Singularity of the product of two rectangular matrices? Let $A$ be an $m \times n$ matrix and $B$ be an $n \times m$ matrix where $m<n$. 
Then can we say that the product $AB_{m \times m}$ is always singular or always non-singular?
Also, can we say that $BA_{n \times n}$ is always singular or non-singular?. Does this change any thing?
I was thinking that since $m<n$ we have Rank$(A) \leq m$ and similarly Rank$(B) \leq n$ and also that Rank$(AB) \leq min($Rank$(A)$,Rank$(B)$), but will that help?
How can I think about this problem?
 A: We have 


*

*$\text{rank}(A)\le m$, since $A$ has $m$ rows.$\\[4pt]$

*$\text{rank}(B)\le m$, since $B$ has $m$ columns.


so
$$\text{rank}(BA)\le\min(\text{rank}(A),\text{rank}(B))\le m < n$$
hence, since $BA$ is an $n{\,\times\,}n$ matrix, $BA$ is singular.

On the other hand, $AB$ may or may not be singular.

All we need is one example of each . . .

To keep it simple, let $m=1,\,n=2$.

For an example where $AB$ ends up being singular, let $A,B$ be given by
$$
A=
\pmatrix
{
1 & 0\cr
}
,\;\;\;
B=
\pmatrix
{
0\cr
1\cr
}
$$
so we have $\text{rank}(A)=\text{rank}(B)=1$, but $AB=0$, which is singular.

For an example where $AB$ ends up being non-singular, let $A,B$ be given by
$$
A=
\pmatrix
{
1 & 0\cr
}
,\;\;\;
B=
\pmatrix
{
1\cr
0\cr
}
$$
so we have $\text{rank}(A)=\text{rank}(B)=1$, and $AB=(1)$, which, as a $1{\,\times\,}1$ matrix, is non-singular.
A: $\textbf{BA is always Singular}$:

Result:  $A$ is $m \times n$ and  $m<n$ implies $Ax=0$ has non zero solution $x_0$.

Proof:

View the matrix $A$ as a linear map $A:\Bbb{F}^n \rightarrow \Bbb{F}^m, x \mapsto Ax.$
By dimension theorem, $$dim\;\Bbb{F}^n=rank\;A+null\;A $$
$$n\leq m+null\;A$$
So, $null\;A \geq n-m >0$,. Hence $Ax=0$ has a non zero solution (say $x_0$). QED

Now $$(BA)x_0=B(Ax_0)=B.0=0$$
Hence $BAx=0$ has non zero solution, concluding $BA$ is singular.
$\textbf{For AB, it may or may not (as in the comment):}$
For example for $m=2$  and $n=3$, Consider
$A=\begin{pmatrix} 1& 0 & 0\\ 0 &1 &0\end{pmatrix}$ and $B=\begin{pmatrix} 1& 0 \\ 0 &1\\ 2 & 2\end{pmatrix}$. Then $AB=I_2$
For the other one, consider $A=\begin{pmatrix} 1& 2 \end{pmatrix}$ and $B=\begin{pmatrix} 2\\ -1 \end{pmatrix}$. Then $AB=0$
