Exercise 4.1.30

When $\mathit{AB}=0$, every column of $B$ is multiplied by $A$ to give zero. So the column space of $B$ is contained in the nullspace of $A$. Therefore the dimension of $C(B)\le$ dimension of $N(A)$. This means $\mathrm{rank}<!--\; FUNCTION\; APPLICATION\; -->(B)\le 4-\mathrm{rank}<!--\; FUNCTION\; APPLICATION\; -->(A)$.