Proof of the Schur complement formula by block factorization

Proof

We factor the block matrix into a lower and an upper block triangular matrix.

1. Let $M = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ be the block matrix of the statement, with $p, q \ge 1$. Let $S = D - C A^{-1} B$.
2. Let $L = \begin{pmatrix} I_p & 0 \\ C A^{-1} & I_q \end{pmatrix}$ and $U = \begin{pmatrix} A & B \\ 0 & S \end{pmatrix}$, where $I_p, I_q$ are identity matrices and $0$ are zero blocks.
3. Block products: by the definition of the matrix product, entry $(r, s)$ of a product is $\sum_k X_{rk} Y_{ks}$.
4. Split this sum into the indices $k \le p$ and $k > p$. This shows that block matrices with the same splitting multiply block by block, like $2 \times 2$ matrices.
5. By step 4, the top blocks of $LU$ are $I_p A + 0 \cdot 0 = A$ and $I_p B + 0 \cdot S = B$.
6. The bottom left block of $LU$ is $C A^{-1} A + I_q \cdot 0 = C (A^{-1} A) = C$, by associativity and the definition of the inverse.
7. The bottom right block of $LU$ is $C A^{-1} B + S = C A^{-1} B + D - C A^{-1} B = D$. So $LU = M$.
8. By the rule for block triangular matrices, $\det U = \det A \cdot \det S$.
9. The transpose of $L$ is $\begin{pmatrix} I_p & (C A^{-1})^T \\ 0 & I_q \end{pmatrix}$, which is upper block triangular.
10. By the rule for the determinant of a transpose, the rule for block triangular matrices and $\det I_p = \det I_q = 1$ by the determinant of the identity, $\det L = \det L^T = 1$.
11. By the product rule and steps 7, 8 and 10, $\det M = \det L \cdot \det U = \det A \cdot \det(D - C A^{-1} B)$.