Determinant of a 2n x 2n antisymmetric matrix

Compute the determinant of the $\displaystyle 2n \times 2n$ matrix $\displaystyle \left(\begin{array} {ccccc} 0 & 1 & 1 & ... & 1 \\ -1 & 0 & 1 & ... & 1 \\ -1 & -1 & 0 & ... & 1 \\ .. \\ -1 & -1 & ... & -1 & 0 \end{array}\right)$

(ie. all diagonal entries equal to zero, all above diagonal entries equal to 1, all below diagonal entries equal to -1)

I am not sure how to go about finding this in general. I have found that the determinants when $\displaystyle n$ is 1 and 2 are both 1, but that doesn't really help me in general, and I don't really have any idea what to start with for the general case. Any tips?

Re: Determinant of a 2n x 2n antisymmetric matrix

Hint: do $\displaystyle c_k\rightarrow c_k-c_{k-1}$ for $\displaystyle k=2,\ldots, 2n$: you will have to compute, after expanding from the first line, the following $\displaystyle (2n-1)\times (2n-1)$ determinant

$\displaystyle \begin{vmatrix}1&1&0&\ldots&0\\1&1&1&\ldots&0\\1&0 &1&\ldots&0\\\vdots&\vdots&\ddots&\ddots&\vdots\\1 &0&0&\ldots&1\end{vmatrix}$.

To compute this, notice that the first $\displaystyle 2n-2$ components of the vector which composes the first column can be written as linear combinaisons of the other columns, namely $\displaystyle c_1 = \sum_{k=0}^{n-1}c_{2k+1}$. After expanding, you only have to compute the determinant of a triangular matrix.

Re: Determinant of a 2n x 2n antisymmetric matrix

Thank you! I wasn't really sure how you ended up with that $\displaystyle 2n-1 \times 2n-1$ matrix, but via a similar procedure I did get the $\displaystyle 2n \times 2n$ matrix $\displaystyle \left( \begin{array} {cccccc} 0 & 1 & 0 & 0 & ... & 0 \\ -1 & 1 & 1 & 0 & ... & 0 \\ -1 & 0 & 1 & 1 & ... & 0 \\ . & . & . & . & . & . \\ -1 & 0 & 0 & 0 & ... & 1 \end{array} \right) $ which, after adding $\displaystyle \sum_{k=2}^{2n}c_k$ to $\displaystyle c_1$ gives $\displaystyle \left( \begin{array} {cccccc} 1 & 1 & 0 & 0 & ... & 0 \\ 1 & 1 & 1 & 0 & ... & 0 \\ 1 & 0 & 1 & 1 & ... & 0 \\ . & . & . & . & . & . \\ 1 & 0 & 0 & 0 & ... & 1 \\ 0 & 0 & 0 & 0 & ... & 1 \end{array} \right) $

Then, with expansion along the first row, you get the determinant to be 1, as it is of the form $\displaystyle 1($*det of triangular matrix with ones on the diagonals*$\displaystyle ) -1( $*det of matrix which can be reduced to a triangular matrix with a zero on the diagonal by adding rows*$\displaystyle )$, which is just $\displaystyle 1(1)-1(0)$. Thanks!

Re: Determinant of a 2n x 2n antisymmetric matrix

Yes, I found $\displaystyle 1$ too.