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.
Best Answer
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 $n$.