Determinant of Matrix Product/Proof 2
Theorem
Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be a square matrices of order $n$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.
Then:
- $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$
That is, the determinant of the product is equal to the product of the determinants.
Proof
Consider two cases:
- $(1): \quad \mathbf A$ is not invertible.
- $(2): \quad \mathbf A$ is invertible.
Proof of case $1$
Assume $\mathbf A$ is not invertible.
Then:
- $\map \det {\mathbf A} = 0$
Also if $\mathbf A$ is not invertible then neither is $\mathbf A \mathbf B$.
Indeed, if $\mathbf A \mathbf B$ has an inverse $\mathbf C$, then $\mathbf A \mathbf B \mathbf C = \mathbf I$, whereby $\mathbf B \mathbf C$ is a right inverse of $\mathbf A$.
It follows by Left or Right Inverse of Matrix is Inverse that in that case $\mathbf B \mathbf C$ is the inverse of $A$.
It follows that:
- $\map \det {\mathbf A \mathbf B} = 0$
Thus:
- $0 = 0 \cdot \map \det {\mathbf B}$
- $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \cdot \map \det {\mathbf B}$
$\Box$
Proof of case $2$
Assume $\mathbf A$ is invertible.
Then $\mathbf A$ is a product of [[elementary row matrices, $\mathbf E$.
Let $\mathbf A = \mathbf E_k \mathbf E_{k - 1} \cdots \mathbf E_1$.
So:
- $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf E_k \mathbf E_{k - 1} \cdots \mathbf E_1 \mathbf B}$
It remains to be shown that for any square matrix $\mathbf D$ of order $n$:
- $\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \cdot \map \det {\mathbf D}$
Let $e_i \paren {\mathbf I} = \mathbf E_i$ for all $i \in \closedint 1 k$, then using Elementary Row Operations as Matrix Multiplications and Effect of Sequence of Elementary Row Operations on Determinant yields:
- $\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E_k \mathbf E_{k - 1} \dotsm \mathbf {E_1} \mathbf D} = \map \det {e_k e_{k - 1} \cdots e_1 \paren {\mathbf D} } = \alpha \map \det {\mathbf D}$
Using Elementary Row Operations as Matrix Multiplications and Effect of Sequence of Elementary Row Operations on Determinant, and Unit Matrix is Unity of Ring of Square Matrices:
- $\map \det {\mathbf E} = \map \det {\mathbf E_k \mathbf E_{k - 1} \cdots \mathbf {E_1} \mathbf I} = \map \det {e_k e_{k - 1} \cdots e_1 \paren {\mathbf I} } = \alpha \map \det {\mathbf I}$
From Determinant of Unit Matrix:
- $\map \det {\mathbf E} = \alpha$
And so $\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \cdot \map \det {\mathbf D}$
$\Box$
Therefore:
- $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$
as required.
$\blacksquare$