I have to find the determinant of an $n \times n$ exchange matrix. After applying values to $n$, I found out the determinant has a periodicity of $4$ and its formula is $(-1)^{n(n-1)/2}$. Literature revision confirms this. However, I don't know how to reach this conclusion formally neither how to prove it.
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Have you tried expansion by minors? – saulspatz Apr 11 '21 at 00:16
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Related – Rodrigo de Azevedo Dec 27 '23 at 12:29
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2Swapping two rows (or columns) of a matrix changes its determinant by a factor of $-1$. So, how many such swaps are needed to change the exchange matrix into the identity matrix? – Chris Lewis Dec 27 '23 at 12:37
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@ChrisLewis And proofs can be found here – Rodrigo de Azevedo Dec 27 '23 at 12:47
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Let $J_n$ denote the exchange matrix of size $n$. It suffices to show the following:
- $\det(J_1) = 1$
- $\det(J_2) = -1$
- For $n \geq 3$, $\det(J_n) = -\det(J_{n-2})$.
The first two statements are easy to prove directly. For the third statement, proceed as follows: $$ \begin{align} \det(J_n) &= \det \pmatrix{0&J_{n-2}\\J_2 & 0} \\ & = \det\pmatrix{J_2 & 0\\0&J_{n-2}} \\ & = \det(J_2)\det(J_{n-2}) = -\det(J_{n-2}). \end{align} $$
Ben Grossmann
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1One: switching rows of a matrix changes the sign of the determinant. Two: the determinant of the block matrix $\pmatrix{A&0\0&B}$ is $\det(A)\det(B)$. – Ben Grossmann Apr 11 '21 at 17:10