If A is an $5 \times 4$ matrix, and B is an $4 \times 3$ matrix, $rankA=3$ and $rankB=2$ then $A \times B$ must be a non null matrix. Why or why not?

Question

At first, I was thinking that since product AB would be a linear combination of columns of A and rows of B, and A and B are not null, then the product has to be non null, but now I am not sure. Now I think that the product can be a null matrix, but I can't explain how in either case.

Answer 1

Think of the matrices $A$ and $B$ as linear transformations $L_A$ and $L_B$ respectively on column vectors. $AB$ can be represented as the composite function $L_A \circ L_B$.

$\mathrm{rank B} = 2$ suggests that $\dim(\mathrm{Im}(L_B)) = 2$. In other words, the image of $L_B$ has two linearly independent column vectors. $L_A$ maps a vector in $\Bbb{F}^4$ to another vector in $\Bbb{F}^5$. $\mathrm{rank A} = 3$ suggests that $\dim(\mathrm{Im}(L_A)) = 3$, so the dimension of the kernel of $L_A$ is $4-\mathrm{rank A}=4-3 = 1$. (i.e. $\dim N(L_A) = 4 - 3=1$)

Can you cotinue from this?

At most one vector from any basis for $\mathrm{Im}(L_B)$ can be in $N(L_A)$, so $\dim(L_{AB}) = \dim{L_A \circ L_B} \ge 2 - 1 = 1$.