Do you know that the rank of a matrix is equal to the rank of the transposed? I.e. for all matrices , we have .
Given that you know the one inequality to be true, you can use this to prove the other.
Every column of AB is a linear combination of the columns of A. Then, Rank (AB) <= Rank (A).
Prove Rank (AB) <= Rank (B).
I see how Rank (AB) <= Rank (A) because the rankk of A is the number of its linearly independent columns, and AB will have the same number of less of linealry independent columns. But how can I prove this for B?
Here's another, perhaps more basic, way of looking at it. For linear transformation A, from vector space V to vector space U, the "rank of A", rank(A), is the dimension of A(V), a subspace of U. If A is invertible, then A(V) has the same dimension as V. If not, then A(V) has smaller dimension than V. That is, . (In fact, where " " is the dimension of the nullspace of A but you don't need to use that here.)
If B is a linear transformation from U to V and A is a linear transformation from V to W, then AB is a linear transformation from U to W. For any vector v, in U, ABv is, by definition, A(Bv) so any vector in AB(U) is A applied to a vector in B(U). The dimension of B(U) is rank(B) and, as we just saw, AB(U) must have dimension no larger than that- .