Question

Every column of AB is a combination of the columns of A. Then the dimensions of the column spaces give rank(AB) ≤ rank(A). Problem: Prove also that rank(AB) ≤ rank(B)

Answer 1

Answer:

The proof is given below:

Step-by-step explanation:

We recall that the rank of a matrix MM is the dimension of the range R(M)R(M) of the matrix MM.

So we have

rank(AB)=dim(R(AB)),rank(A)=dim(R(A)).rank(AB)=dim(R(AB)),rank(A)=dim(R(A)).

In general, if a vector space VV is a subset of a vector space WW, then we have

dim(V)≤dim(W).dim (V)≤dim(W).

Thus, it suffices to show that the vector space R(AB)R(AB) is a subset of the vector space R(A)R(A).

Consider any vector y∈R(AB)y∈R(AB). Then there exists a vector x∈Rlx∈Rl such that y=(AB)xy=(AB)x by the definition of the range.

Let z=Bx∈Rnz=Bx∈Rn.

Then we have

y=A(Bx)=Azy=A(Bx)=Az

and thus the vector yy is in R(A)R(A). Thus R(AB)R(AB) is a subset of R(A)R(A) and we have

rank(AB)=dim(R(AB))≤dim(R(A))=rank(A) [Proved]