- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
3.10 $\quad$ The determinant of a block-diagonal matrix
$\quad$ A square matrix $C$ of the form \begin{align*} C &= \begin{bmatrix} A & O \\ O & B \end{bmatrix}, \end{align*} where $A$ and $B$ are square matrices and each $O$ denotes a matrix of zeros, is called a block-diagonal matrix with two diagonal blocks $A$ and $B.$ An example is the $5 \times 5$ matrix \begin{align*} C &= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 4 & 5 & 6 \\ 0 & 0 & 7 & 8 & 9 \\ \end{bmatrix}. \end{align*} The diagonal blocks in this case are \begin{align*} A &= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}. \end{align*}
$\quad$ The next theorem shows that the determinant of a block-diagonal matrix is the product of the determinants of its diagonal blocks.
$\quad$ Theorem 3.7. $\quad$ For any two square matrices $A$ and $B,$ we have \begin{align*} \det \begin{bmatrix} A & O \\ O & B \end{bmatrix} &= (\det A)(\det B). \end{align*}
$\quad$ Proof. $\quad$ Assume $A$ is $n \times n$ and $B$ is $m \times m.$ We can express the block-diagonal matrix as a product of the form \begin{align*} \begin{bmatrix} A & O \\ O & B \end{bmatrix} &= \begin{bmatrix} A & O \\ O & I_m \end{bmatrix}\begin{bmatrix} I_n & O \\ O & B \end{bmatrix} \end{align*} where $I_m$ and $I_n$ are the $m \times m$ and $n \times n$ identity matrices, respectively. Then, by the product formula for the determinant, we have \begin{align*} \det \begin{bmatrix} A & O \\ O & B \end{bmatrix} &= \det \begin{bmatrix} A & O \\ O & I_m \end{bmatrix} \det \begin{bmatrix} I_n & O \\ O & B \end{bmatrix}. \end{align*} But because the upper right-hand block of $\begin{bmatrix} A & O \\ O & I_m \end{bmatrix}$ contains zeros, we can treat $\det \begin{bmatrix} A & O \\ O & I_m \end{bmatrix}$ as a function of the $n$ rows of $A.$ We can verify that this function satisfies the axioms of a determinant of order $n,$ and thus by the uniqueness of the determinant, we have $\det \begin{bmatrix} A & O \\ O & I_m \end{bmatrix} = \det A.$ And by a similar argument, we have $\begin{bmatrix} I_n & O \\ O & B \end{bmatrix} = \det B,$ hence $ \det \begin{bmatrix} A & O \\ O & B \end{bmatrix} = (\det A)(\det B). \quad \blacksquare$