# Determinant of 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:

- $\ds \mathbf A = \begin {bmatrix} \mathbf A_1 & 0 & \cdots & 0 \\ 0 & \mathbf A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf A_k \end {bmatrix}$

Then the determinant $\map \det {\mathbf A}$ of $\mathbf A$ satisfies:

- $\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^k \map \det {\mathbf A_i}$

## Proof

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To prove this fact, we need to prove additional helper propositions.

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$\textbf{Claim 1}$

The determinant of the block-diagonal matrix of type $M = \begin{pmatrix} A & 0 \\ 0 & I \\ \end{pmatrix}$ or $M = \begin{pmatrix} I & 0 \\ 0 & A \\ \end{pmatrix}$ equals $\map \det A$.

$\textbf{Proof}$

For this we utilize mathematical induction.

$\textbf{Base case}$ - $k = 1$ and $I = I_k = 1$.

Then, $\map \det M = 1 \map \det A + 0 + \ldots + 0$, because the first column (or last, depending on the location of $I$ block) has all zeros except the first (or last) element.

$\textbf{Induction step}$

By the same reason - all except one elements of the column are zeros - we have $\map \det {M_k} = 1 \map \det {M_{k - 1} } + 0 + \ldots + 0$, where $M_k$ is the block matrix discussed above wiith $I_k$ block.

Thus, by induction:

- $\map \det {M_k} = \map \det {M_{k - 1} } = \ldots = \map \det {M_1} = \map \det A$

$\Box$

$\textbf{Claim 2}$ about the determinant of upper-triangular block matrix

$M = \begin{pmatrix} A & B \\ 0 & D \\ \end{pmatrix}$, $\map \det M = \map \det A \map \det D$

$\textbf{Proof}$

$M = \begin{pmatrix} A & B \\ 0 & D \\ \end{pmatrix} = \begin{pmatrix} I & 0 \\ 0 & D \\ \end{pmatrix} \begin{pmatrix} I & B \\ 0 & I \\ \end{pmatrix} \begin{pmatrix} A & 0 \\ 0 & I \\ \end{pmatrix}$. Thus, using the properties $\map \det {A B} = \map \det A \map \det B$, $\map det I = 1$ and $\det \begin{pmatrix} I & B \\ 0 & I \\ \end{pmatrix} = 1$, because this is just triangular matrix with all ones on the diagonal.

So, we get $\map \det M = \map \det D \map \det A$

$\Box$

From the above propositions one can see that for $ A = \begin{pmatrix}
A_1 & 0 \\
0 & A_2 \\
\end{pmatrix}$, which is the special case of the upper-triangular matrix, $\map \det A = \map \det {A_1} \map \det {A_2}$.

Since $A$ is diagonal, $A_1$ and $A_2$ are also diagonal and their determinants equal to the product of corresponding diagonal blocks.

Thus:

- $\map \det {A_1} = \map \det {A_{1, 1} } \map \det {A_{1, 2} }$

and:

- $\map \det {A_2} = \map \det {A_{2, 1} } \map \det {A_{2, 2} }$

which imply:

- $\map \det A = \map \det {A_{1, 1} } \map \det {A_{1, 2} } \map \det {A_{2, 1} } \map \det {A_{2, 2} }$

Following this recursion argument, $\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^k \map \det {\mathbf A_i}$

$\blacksquare$

## Also see

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