Question: I'm having some trouble with a problem in linear algebra:

Let $A$ be a matrix with dimensions $m \times n$ and $B$ also a matrix but with dimensions $n \times m$ which is **not** a null matrix. (That's all that's written - I assume A may or may not be a null matrix).

Given that $AB=0$:

1. Prove there is a non-trivial solution to the system of equations $Ax=0$
2. Assume $A\neq0$ . Does the system $Bx=0$ also have a non-trivial solution? If so, prove the argument. If not, provide a contradictory example.

There's a third part to the question but I managed to solve it and its content isn't really relevant here because it provided us a defined $A$ of real numbers, but I'm pretty lost with the first two arguments - I'm having trouble putting what I think into words. Can anyone help with this? Thanks!

EDIT:

Okay so I think I'm supposed to deal with the different cases of $m$ and $n$:

If $n > m$ obviously the system $Ax=0$ has infinite solutions because we'll have more variables than equations.

What I haven't quite figured out is how to prove that:

If $AB=0$ and $m=n$ or $m > n$, then it immediately follows that $Rank(A) < n$ .

Any help with this would be greatly appreciated.

Answer: 
Solved it.

1. If $AB=0$ that means that matrix $A$ multiplied by any column vector $b$ in $B$ will be equal to the zero vector. Since we know that $B\neq 0$, there must be at least one column vector $b$ in $B$ that **isn't** the zero vector. So to summarize, since $A$ multiplied by any column vector in $B$ returns 0 and we know there is a non-zero column vector in $B$, the system of equations $Ax=0$ has at least one non-trivial solution, where $x$ can be the zero vector or a non-zero vector from $B$.
2. I have found a contradictory example. Basically to disprove the argument I need to find matrices $A,B$ that meet the following criteria:

	* $A\_{m\times n}, B\_{n\times m}$
	* $A,B\neq0$
	* $Bx=0$ **only** has a trivial solution

Here they are:

$$
A=\begin{bmatrix}
0&0&0&1\\
0&0&0&0\\
0&0&0&0
\end{bmatrix}\_{3\times 4}\ ,\ 
B=
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
0&0&0
\end{bmatrix}
\\
$$
$$
A\_{m\times n}, B\_{n\times m}\ \ \checkmark
\\
A,B\neq 0\ \ \checkmark
\\
AB=0\ \ \checkmark
\\
Bx=0\rightarrow\ one\ solution\ \ \checkmark 
$$

And there's a perfect contradictory example to the argument.
1. A has to be linearly dependent, since it is the case that $AB = 0$. If B is not the zero matrix, then it has to be the case that there is some row of $A$ in Reduce Row Echelon Form that is a zero row. Because of this, it is linearly dependent. It therefore does not matter if $A$ is the zero matrix or not, since the result is the same. By definition, a linearly dependent matrix has non-trivial solutions to $Ax=0$ (since its nullspace is not $\{0\})$.