- 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
3.11. $\quad$ Exercises
2. (a) $\quad$ Extend Theorem 3.7 to block-diagonal matrices with three diagonal blocks:
$$\det \begin{bmatrix} A & O & O \\ O & B & O \\ O & O & C \end{bmatrix} = (\det A)(\det B)(\det C).$$
(b) $\quad$ State and prove a generalization for block-diagonal matrices with any number of diagonal blocks.
Solution. $\quad$
(a) $\quad$ Extend Theorem 3.7 to block-diagonal matrices with three diagonal blocks: $$\det \begin{bmatrix} A & O & O \\ O & B & O \\ O & O & C \end{bmatrix} = (\det A)(\det B)(\det C).$$ Proof. $\quad$ We can express the upper left-hand corner of the matrix as $A' = \begin{bmatrix} A & O \\ O & B \end{bmatrix}.$ Accordingly, we can rewrite the three-block diagonal matrix as a two-block diagonal matrix: \begin{align*} \begin{bmatrix} A & O & O \\ O & B & O \\ O & O & C \end{bmatrix} &= \begin{bmatrix} A' & O \\ O & C \end{bmatrix}. \end{align*} Applying Theorem 3.7, we find that \begin{align*} \det \begin{bmatrix} A & O & O \\ O & B & O \\ O & O & C \end{bmatrix} &= \det \begin{bmatrix} A' & O \\ O & C \end{bmatrix} \\ &= (\det A')(\det C) \end{align*} But because $A' = \begin{bmatrix} A & O \\ O & B \end{bmatrix},$ is a block-diagonal matrix, we can apply Theorem 3.7 once more to get $\det A' = (\det A)(\det B),$ giving us the extended theorem: $$\det \begin{bmatrix} A & O & O \\ O & B & O \\ O & O & C \end{bmatrix} = (\det A)(\det B)(\det C). \quad \blacksquare$$
(b) $\quad$ State and prove a generalization for block-diagonal matrices with any number of diagonal blocks.
For a square matrix with diagonal blocks $A_1, \dots, A_n,$ where $n$ is an arbitrary integer $\geq 1,$ we have
\begin{align*}
\det \begin{bmatrix}
A_1 & O & \cdots & O \\
O & A_2 & \cdots & O \\
\vdots & \vdots & \ddots & \vdots \\
O & O & \cdots & A_n
\end{bmatrix} &= (\det A_1)(\det A_2)\cdots(\det A_n)
\end{align*}
The proof of this generalized theorem is an extension of that used in part (a). We first denote the $n - 1 \times n - 1$ array containing the diagonal blocks $A_1, \dots, A_{n-1}$ as $A'_{n-1}$ and apply Theorem 3.7 to get $(\det A'_{n-1})(\det A_n).$ Applying this process once more, we get $(\det A'_{n-2})(\det A_{n-1})(\det A_n).$ Then, applying the process an additional $n - 1$ times, we reach our generalization $(\det A_1)(\det A_2)\cdots(\det A_n). \quad \blacksquare$