How can I show that [if an $m\times m$ matrix $A$ is not invertible,] there is an $m\times m$ matrix $B$ where $AB=0$ but $B$ is not equal to $0$?
I am including the following, in case you have not yet encountered the definitions or concepts of "null space", "rank/dimension", or "kernel":
You answered your own question in that you need only show that when $A$ is not invertible, then there is some (indeed many) non-zero matrix $B$ for which $AB = 0$. To answer this question, it suffices to show existence of a non-zero matrix $B$ such that $AB \neq 0$. That is, you thereby show that $B$ need not be identically zero.
You can prove this by induction on $m$,
- Show the proposition is true for the base case $m = 1$.
- Assume the proposition holds for $(m-1)\times (m-1)$ matrices $A$ and $B$.
- Show that this assumption implies that the proposition must be true for $m\times m$ matrices, as well.
- Conclusion: for all $m\times m$ non-invertible matrices $A$, there is an $m \times m$ matrix $B$ such that $B \neq 0$, but $AB = 0$.
You can also use the property of the determinant to get more intuition about this problem (property: $\text{det}\,(AB) = \text{det}\,(A)\,\text{det}\,(B)$):
Taking the determinant of each side of the equation $AB = 0$, and knowing that the determinant of an non-invertible matrix $A$ is equal to $0$, we have:
$\text{det}(AB) = \text{det}(0)$ $\iff \text{det}A \;\text{det}B = 0\cdot \text{det} B = 0$ for all matrices $B$. We know that for every invertible matrix $B$, $B\not \equiv 0$, and we also know that for an invertible $B$, $\text{det}B \neq 0$, but nonetheless $\text{det} A \;\text{det} B = 0\cdot \text{det} B = 0$.