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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.

hallo007
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  • Yess! I looked again and saw that I could try to diagonalize the matrix from the form of $ \begin{pmatrix}a&b\b&a\end{pmatrix} $ where $a,b$ are "just" numbers. With this analogy I also got the same decomposition. – hallo007 Jan 20 '22 at 04:39

2 Answers2

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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.

Golden_Ratio
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balddraz
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This follows elementarily from simple row operations, which are valid in the block matrix setting. \begin{align}&\left|\begin{matrix}A &B\\B&A\end{matrix} \right|&\text{replace Row2 with Row2 }-\text{Row1}\\ &=\left|\begin{matrix}A &B\\B-A&A-B\end{matrix} \right|&\text{replace Col1 with Col1 }+\text{Col2}\\ &=\left|\begin{matrix}A+B &B\\0&A-B\end{matrix} \right|&\text{determinant of block triangular matrix}\\ &=|A+B||A-B|\end{align} One can find the last step here: The determinant of block triangular matrix as product of determinants of diagonal blocks .

Calvin Khor
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