725●6●12

S.A.A · Answer 2 · 2012-08-30 00:02:33Z UTC

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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: $Ker A^* \perp Im A$.

answered Aug 30 at 0:02

S.A.A
237●7

Rank of ABA where B is positive definite

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