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A friend of mine asked me to calculate the value of a perticular determinant, but after spending quite some time on it I'm unable to find a solution to it. The given determinant is:

$$ \begin{vmatrix} 0 & 1 & \cdots &n-2&n-1 \\ n-1 & 0 & \cdots &n-3&n-2 \\ \vdots & \vdots & \ddots & \vdots &\vdots\\ 1 & 2 & \cdots & n-1 &0 \\ \end{vmatrix} $$

Also my friend is supposed to solve this determinant by more or less elementary methods, like elementary row operations, transpose... and no eigenvalues, nor adjoints.


First of all it seems like the value of the determinant is:

$$ \frac{1}{2n}\begin{vmatrix} 0 & n & \cdots &n&n \\ n & 0 & \cdots &n&n \\ \vdots & \vdots & \ddots & \vdots &\vdots\\ n & n & \cdots & n &0 \\ \end{vmatrix} $$

Anyway I tried adding all the rows to the last one and by taking $\frac{n(n-1)}{2}$ in front to have a row of 1s, but it didn't helped me much.

Then I tried to use the fact that:

$$ \begin{vmatrix} A & C \\ 0 & B \\ \end{vmatrix} = \det(A) \cdot \det(B) $$, where $A,B$ are square matrices. By subtracting the $(\frac{n}{2} + i)$-th column from the $i$-th ($i=0,1,2...$) and then adding the $i$-th row to the $(\frac{n}{2} + i)$-th row, one can obtain the wanted form, but again nothing more. Also this method pretty much fails when $n$ is odd.

My best attepmt was to notice that $A_{ij} + A_{ji} = n$ and using $\det(A^{T}) = \det(A)$, but unfortunatelly

$$\begin{vmatrix} 0 & n & \cdots &n&n \\ n & 0 & \cdots &n&n \\ \vdots & \vdots & \ddots & \vdots &\vdots\\ n & n & \cdots & n &0 \\ \end{vmatrix} \not = \begin{vmatrix} 0 & 1 & \cdots &n-2&n-1 \\ n-1 & 0 & \cdots &n-3&n-2 \\ \vdots & \vdots & \ddots & \vdots &\vdots\\ 1 & 2 & \cdots & n-1 &0 \\ \end{vmatrix} + \left( \begin{vmatrix} 0 & 1 & \cdots &n-2&n-1 \\ n-1 & 0 & \cdots &n-3&n-2 \\ \vdots & \vdots & \ddots & \vdots &\vdots\\ 1 & 2 & \cdots & n-1 &0 \\ \end{vmatrix}\right)^T$$ I tried to decompose the LHS by using the $n-$linearity of the determinants, by then one gets $2^n$ determinants on the RHS similar to the first one, but things are messy.

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    $\begingroup$ I don't know enough to speak with authority but this looks like a circulant matrix which has a well known solution to their eigenvectors and eigenvalues. Given the form of the eigenvalues. it looks pretty difficult to solve this using overly elementary methods. $\endgroup$ – JessicaK Nov 6 '16 at 20:57
  • $\begingroup$ @JessicaK Unfortunately my friend isn't supposed to use eigenvalues. As I've mentioned my last attempt seems like the best idea, as if I can bring the determinant to the form I mentioned firstly I can solve by using elementary operations $\endgroup$ – Stefan4024 Nov 6 '16 at 21:03
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    $\begingroup$ With an aid of computer \begin{align*} a_{ij} &\equiv j-i\pmod{n} \\ \det (a_{ij}) &= \frac{(n-1)(-n)^{n-1}}{2} \end{align*} $\endgroup$ – Ng Chung Tak Nov 6 '16 at 21:26
  • $\begingroup$ @NgChungTak I've noticed both things, but how do we get that last line? I can't find any way to prove that result. $\endgroup$ – Stefan4024 Nov 6 '16 at 21:29
  • $\begingroup$ @Stefan4024 I've just tried small $n$ and observing the pattern of factors. $\endgroup$ – Ng Chung Tak Nov 6 '16 at 21:30
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I take it this is a Toeplitz matrix (diagonals are constant).

Subtract row $n-1$ from row $n$, then row $n-2$ from row $n-1$, ... row $1$ from row $2$. Then subtract column $1$ from each other column. Then add $1/n$ times the second, third, ..., last columns to the first column. The result should be upper triangular.

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  • $\begingroup$ Quite elegant solution. One little note is that maybe we can shift each column to the right and then do the algorithm you've mentioned, as the calculations are better and the general form too. $\endgroup$ – Stefan4024 Nov 6 '16 at 22:22

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