If $A\in M_{5,4}$, $\operatorname{rank}A=3$ and $B\in M_{4,3}$, $\operatorname{rank}B=2$ then $AB$ must be a non-zero matrix.

Question

Answer with TRUE or FALSE and give a proper explanation.

If $A\in M_{5,4}$, $\operatorname{rank}A=3$ and $B\in M_{4,3}$, $\operatorname{rank}B=2$ then $AB$ must be a non-zero matrix.

My approach:

From the information given above we can describe $A$,$B$ and $AB$ like this:

$A:\mathbb{R}^4 \rightarrow \mathbb{R}^5$

$B:\mathbb{R}^3 \rightarrow \mathbb{R}^4$

$AB:\mathbb{R}^3 \rightarrow \mathbb{R}^5$

We know that $\operatorname{rank}A + \dim N(A)=\dim\mathbb{R}^4 \Rightarrow \dim N(A)=1$

$R(B)$,$N(A) \in \mathbb{R^4}$.

Since $\operatorname{rank}B \gt \dim N(A)$ there exists an vector $y$ such that $y\in R(B),y\notin N(A)$. So $y=Bx$ and $Ay\neq0 \Rightarrow ABx\neq 0$.

Now I don't know if the statement is TRUE or FALSE because I've found an vector such that $AB\neq0$ but what if there exists an vector that is both in $R(B)$ and in $N(A)$? That would mean that $AB=0$ and so the answer would be FALSE, but I'm not sure does such an vector exists?

Answer 1

Since $R(B)$ has dimension $2$ and $N(A)$ has dimension $1$ (by the Rank-Nullity Theorem), it must be that $R(B)\nsubseteq N(A)$ and thus $AB\neq 0$.

Answer 2

This is a direct application of the Sylvester's rank inequality: If $A \in M_{m \times n}$, $B \in M_{n \times k}$, then $$\text{rank}(AB) \geq \text{rank}(A) + \text{rank}(B) - n.$$

In your case, you have $\text{rank}(AB) \geq 3 + 2 - 4 = 1 > 0$, so the statement is true.