$AB$ is not invertible Is it true that if $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix, with $m > n $ then $\det(AB)=0$?
 A: Since $rank(AB)\le rank(A)\le \min\{m,n\}=n<m$ and $AB$ is an $m\times m$ matrix,
$AB$ is not invertible and therefore $\det(AB)=0$.
A: Since $m>n$, we must have that $\ker B$ is non-trivial (look at row canonical form, for example). Hence $Bv=0$ for some $v \neq 0$, and so $ABv=0$. It follows that $AB$ is singular and that $\det (AB) = 0$.
A: If
$$
A=(a_{ij}) \in \mathbb{R}^{m\times n}, \ B=(b_{ij})\in \mathbb{R}^{n\times m},
$$
with $m>n$, then
$$
C:=AB=(c_{ij}) \in \mathbb{R}^{m\times m},
$$
with
$$
c_{ij}=\sum_{k=1}^na_{ik}b_{kj}.
$$
For $j=1,\ldots,m$ let 
$$
\mathbf{c}_j=Ce_j \in \mathbb{R}^m,
$$
where $(e_1,\ldots, e_m)$ stands for the canonical basis of $\mathbb{R}^m$.
Then we have
$$
\mathbf{c}_j=\left[
\begin{array}{c}
\sum_{k=1}^na_{1k}b_{kj}\\
\vdots\\
\sum_{k=1}^na_{mk}b_{kj}
\end{array}
\right]=\sum_{k=1}^nb_{kj}
\left[
\begin{array}{c}
a_{1k}\\
\vdots\\
a_{mk}
\end{array}
\right]=
\sum_{k=1}^nb_{kj}\mathbf{a}_k,
$$
with
$$
\mathbf{a}_j=\left[
\begin{array}{c}
a_{1j}\\
\vdots\\
a_{mj}
\end{array}
\right] \in \mathbb{R}^n.
$$
We now have
\begin{eqnarray}
\det(C)&=&\det(\mathbf{c}_1,\ldots,\mathbf{c}_m)=\sum_{k_1=1}^n\sum_{k_2=1}^n\ldots\sum_{k_m=1}^nb_{k_1,1}b_{k_2,2}\ldots b_{k_m,m}\det(\mathbf{a}_{k_1},\ldots,\mathbf{a}_{k_m})
\end{eqnarray}
Since there are exactly $n$ vectors $\mathbf{a}_j$, and the determinant has $m>n$ entries we have
$$
\det(\mathbf{a}_{k_1},\ldots,\mathbf{a}_{k_m})=0 \quad \forall k_1,\ldots,k_m \in \{1,\ldots,n\}
$$
because at least two entries must be equal. Hence
$$
\det(C)=0.
$$
A: Twink, there is another way of looking at your statement.
Take for example a 3x2 matrix in which you choose your numbers. Then take a 2x3 matrix in which you put letters for your entries, say a,b,c,d,e,f
Now perform matrix multiplication. You will end up with a 3 by 3 matrix, but more interestingly, examine its rows. All your coefficients are the same when you compare the entries! In fact, no matter what you ;pick for your a,b,c,... You will have in each row the same value!. A simple row opration then shows that there is linear dependence. Now my approach is NOT a rigorous prove like colleagues above have done. It's just a different approach as to see how your statement "works"
A: Basically the same as cooper.hat's answer with lower technology.
Consider the system $$B\mathbf x = 0.$$ Since $B$ is an $n\times m$ matrix and $n\lt m$, this system has infinite solutions. So there is one non trivial solution say $\mathbf y$. But then $\mathbf y$ is also a nontrivial solution to the system $$(AB)\mathbf x = 0$$
and so $AB$ is not invertible.
So $\det(AB)=0$.
