Show the determinant of a Product of Non-Square Matrices equals Zero if Multiplied so that the Larger Matrix is Output Suppose I have a $n \times m$ matrix $A$ and an $m \times n$ matrix $B$. 
Also suppose that $m > n$. 
If I multiply $A \cdot B$, I will obtain an $n \times n$ square matrix $C_{Small}$
If I multiply $B \cdot A$, I will obtain an $m \times m$ square matrix $C_{Big}$
As $m>n$ this matrix will be larger in size.
Is it possible to prove that the determinant of matrix $C_{Big}$ will always be equal to zero while that of matrix $C_{Small}$ will not always be zero.
Effectively I need to show that the matrix $C_{Big}$ is not linearly independent. 
And I would like to see that this does not necessarily follow when I calculate the determinant of the $C_{Small}$ matrix.
Thanks.
 A: You can use the rank of a matrix, and how it varies with the product of two matrices (which we denote by $X$ and $Y$):-
$$\text{rank}(XY)\le \min(\text{rank}(X),\text{rank}(Y))$$
For the matrices in question, the big matrix product $C_{Big}$ will always have a rank less than its dimension of $m$ (a maximum value of $n$, as matrices $A$ and $B$ will have a maximum rank of $n$ each), and so will have linearly dependent rows/columns. Its determinant will be zero.  
On the other hand the small product $C_{Small}$ will have a maximum rank of $n$, which equals its dimension, so it can have a non-zero determinant. For this to hold, a necessary (but not sufficient) condition is that both $A$ and $B$ have ranks equal to $n$.
A: It's simple to prove that $\det(C_{Big}) = 0$:
$$\det(C_{Big}) = \det(B_{m \times n}\ A_{n \times m}) = \det\left(
\begin{bmatrix}
    b_{11}       & \dots  & b_{1n} \\
    \vdots & & \vdots\\
    b_{m1}       & \dots & b_{mn}
\end{bmatrix}_{m \times n}
\begin{bmatrix}
    a_{11}       & \dots  & a_{1m} \\
    \vdots & & \vdots\\
    a_{n1}       & \dots & a_{nm}
\end{bmatrix}_{n \times m}
\right) \stackrel{*}{=} \\ 
\stackrel{*}{=} \det\left(
\begin{bmatrix}
    b_{11}       & \dots  & b_{1n} & 0_{1(n+1)} & \dots & 0_{1m}\\
    \vdots & & & & & \vdots\\
    b_{m1}       & \dots & b_{mn} & 0_{m(n+1)} & \dots & 0_{mm}
\end{bmatrix}_{m \times m}
\begin{bmatrix}
    a_{11}       & \dots  & a_{1m} \\
    \vdots & & \vdots\\
    a_{n1}       & \dots & a_{nm} \\
    0_{(n+1)1}       & \dots & 0_{(n+1)m} \\
    \vdots & & \vdots\\
    0_{m1}       & \dots & 0_{nm}
\end{bmatrix}_{m \times m}
\right) = \\ 
= \det
\begin{bmatrix}
    b_{11}       & \dots  & b_{1n} & 0_{1(n+1)} & \dots & 0_{1m}\\
    \vdots & & & & & \vdots\\
    b_{m1}       & \dots & b_{mn} & 0_{m(n+1)} & \dots & 0_{mm}
\end{bmatrix}_{m \times m}\
\det\begin{bmatrix}
    a_{11}       & \dots  & a_{1m} \\
    \vdots & & \vdots\\
    a_{n1}       & \dots & a_{nm} \\
    0_{(n+1)1}       & \dots & 0_{(n+1)m} \\
    \vdots & & \vdots\\
    0_{m1}       & \dots & 0_{nm}
\end{bmatrix}_{m \times m} = \\= 0 \cdot 0 =0$$
I marked an equation with an $*$ icon. This equation holds even without the $\det$ sign on both sides, since mutliplying the smaller matrices gives the same result as multiplying the ones with extra $0$-coloumns and $0$-rows. Those coloumns and rows simply don't get added to the resulting matrix's values:
$$[B_{m \times n}\ A_{n \times m}]_{ij} = b_{i1}a_{1j} + \dots + b_{in} a_{nj}   = \\ = b_{i1}a_{1j} + \dots + b_{in}a_{nj} + (0_{i(n+1)}0_{(n+1)j} + \dots + 0_{im}0_{mj}) = [B_{m \times m}\ A_{m \times m}]_{ij}$$
Since $\det(AB) = \det(A)\det(B)$, and the resulting 2 matrices either have $0$-coloumn(s) or $0$-row(s), they both, and resulting matrix also has determinant $0$.
As for $\det(C_{Small})$, it can be $0$, is specific cases, but in most cases it isn't, for example:
$$\det\left(\begin{bmatrix}
1 & 2
\end{bmatrix}
\begin{bmatrix}
3 \\
4
\end{bmatrix}\right) = \det([11]) = 11$$
But:
$$\det\left(\begin{bmatrix}
4 & -6
\end{bmatrix}
\begin{bmatrix}
9 \\
6
\end{bmatrix}\right) = \det([0]) = 0.$$
