Determinant of Matrix Product

Theorem

Let $\mathbf A = \left[{a}\right]_n$ and $\mathbf B = \left[{b}\right]_n$ be a square matrices of order $n$.

Let $\det \left({\mathbf A}\right)$ be the determinant of $\mathbf A$.

Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.

Then:

$\det \left({\mathbf A \mathbf B}\right) = \det \left({\mathbf A}\right) \det \left({\mathbf B}\right)$

That is, the determinant of the product is equal to the product of the determinants.

Proof 1

Let $\mathbf C = \left[{c}\right]_n = \mathbf A \mathbf B$.

Thus:

$\displaystyle \forall i, j \in \left[{1 .. n}\right]: c_{i j} = \sum_{l=1}^n a_{i l} b_{l j}$

Then:

	\(\displaystyle \)	\(\displaystyle \det \left({\mathbf C}\right)\)	\(=\)	\(\displaystyle \sum_{\lambda} \left({\operatorname {sgn} \left({\lambda}\right) \prod_{k=1}^n c_{k \lambda \left({k}\right)} }\right)\)	\(\displaystyle \)	by definition of determinant
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{\lambda} \left({\operatorname {sgn} \left({\lambda}\right) \prod_{k=1}^n \left({\sum_{l=1}^n a_{k l} b_{l \lambda \left({k}\right)} }\right)}\right)\)	\(\displaystyle \)	by definition of matrix product

This gets messy very quickly, so we can try another approach.

From Square Matrix Row Equivalent to Triangular Matrix, you can turn a square matrix into a triangular matrix by using elementary row operations that, from Effect of Elementary Row Operations on Determinant, do not change either the value or the sign of its determinant.

So suppose:

• $\mathbf A$ is row equivalent to the upper triangular matrix $\mathbf T_A$
• $\mathbf B$ is row equivalent to the upper triangular matrix $\mathbf T_B$

such that:

• $\det \left({\mathbf A}\right) = \det \left({\mathbf T_A}\right)$
• $\det \left({\mathbf B}\right) = \det \left({\mathbf T_B}\right)$

From Determinant of a Triangular Matrix, we have that $\det \left({\mathbf T_A}\right)$ and $\det \left({\mathbf T_B}\right)$ are equal to the product of their diagonal elements.

From Product of Triangular Matrices, it also follows that $\mathbf T_A \mathbf T_B$ is an upper triangular matrix whose diagonal elements are the products of the diagonal elements of $\mathbf T_A$ and $\mathbf T_B$.

Thus it follows that:

$\det \left({\mathbf T_A \mathbf T_B}\right) = \det \left({\mathbf T_A}\right) \det \left({\mathbf T_B}\right)$

Thus it follows that:

$\det \left({\mathbf A \mathbf B}\right) = \det \left({\mathbf A}\right) \det \left({\mathbf B}\right)$

$\blacksquare$

Proof 2

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 $\det \left({\mathbf A}\right) = 0$.

Also if $\mathbf A$ is not invertible then neither is $\mathbf A \mathbf B$, and so

$\det\left({\mathbf A \mathbf B}\right) = 0$

Thus:

$0 = 0 \cdot \det\left({\mathbf B}\right)$

$\det \left({\mathbf A\mathbf B}\right) = \det \left({\mathbf A}\right) \cdot \det\left({\mathbf B}\right)$

Proof of case $(2)$:

Assume $\mathbf A$ is invertible.

Then $\mathbf A$ is a product of elementary matrices, $\mathbf E$.

Let $\mathbf A = \mathbf E^{k} \mathbf E^{k-1} \cdots \mathbf E^{1}$.

So:

$\det \left({\mathbf A\mathbf B}\right) = \det \left({\mathbf E^{k}\mathbf E^{k-1} \cdots \mathbf E^{1} \mathbf B}\right)$

But for any matrix $\mathbf D$:

$\det \left({\mathbf E \mathbf D}\right) = \det \left({\mathbf E}\right) \cdot \det\left({\mathbf D}\right)$

by Effect of Elementary Row Operations on Determinant.

Therefore:

$\det \left({\mathbf A \mathbf B}\right) = \det \left({\mathbf E^{k}}\right) \det\left({\mathbf E^{k-1}}\right) \cdots \det\left({\mathbf E^{1}}\right) \det\left({\mathbf B}\right)$

$\det\left({\mathbf A \mathbf B}\right) = \det\left({\mathbf E^{k} \mathbf E^{k-1} \ldots \mathbf E^{1}}\right) \det \left({\mathbf B}\right)$

$\det \left({\mathbf A \mathbf B}\right) = \det\left({\mathbf A}\right) \cdot \det\left({\mathbf B}\right)$

as required.

Sources

• John F. Humphreys: A Course in Group Theory (1996): $\text A.2$: Theorem $\text A.9 \ (1)$