Thread: rank proof

Originally Posted by hayter221

If B and C are invertible then prove rank(BA)=rank(AC)=rank(A).

Whether you mean matrices or linear maps is the same: as B,C are invertible they are 1-1 onto maps (matrices) so BA is an map ONTO the image of A ==> rk(BA) = rk(A), and the same with C.

Tonio

Hi -Based on this post, I was trying to look if there is generic rule for finding rank(AB) based on rank(A) and rank(B). I was trying to work that out but not going anywhere.

All I could find was rank(AB)=< Min[rank(A),rank(B)]

Can we say more?

hey tonio, can you rephrase your response solely in terms of matrices, or would the statement be the same?

Originally Posted by hayter221

hey tonio, can you rephrase your response solely in terms of matrices, or would the statement be the same?

It'd be the same using matrix language: instead of dimension of image use rank, instead of kernel use nullity, etc.

Tonio