Given $A_{m\times n}$ and $B_{n \times m} (mI have the following question which I can't seem to wrap my head around. I don't see how we can determine the desired just from the given info.

Given $A_{m\times n}$ and $B_{n \times m}$ $(m<n)$. prove that AB is not singular (for every A) and BA is singular 

Thanks for the help.
EDIT: Question in another form (closer to original text):

Given $A_{m\times n}$ and $B_{n \times m} (m<n)$. 
1) Is AB Singular? 
2) Is BA Singular? 

My answers say that 1 is false and 2 is true.
 A: If $A$ is $ m \times n $ for $ m < n $, then the columns of $A$ are vectors in $\mathbb{R}^m$ (or whatever the field is). Since there are more than $ m $ of them, they must be linearly dependent and so there are nontrivial solutions to $A\mathbf{x} = \mathbf{0}$. 
Multiplying by $B$ on the left of both sides of this equation gives $BA \mathbf{x} = B\mathbf{0}$ so $BA \mathbf{x} = \mathbf{0}$ (albeit a different zero vector). Since $A\mathbf{x} = \mathbf{0}$ had a nontrivial solution, so must $(BA)\mathbf{x} = \mathbf{0}$ and hence $BA$ is singular.
It is possible that $AB$ is singular or non-singular. For example, if $A$ is a zero matrix then $AB$ is also a zero matrix (of a different size) and so must be singular. However, for example, if $A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{pmatrix}$ then $AB = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $AB$ is nonsingular.
In general, if the columns of $B$ define combinations of the columns of $A$ that are independent, then $AB$ will be nonsingular.
