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:
- $\displaystyle \mathbf A = \begin {bmatrix} \mathbf A_1 & 0 & \cdots & 0 \\ 0 & \mathbf A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf A_k \end {bmatrix}$
Then the determinant $\map \det {\mathbf A}$ of $\mathbf A$ satisfies:
- $\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^k \map \det {\mathbf A_i}$
Proof
To prove this fact, we need to prove additional helper propositions.
$\textbf{Claim 1}$
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 $\det(A)$.
$\textbf{Proof}$
For this we utilize mathematical induction.
$\textbf{Base case}$ - $k = 1$ and $I = I_k = 1$.
Then, $\det(M) = 1\det(A) + 0 + \ldots + 0$, 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 $\det(M_k) = 1\det(M_{k-1}) + 0 + \ldots + 0$, where $M_k$ is the block matrix discussed above wiith $I_k$ block.
Thus, by induction:
- $\det(M_k) = \det(M_{k-1}) = \ldots = \det(M_1) = \det(A)$
$\Box$
$\textbf{Claim 2}$ about the determinant of upper-triangular block matrix
$M = \begin{pmatrix} A & B \\ 0 & D \\ \end{pmatrix}$, $\det(M) = \det(A)\det(D)$
$\textbf{Proof}$
$M = \begin{pmatrix} A & B \\ 0 & D \\ \end{pmatrix} = \begin{pmatrix} I & 0 \\ 0 & D \\ \end{pmatrix} \begin{pmatrix} I & B \\ 0 & I \\ \end{pmatrix} \begin{pmatrix} A & 0 \\ 0 & I \\ \end{pmatrix}$. Thus, using the properties $\det(AB) = \det(A)\det(B)$, $det(I) = 1$ 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 $\det(M) = \det(D)\det(A)$
$\Box$
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, $\det(A) = \det(A_{1})\det(A_{2})$.
Since $A$ is diagonal, $A_{1}$ and $A_{2}$ are also diagonal and their determinants equal to the product of corresponding diagonal blocks.
Thus:
- $\det(A_{1}) = \det(A_{1,1})\det(A_{1,2})$
and:
- $\det(A_{2}) = \det(A_{2,1})\det(A_{2,2})$
which imply:
- $\det(A) = \det(A_{1,1})\det(A_{1,2})\det(A_{2,1})\det(A_{2,2})$
Following this recursion argument, $\ds \det \left({\mathbf A}\right) = \prod_{i \mathop = 1}^k \det \left({\mathbf A_i}\right)$
$\blacksquare$
Also see
- Determinant of Diagonal Matrix, a special case of this theorem.