Prove that det(BA) = 0 under some circumstances How to prove that:
$ det(BA) = 0 $  Assuming:
$ m < n, A \in M_{mxn}, B\in M_{nxm}$?
 A: rank$(BA)\le $min(rank(A),rank (B))$= m<n$
Now as $AB$ is a $n\times n$ matrix and as its rank is less than $n$ its rows(columns) are not independent vectors implying $det(BA)=0$
Proof of the fact: rank$(BA)\le $min(rank(A),rank (B))$
$\mbox{rank } A\le m$ and $\text{rank }B\le m$(Obvious fact as rank (A)=dimension of the columnspace of A=dimension of the row space of A)
Let $E_{n\times n}B$ be the row echelon form of $B$ and let $AE_{m\times m} $ be the column echelon form of $A$.($E_{n\times n} ,E_{m\times m}$ are elementary matrices)
We know $rank (BA)=rank(E_{n\times n}BAE_{m\times m} )$ 
But $E_{n\times n}BAE_{m\times m} =\begin{pmatrix}
L&0\\
0&0\\
\end{pmatrix}$
where $L$ is an $k\times l$ matrix with $k\le rank (B),l\le rank(A)$.
so rank $(E_{n\times n}BAE_{m\times m} )$=rank $\begin{pmatrix}
L&0\\
0&0\\
\end{pmatrix}\le \min\{k,l\}\le \min\{\mbox{rank } A,\mbox{rank }B\}$
A: I'm assuming that your convention is so that $BA \in M_{n \times n}$, as it makes your claim true.
I would think in terms of linear transformations and the geometric interpretation of the determinant. Since the image of A has dimension at most $m$, the image of BA also has dimension at most $m$. Since $m<n$, this means the image of a unit $n$-cube under BA has zero $n$-dimensional volume; i.e. $\text{det} (BA) = 0$.
A: Using the fact that rank$(BA)\leq\min \lbrace\mbox{ rank}(B),\mbox{ rank}(A)\rbrace$ we have $$\mbox{ rank}(BA)\leq\min\lbrace m,n\rbrace=m<n$$ Now note that $BA\in M_{n\times n}$ and hence $\det (BA)=0.$
