Determinant of a Block Diagonal Matrix

Theorem

Let $\mathbf A$ be a block diagonal matrix of order $n$.

Let $\mathbf{A}_1,\ldots,\mathbf{A}_k$ be the square matrices on the diagonal, i.e.:

$\displaystyle \mathbf A = \begin{bmatrix} \mathbf{A}_1 & 0 & \cdots & 0 \\ 0 & \mathbf{A}_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf{A}_k \end{bmatrix}$

Then the determinant of $\mathbf A$, $\det \left({\mathbf A}\right)$, satisfies:

$\displaystyle \det \left({\mathbf A}\right) = \prod_{i=1}^k \det \left({\mathbf{A}_i}\right)$

See also

• Determinant of a Diagonal Matrix, a special case of this theorem.