TITLE: Multiplying matrices / corresponding systems of equations
QUESTION [0 upvotes]: 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 $AB=0$:

Prove there is a non-trivial solution to the system of equations $Ax=0$
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.

REPLY [1 votes]: Solved it.

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 $B$. 
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.
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