If B and C are invertible then prove rank(BA)=rank(AC)=rank(A).
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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.Quote:
Originally Posted by hayter221
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?
Quote:
Originally Posted by hayter221
It'd be the same using matrix language: instead of dimension of image use rank, instead of kernel use nullity, etc.
Tonio