Determinant of a special block matrix $\big(\begin{smallmatrix}A&B\\B&A\end{smallmatrix}\big)$ Let be $ A,B\in \mathbb{K}^{n,n} $ arbitrary matrices. Then
$$ \det\Bigg(\underbrace{\begin{pmatrix}A&B\\B&A\end{pmatrix}}_{=:L}\Bigg)=\det(A+B)\cdot \det(A-B) $$
My idea: I consider
$$ \begin{align}&\det(A+B)\cdot \det(A-B)\\[10pt]&=\det(A+B)\cdot 1\cdot \det(A-B)\cdot 1\\[10pt]&=\Big(\det(A+B)\cdot \det(I_n)\Big)\cdot \Big(\det(A-B)\cdot \det(I_n)\Big)\\[10pt]&=\det\left(\begin{pmatrix}A+B&0\\0&I_n\end{pmatrix}\right)\cdot \det\left(\begin{pmatrix}I_n&0\\0&A-B\end{pmatrix}\right)\\[10pt]&=\det\Bigg(\underbrace{\begin{pmatrix}A+B&0\\0&A-B\end{pmatrix}}_{=:M}\Bigg) \end{align}$$
So far so good. No I tried to find a matrix $X\in \mathbb{K}^{n,n}$ such that $ L=X^{-1}\cdot M\cdot X $ but I couldn't.

You can use this decomposition:
$$
\begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix}^{-1} \begin{pmatrix} A &B \\ B &A \end{pmatrix} \begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix} = \begin{pmatrix} A+B &0 \\ 0 &A-B \end{pmatrix}
$$ You can see the motivation for this by thinking of $A, B$ as real numbers and $I_n$ as $1$.
The only thing the careful might ask is whether $\begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix}^{-1}$ makes sense and indeed it does. You can check that $\begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix}^{-1} = \frac{1}{2}\begin{pmatrix} I_n &I_n \\ -I_n &I_n \end{pmatrix}$ by manually multiplying it with $\begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix}$ to get the $2n \times 2n$ identity.
