Determinant of matrix obtained by commuting matrices

Question

The Question is to prove that :

For Commuting $n\times n$ matrices $A,B,C,D$ over a field $F$,

Determinant of $\left(\begin{array}{cccc} A & B \\ C & D \\ \end{array} \right)$ is given by $\det(AD-BC)$

I have no idea how to proceed for this except at the case of $n=1$ where $\left(\begin{array}{cccc} A & B \\ C & D \\ \end{array} \right)$=$\begin{pmatrix} a& b \\ c & d \\ \end{pmatrix}$ for some $a,b,c,d\in F$ and I know $\det\left(\begin{array}{cccc} a& b \\ c & d \\ \end{array} \right)=ad-bc=\det(ad-bc)=\det(AD-BC)$

So, for $n=1$ we have $\det\left(\begin{array}{cccc} A & B \\ C & D \\ \end{array} \right)=\det(AD-BC)$

I have no idea how to proceed for general $n$ not even when $n=2$

Do I need to proceed by induction?I doubt that it may not work..

please provide some hints to prove this case...

Thank You.

Answer 1

This is a partial duplicate of the thread "block matrices problem". You question is addressed in the last paragraph of my answer.

More importantly, note that if only a pair of matrices (but not four of them) on the same row or same column commute, the order of matrices matters. In short, we have $$ \det\pmatrix{A&B\\ C&D}= \begin{cases} \det(AD-BC) & \text{ if } CD=DC,\\ \det(DA-CB) & \text{ if } AB=BA,\\ \det(DA-BC) & \text{ if } BD=DB,\\ \det(AD-CB) & \text{ if } AC=CA. \end{cases} $$

Answer 2

Here is a method that works if $F = \mathbb{C}$. Let $\mathcal{F} = \{A, B, C, D\}$ and let $P\in M_{n}(\mathbb{C})$ such that $P^{-1}XP$ is upper triangular for each $X\in \mathcal{F}$. Then \begin{equation*} \det\begin{pmatrix} A & B\\ C & D \end{pmatrix} = \det\begin{pmatrix} P^{-1} & 0\\ 0 & P^{-1} \end{pmatrix}\begin{pmatrix} A & B\\ C & D \end{pmatrix}\begin{pmatrix} P & 0\\ 0 & P \end{pmatrix} = \det\begin{pmatrix} P^{-1}AP & P^{-1}BP\\ P^{-1}CP & P^{-1}DP \end{pmatrix}. \end{equation*} Further, $\det(AD-BC) = \det(P^{-1}(AD-BC)P) = \det((P^{-1}AP)(P^{-1}DP)-(P^{-1}BP)(P^{-1}CP))$. Thus it suffices to assume that $A$, $B$, $C$ and $D$ are all upper triangular. We prove the result by induction. For $n = 1$ there is nothing to prove. Expanding the determinant about the first column we have \begin{align*} \det\left(\begin{array}{ccc|ccc} a_{11} & & \ast & b_{11} & & \ast\\ & \ddots & & & \ddots & \\ & & a_{nn} & & & b_{nn}\\ \hline c_{11} & & \ast & d_{11} & & \ast\\ & \ddots & & & \ddots & \\ & & c_{nn} & & & d_{nn} \end{array}\right) =& (-1)^{1+1}a_{11}\det\left(\begin{array}{ccc|cccc} a_{22} & & \ast & 0 & b_{22} & & \ast\\ & \ddots & & \vdots & & \ddots & \\ & & a_{nn} & 0 & & & b_{nn}\\ \hline c_{12} & \dots & c_{1n} & d_{11} & & & \ast\\ c_{22} & & \ast & & \ddots & & \\ & \ddots & & & & \ddots & \\ & & c_{nn} & & & & d_{nn} \end{array}\right)\\ &+(-1)^{1+n+1}c_{11}\det\left(\begin{array}{ccc|cccc} a_{12} & \dots & a_{1n} & b_{11} & & & \ast\\ a_{22} & & \ast & & \ddots & & \\ & \ddots & & & & \ddots & \\ & & a_{nn} & & & & b_{nn}\\ \hline c_{22} & & \ast & 0 & d_{22} & & \ast\\ & \ddots & & \vdots & & \ddots & \\ & & c_{nn} & 0 & & & d_{nn} \end{array}\right). \end{align*} Expanding both terms about the $n^{\text{th}}$ column we get \begin{equation*} \det\left(\begin{array}{c|c} A & B\\ \hline C & D \end{array}\right) = (a_{11}d_{11}-b_{11}c_{11})\det\left(\begin{array}{ccc|ccc} a_{22} & & \ast & b_{22} & & \ast\\ & \ddots & & & \ddots & \\ & & a_{nn} & & & b_{nn}\\ \hline c_{22} & & \ast & d_{22} & & \ast\\ & \ddots & & & \ddots & \\ & & c_{nn} & & & d_{nn} \end{array}\right). \end{equation*} Conclude using the inductive hypothesis.