Determinant of a 2n x 2n antisymmetric matrix

Compute the determinant of the matrix

(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 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 for : you will have to compute, after expanding from the first line, the following determinant

.

To compute this, notice that the first components of the vector which composes the first column can be written as linear combinaisons of the other columns, namely . 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 matrix, but via a similar procedure I did get the matrix which, after adding to gives

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

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

Yes, I found too.