Proof of: $AB=0 \Rightarrow Rank(A)+Rank(B) \leq n$

Question

As the title says, am searching for a proof of

If $A,B \in \mathbb{R}^{n\times n}$ and $AB=0$ then $\mathrm{rank}(A)+\mathrm{rank}(B) \leq n$

I am doing this as preparation for an upcoming exam and can't figure a way to start. Please just post small hints as answers. I will try to go from there.

Thank you

ftiaronsem

Answer 1

By the Rank-Nullity Theorem, $\mathrm{rank}(A)+\mathrm{nullity}(A)=n$. The problem would be solved if you could show that $\mathrm{rank}(B)\leq\mathrm{nullity}(A)$. Presumably, $AB=0$ will play a role in that, since the result is false otherwise (if $A$ and $B$ were both invertible, for example, then $AB\neq 0$, and $\mathrm{rank}(A)+\mathrm{rank}(B) = 2n\gt n$).