# Determinant of Diagonal Matrix

## Theorem

Let $\mathbf A = \begin{bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end{bmatrix}$ be a diagonal matrix.

Then the determinant of $\mathbf A$ is the product of the elements of $\mathbf A$.

That is:

$\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^n a_{ii}$

## Proof

As a diagonal matrix is also a triangular matrix (both upper and lower), the result follows directly from Determinant of Triangular Matrix.

$\blacksquare$