How to prove $\det \begin{bmatrix}A & B \\ C & D \end{bmatrix} =\det(AD-BC)$, where $A, B, C,$ and $D$ are $n\times n$ and commuting.

Question

I can solve for the cases in which $C=0$ or $B=0$. But I cannot find any idea for the case in which both $B$ and $C$ are nonzero matrices.

http://www.ee.iisc.ac.in/new/people/faculty/prasantg/downloads/blocks.pdf — Alex R., May 20 '18 at 21:16

Eclipse Sun · Accepted Answer · 2023-07-06T05:46:21.537

3

I assume that $A$ is an $n\times n$ matrix.

First suppose that $A$ is invertible. By Gauss elimination, $$\begin{pmatrix}I & 0 \\ -CA^{-1} & I\end{pmatrix}\begin{pmatrix}A & B \\ C & D\end{pmatrix}=\begin{pmatrix}A & B \\ 0 & D-CA^{-1}B\end{pmatrix}.$$ Hence, by what you already solved, the determinant is $\det(A)\det(D-CA^{-1}B)=\det(AD-ACA^{-1}B)=\det(AD-CB)$, since $AC=CA$.

For $A$ not invertible, use approximation.

edited Jul 06 '23 at 05:46

answered May 20 '18 at 21:22

Eclipse Sun

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Could you please explain what did you mean by "use approximation"? – Majid May 22 '18 at 22:24
If $\det(A)=0$, then you can show that for small $t>0$, $\det(A+tI)\ne 0$. The argument above shows that $\det\begin{pmatrix}A+tI & B \ C & D \end{pmatrix}=\det((A+tI)D-CB)$. Then let $t\to 0$. – Eclipse Sun May 23 '18 at 18:04
I think the $CA^{-1}$ should have a negative sign in the first matrix? – WillG Jun 28 '23 at 07:48
You are right. Fixed! – Eclipse Sun Jul 06 '23 at 05:48

How to prove $\det \begin{bmatrix}A & B \\ C & D \end{bmatrix} =\det(AD-BC)$, where $A, B, C,$ and $D$ are $n\times n$ and commuting.

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