Determinants of block matrices with commuting blocks

Question

On Wikipedia, I read that

$$\det \begin{pmatrix} A & B\\B& A\end{pmatrix} = \det(A+B) \, \det(A-B)$$

if $A$ and $B$ commute. Does this hold even if $ A $ and $ B$ are not invertible?

I spent a short amount of time trying to come up with a counter example, but it worked for the noninvertible matrices I picked. I'm curious if this is in fact true. Does anyone have a proof of it?

Answer 1

We need no assumptions about $A,B$ except that they have same dimension. Then apply basic operation of subtracting a set of rows from set of the other rows and also summing a set of columns with set of other columns, which does not change the determinant. You will simply get:

$$ \det \begin{pmatrix} A & B\\B& A\end{pmatrix}=\det \begin{pmatrix} A & B\\B-A& A-B\end{pmatrix}\\ =\det \begin{pmatrix} A+B & B\\0& A-B\end{pmatrix} =\det(A-B)\det(A+B) $$

Answer 2

From a more general perspective, since $A$ and $B$ commute, your determinant can be evaluated by the block-matrix determinant formula to $\det(A^2-B^2)$. Again, since $A$ and $B$ commute, we have $A^2-B^2=(A+B)(A-B)$ and the result follows.