# Proof: $\text{rank}(A) + \text{rank}(B) \le4$, if matrices multiplication $A\times B = 0$

Problem :

Suppose $A$ is a $3$ by $4$ and $B$ is $4$ by $5$ and $AN=0$. So $N(A)$ contains $C(B)$. Prove from the dimensions of $N(A)$ and $C(B)$ that $\text{rank}(A)+\text{rank}(B)\le4$

This is my solution:

Because matrix A is $3\times4$, then $\max(\text{rank}(A)) = 3$ by definition.

We know that $N(A)$ contains $C(B)$. We can consider $AB = 0$ as $Ax = 0$, where $x = B$. As we know, the dimension of Null space is $n - r$, where $n$ is number of columns and $r$ is a rank of a matrix. So, if matrix $A$ has maximum $\text {rank}(3)$, than dimension of $N(A) = 4-3 = 1$. When matrix $A$ has minimum $\text{rank}(1)$, then dimension of $N(A) = 4-1 = 3$. As we know, $C(B)$ lies in the $N(A)$ (because of $AB = 0$).

Be definition, dimension of a space = number of vectors in every basis. It means, that matrix$(B) = N(A)$ has rank number of columns of $A$ - rank of $A$.

So rank of $A$ varies from $1$ to $3 \implies$ rank of matrix B varies from $3$ to $1$.

As a result $\text{rank}(A) + \text{rank}(B) \le4$

Is my logic(proof) correct? Can you check it please.

$\operatorname{rk}A + \dim \ker A = 4$.
Since ${\cal R}B \subset \ker A$, we have $\operatorname{rk} B \le \dim \ker A$, hence $\operatorname{rk}A + \operatorname{rk}B \le 4$.
• What is $\mathcal R$ here? I've never seen this notation before in linear algebra. Sep 26 '16 at 15:24