Let \(A\) be an \(n \times n\) matrix over \(K\), a commutative ring with identity. Suppose \(A\) has the block form \[ A = \begin{bmatrix} A_1 & 0 & \cdots & 0 \\ 0 & A_2 & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & A_k \end{bmatrix} \] where \(A_j\) is an \(r_j \times r_j\) matrix. Prove \[ \det A = (\det A_1)(\det A_2) \cdots (\det A_k). \]
Solution
1. Two-block case:
1.1. Let \(A = \begin{pmatrix} B & 0 \\ 0 & C \end{pmatrix}\), where \(B \in M_r(K)\) and \(C \in M_s(K)\) with \(r + s = n\).By the Leibniz formula for the determinant of \(A = (a_{ij})\):\[ \det A = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n a_{i, \sigma(i)}. \]
definition of determinant.
1.2. The matrix entries satisfy \(a_{i, j} = 0\) whenever \(1 \le i \le r < j \le n\) or \(1 \le j \le r < i \le n\).Thus a product \(\prod_{i=1}^n a_{i, \sigma(i)}\) is non-zero only if \(\sigma(\{1, \dots, r\}) = \{1, \dots, r\}\) and \(\sigma(\{r+1, \dots, n\}) = \{r+1, \dots, n\}\).
block-diagonal structure has zero off-diagonal blocks.
1.3. Such permutations \(\sigma \in S_n\) are precisely the direct product \(\sigma = \tau \times \pi\) where \(\tau \in S_r\) and \(\pi \in S_s\).Because there are no inversions between \(\{1, \dots, r\}\) and \(\{r+1, \dots, n\}\) (since \(\tau(i) \le r < \pi(j)\) for all \(i \le r < j\)), the sign decomposes as:\[ \operatorname{sgn}(\sigma) = \operatorname{sgn}(\tau) \operatorname{sgn}(\pi). \]
parity of disjoint block permutations.
1.4. Factoring the summation:\[ \begin{aligned} \det A &= \sum_{\tau \in S_r} \sum_{\pi \in S_s} \operatorname{sgn}(\tau)\operatorname{sgn}(\pi) \left( \prod_{i=1}^r B_{i, \tau(i)} \right) \left( \prod_{j=1}^s C_{j, \pi(j)} \right) \\ &= \left( \sum_{\tau \in S_r} \operatorname{sgn}(\tau) \prod_{i=1}^r B_{i, \tau(i)} \right) \left( \sum_{\pi \in S_s} \operatorname{sgn}(\pi) \prod_{j=1}^s C_{j, \pi(j)} \right) \\ &= (\det B)(\det C). \end{aligned} \]
distributivity in commutative ring \(K\).
2. General \(k\)-block case by induction:
2.1. For \(k = 1\), the statement is the identity \(\det(A_1) = \det(A_1)\).
base case.
2.2. Assume the result holds for \(k - 1\) diagonal blocks.Write \(A = \operatorname{diag}(A_1, A_2, \ldots, A_k)\) as a \(2 \times 2\) block-diagonal matrix:\[ A = \begin{pmatrix} A_1 & 0 \\ 0 & \widetilde{A} \end{pmatrix}, \qquad \text{where } \widetilde{A} = \operatorname{diag}(A_2, \ldots, A_k). \]Applying 1 and the induction hypothesis:\[ \det A = (\det A_1)(\det \widetilde{A}) = (\det A_1) \prod_{j=2}^k \det A_j = \prod_{j=1}^k \det A_j. \]
induction step using 1.
3. Conclusion:\(\det A = (\det A_1)(\det A_2) \cdots (\det A_k)\). Q.E.D.
1 and 2.