How do I prove that the determinant of an nxn, skew-symmetric matrix is always zero if n is odd?

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- Oct 20th 2010, 02:42 PMmiatchguySkew-symmetric matrix determinant question
How do I prove that the determinant of an nxn, skew-symmetric matrix is always zero if n is odd?

- Oct 20th 2010, 03:27 PMAckbeet
Well, you know that $\displaystyle \det(-A)=(-1)^{n}\det(A).$ Does that help?

- Oct 20th 2010, 04:04 PMmiatchguy
- Oct 21st 2010, 02:42 AMAckbeet
And what's the definition of a skew-symmetric matrix? Can you tie that in with the equation I just gave you?

- Oct 21st 2010, 03:36 AMmiatchguy
I got it already. Thanks alot for the help. I just proved that an inverse retains the upper triangularity of the original matrix, and I the determinant scalar can't affect the zero entries.

- Oct 21st 2010, 03:57 AMAckbeet
There's a problem: not all skew-symmetric matrices are invertible.

- Apr 5th 2013, 08:15 PMmathguy25Re: Skew-symmetric matrix determinant question
Basic facts:

(*) det(A) = det(A^T)

(**) det(kA) = k^n det(A) where A is nxn matric and k is scalar

Let n be odd and let A be a skew-symmetric nxn matrix. Then A^T = -A. Then det(A^T) = det(-A). Note that by (**) where k = -1, we see that det(-A) = (-1)^n det(A). Then since n is odd, (-1)^n = -1. Then det(-A) = -det(A). Then by (*) we see that det(A) = det(A^T) = det(-A) = -det(A). Thus, det(A) = -det(A). Then 2*det(A) = 0. Then det(A) = 0.