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)

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]