Determinant of Matrix Product/General Case
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Theorem
Let $\mathbf A_1, \mathbf A_2, \cdots, \mathbf A_n$ be square matrices of order $n$, where $n > 1$.
Then:
- $\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_n} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_n}$
where $\det \mathbf A$ denotes the determinant of $\mathbf A$.
Proof
Proof by induction:
Basis for the Induction
$n = 2$ holds by Determinant of Matrix Product.
So shown for base case.
Induction Hypothesis
This is our induction hypothesis:
- $\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k}$
Now we need to show true for $n = k + 1$:
- $\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k \mathbf A_{k + 1} } = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k} \map \det {\mathbf A_{k + 1} }$
Induction Step
This is our induction step:
\(\ds \map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k \mathbf A_{k + 1} }\) | \(=\) | \(\ds \map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k} \map \det {\mathbf A_{k + 1} }\) | Determinant of Matrix Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k} \map \det {\mathbf A_{k + 1} }\) | Induction hypothesis |
The result follows by induction:
- $\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_n} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_n}$
$\blacksquare$