# Thread: Rank of a Matrix

1. ## Rank of a Matrix

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?

2. Do you know that the rank of a matrix is equal to the rank of the transposed? I.e. for all matrices $A$, we have $\text{Rank}(A) = \text{Rank}(A^T)$.

Given that you know the one inequality to be true, you can use this to prove the other.

3. 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, $rank(A)\le dim(V)$. (In fact, $Rank(A)+ nullity(A)= dim(V)$ where " $nullity(A)$" 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- $rank(AB)\le rank(B)$.