Calculating sign of a permutation of unknown size, but with a pattern

Question

I'm trying to calculate the determinant of the following matrix, using the Leibniz formula: $$ \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 & 1 \\ 1 & 0 & 1 & \ddots & \ddots & 0\\ 0 & 1 & \ddots & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & \ddots & 1 & 0\\ 0 & \ddots & \ddots & 1 & 0 & 1\\ 1 & 0 & 0 & 0& 1 & 0 \end{pmatrix}. $$

So, the only permutation that are relevant (non-zero product) are:

$\sigma_1 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & ... & n \\ 2 & 3 & 4 & 5 & 6 & 7 & ... & 1 \end{pmatrix}$

$\sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & ... & n \\ n & 1 & 2 & 3 & 4 & 5 & ... & 1 \end{pmatrix} $

$\sigma_3 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & ... & n-1 & n \\ 2 & 1 & 4 & 3 & 6 & 5 & ... & n & n-1 \end{pmatrix} $

$\sigma_4 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & ... & n-2 & n-1 & n \\ n & 3 & 2 & 5 & 4 & 7 & ... & n-1 & n-2 & 1 \end{pmatrix}$

How would I proceed to determine $\operatorname{Sgn}(\sigma)$ based on the value of $n$? Thanks!

Answer 1

$\mathrm{sgn}(\sigma)=(-1)^{\mathrm{inv}(\sigma)}$, where $\mathrm{inv}(\sigma)$ is the number of inversions of $\sigma$, that is to say, the number of pairs of positions $(i,j)$ where $i<j$ but $\sigma(i)>\sigma(j)$ (a larger value is to the left of a smaller value).

For the permutations you listed, the numbers of inversions are $$ \begin{split} \mathrm{inv}(\sigma_1)&=n-1,\\ \mathrm{inv}(\sigma_2)&=n-1,\\ \mathrm{inv}(\sigma_3)&=\left\lfloor\frac{n}{2}\right\rfloor,\\ \mathrm{inv}(\sigma_4)&=\left(\left\lfloor\frac{n}{2}\right\rfloor-1\right)+(n-2)+(n-2)+1=\left\lfloor\frac{n}{2}\right\rfloor+2n-4. \end{split} $$

In particular, the parity of the number of inversions is the same for $\sigma_1$ and $\sigma_2$ as well as for $\sigma_3$ and $\sigma_4$, so $\mathrm{sgn}(\sigma_1)=\mathrm{sgn}(\sigma_2)$ and $\mathrm{sgn}(\sigma_3)=\mathrm{sgn}(\sigma_4)$. Thus, the determinant can only take values $-4$, $0$, $4$.

I also see that you implicitly assume that $n$ is even. In that case, $n-1$ is odd, so $\mathrm{sgn}(\sigma_1)=\mathrm{sgn}(\sigma_2)=-1$, so the determinant is $-4$ if $\frac{n}{2}$ is odd, and $0$ if $\frac{n}{2}$ is even.