# Prove $\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = \det(A-B)\det(A+B)$, even when $A$ and $B$ are not commutative.

I am aware of the following identity

$$\det\begin{bmatrix}A & B\\ C & D\end{bmatrix} = \det(A)\det(D - CA^{-1}B)$$

When $$A = D$$ and $$B = C$$ and when $$AB = BA$$ the above identity becomes

$$\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = \det(A)\det(A - BA^{-1}B) = \det(A^2 - B^2) = \det(A-B)\det(A+B)$$.

However, I couldn't prove this identity for the case where $$AB \neq BA$$.

EDIT: Based on @Trebor 's suggestion.

I think I could do the following.

$$\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = \det\begin{bmatrix}A & B\\ B-A & A-B\end{bmatrix} = \det(A^2-B^2) = \det(A-B)\det(A+B)$$.

• Hint: Row transformations. Oct 23, 2021 at 16:16
• – glS
Feb 19, 2023 at 17:23

$$\begin{pmatrix} A & B \\ B & A \end{pmatrix}\xrightarrow{\text{row1 -= row2}}\begin{pmatrix} A-B & B-A \\ B & A\end{pmatrix}\xrightarrow{\text{col2 += col1}}\begin{pmatrix} A-B & O \\ B & A+B\end{pmatrix}$$

Let's say $$A, B$$ are $$n \times n$$ matrices with entries from a field with characteristic $$\ne 2{}^{\color{blue}{[1]}}$$.

Let $$I$$ be the $$n \times n$$ identity matrix and $$J = \begin{bmatrix}I & I \\ -I & I\end{bmatrix}$$. Since $$\det J = 2^n \ne 0$$, $$J$$ is invertible.

Notice $$J \begin{bmatrix}A & B \\ B & A\end{bmatrix} = \begin{bmatrix}A+B & A+B \\ B-A & A-B\end{bmatrix} = \begin{bmatrix}A+B & 0 \\ 0 & A-B\end{bmatrix} J$$ We have $$\det\begin{bmatrix}A & B \\ B & A\end{bmatrix} = \det\begin{bmatrix}A+B & 0 \\ 0 & A-B\end{bmatrix} = \det(A+B)\det(A-B)\tag{*1}$$

Notes

• $$\color{blue}{[1]}$$ - As demonstrated by @Just a user's answer, the requirement that entries from a field with characteristic $$\ne 2$$ can be dropped. $$(*1)$$ continues to work when entries of $$A,B$$ take values from any commutative ring.

Aside from using row/column operations as in @Just a user's answer, we can use the fact that LHS and RHS of $$(*1)$$ are polynomials with integer coefficients in entries of $$A,B$$. Since they are equal as a polynomial, it remains equal when we substitute the entries by elements from any commutative ring.

Your second attempt almost works. You may use the fact that $$\det\pmatrix{A&B\\ C&D}=\det(AD-BC)$$ whenever $$A,B,C,D$$ are square matrices of the same sizes and $$CD=DC$$: $$\det\pmatrix{A&B\\ B&A} =\det\pmatrix{A&B\\ B-A&A-B} =\det\left(A(A-B)-B(B-A)\right) =\det\left((A+B)(A-B)\right).$$