Rank and matrix dimensions

Question

Let $A$ be $m \times n$ and $B$ be $n \times p$, and suppose $AB = 0$.
Explain why rank$A$ + rank$B$ $\leq n$.

So we know the resulting matrix will be of dimensions $ m \times p$, but not sure how rank is involved with $AB$ equaling $0$.

Answer 1

If AB = 0 then the column space of B is a subspace of the nullspace of A.

So rank(B) = dim(column space of B).

By the definition of dimension dim(Null(A)) ≥ dim(rank(B)).We also know dim(Null(A)) + dim(rank(A)) = n and dim(Null(B)) + dim(rank(B)) = n.

Upon substitution we have n−dim(rank(A)) ≥ dim(rank(B)). Thus n ≥ dim(rank(A)+dim(rank(B)).

Answer 2

$AB=0$ implies the column space of $B$ is contained in the nullspace of $A$.

Thus $rank(B) \le null(A) = n - rank(A)$, where the last equality is by the rank-nullity theorem.

Answer 3

Image of $B$ is subspace of kernel of $A$. So if image of $n$ dimensional, rank of $A$ will $n$ less than dimension of $A$.