Question

Can somebody help me out with the following problem:
Every column of AB is a combination of the columns of A. Then the dimension of the column space give rank(AB) <= rank(A). Problem: Prove also that rank(AB) <= rank(B).

Answer 1

I've got 2 ways of looking at this; first, if you're happy with the fact that rank(A')=rank(A), you can use the result rank(AB)<=rank(A) to infer rank(B'A')<rank(B'), then you get the result you want by rank(B'A')=rank((B'A')')=rank(AB) and rank(B')=rank(B).

Alternatively, reorder your variables so that the dependent columns of B all come at the end; now you can write your (m,n) matrix B as [P Q] where P is (m,r) and Q is (m,n-r), where r is rank(B). We know that Q can be expressed as a linear combination of columns of P, so Q=PW where W, which is (r,n-r), gives the linear combinations; thus B=[P PW], so AB=[AP APW]; this shows that the last n-r columns of AB are linear combinations of the first r columns, so rank(AB)<=r.