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

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

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 $\mathrm{rank}(A)+\mathrm{rank}(B) = 2n\gt n$).

By the definition of matrix multiplication, every column in the matrix $B$ is a solution to the homogeneous equation $A{\bf x} = {\bf 0}$, since $AB = {\bf 0}$. In other words you can say that the span of columns of $B$ is a subset of the span of the solutions of the equation $A{\bf x} = {\bf 0}$.

To clear it up, $W_B^c \subseteq P(A)$, when $W_B^c$ is the span of the columns of $B$, and $P(A)$ is the span of the solutions of the equation $A{\bf x} = {\bf 0}$.

From here you know that the dimension of $W_B^c$ is smaller or equal to the dimension of $P(A)$, $\dim(W_B^c) \leq \dim(P(A))$. Also, the span of the columns is the rank of columns of $B$ in other words. Combine that with $\dim(P(A)) = n - \rho(A)$ and you get $\rho(B) \leq n - \rho(A)$ which is what you were looking for $\rho(A) +\rho(B) \leq n$.