Show the determinant of a Product of Non-Square Matrices equals Zero if Multiplied so that the Larger Matrix is Output

Question

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.

Answer 1

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 $n$.

Answer 2

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: $$\det\left(\begin{bmatrix} 4 & -6 \end{bmatrix} \begin{bmatrix} 9 \\ 6 \end{bmatrix}\right) = \det([0]) = 0.$$