Let \(A\), \(B\), \(C\), and \(D\) be \(n \times n\) complex matrices which commute. Let \(E\) be the \(2n \times 2n\) matrix \[ E = \begin{bmatrix} A & B \\ C & D \end{bmatrix}. \] Prove that \(\operatorname{det} E = \operatorname{det}(AD - BC)\).
Solution
1. Prove the formula when the upper-left block is invertible.
Assume first that \(A\) is invertible. Since \(A\) commutes with \(B,C,D\), its inverse also commutes with them. Multiply \(E\) on the left by \[ L=\begin{bmatrix}I&0\\-CA^{-1}&I\end{bmatrix}. \] This block lower-triangular matrix has determinant \(1\), and \[ LE = \begin{bmatrix} A&B\\ 0&D-CA^{-1}B \end{bmatrix}. \] Hence \[ \det E =\det A\,\det(D-CA^{-1}B). \] Because all relevant factors commute, \[ A(D-CA^{-1}B)=AD-CB=AD-BC. \] Taking determinants and using multiplicativity gives \[ \det E =\det\bigl(A(D-CA^{-1}B)\bigr) =\det(AD-BC). \] Thus the formula holds whenever \(A\) is invertible [@HK71].
2. Perturb \(A\) by a scalar to remove the invertibility hypothesis.
For \(t\in\mathbb C\), define \[ E(t)= \begin{bmatrix} A+tI&B\\ C&D \end{bmatrix}. \] The four blocks still commute. The polynomial \[ q(t)=\det(A+tI) \] is monic of degree \(n\), so it has only finitely many zeros. Therefore \(A+tI\) is invertible for infinitely many complex numbers \(t\). For every such \(t\), step 1 gives \[ \det E(t) =\det\bigl((A+tI)D-BC\bigr). \]
Both sides are polynomials in \(t\): determinants are polynomial expressions in their matrix entries, and every entry of the two displayed matrices is affine in \(t\). Since two complex polynomials agreeing at infinitely many values are identical, the equality holds for every \(t\in\mathbb C\). Setting \(t=0\) yields \[ \boxed{ \det\begin{bmatrix}A&B\\C&D\end{bmatrix} =\det(AD-BC).} \]