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There is the standard proof using $$\det(A)=\det( A^{T} ) = \det(-A)=(-1)^n \det(A)$$ I would like a proof that avoids this. Specifically, there is the proof that for $A$ a $\bf{real} $ matrix, the transpose is the same as the adjoint, which gives (using the complex inner product) $\lambda \|x\|^2 =\langle Ax, x \rangle= \langle x, -Ax \rangle=-\overline{\lambda } \|x\|^{2}$, so any eigenvalue is purely imaginary. Then we conclude that, since any odd-dimensional real matrix has a real eigenvalue, that eigenvalue must be zero. This argument doesn't work for a general complex skew-symmetric matrix. Is there something I'm missing, is there a way to modify this argument to get that zero is an eigenvalue for the complex case? Also, can somebody please give a geometric reason why odd-dimensional skew-symmetric matrices have zero determinant (equiv., a zero eigenvalue)?

Thanks!

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  • $\begingroup$ What do you mean by skew-symmetric? Literally skew-symmetric, or antihermitian? $\endgroup$
    – tomasz
    Commented Jul 1, 2012 at 0:57
  • $\begingroup$ Literally skew-symmetric, the transpose equals negative. $\endgroup$
    – FGerard
    Commented Jul 1, 2012 at 5:11

2 Answers 2

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Here is a "geometric" argument for the real case. The skew-symmetric condition is equivalent to $\langle Av, v\rangle =0$ for all vectors $v$. Geometrically, $Av$ is orthogonal to $v$. If $Av$ is never zero for $v\ne 0$, then we get a nonvanishing vector field on the unit sphere, contradicting the Hairy Ball theorem.

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  • $\begingroup$ That's very nice - I've never seen that connection before! $\endgroup$ Commented Jul 1, 2012 at 1:01
  • $\begingroup$ This is great, pretty much along the lines of what I wanted for a geometric idea. Do you have an idea of how get the result for the complex case without using determinant (or using it in a different way)? $\endgroup$
    – FGerard
    Commented Jul 1, 2012 at 5:12
  • $\begingroup$ @FGerard I don't. Do you have a geometric interpretation of the condition $A^T=-A$ in the complex case? $\endgroup$
    – user31373
    Commented Jul 1, 2012 at 19:53
  • $\begingroup$ @LeonidKovalev I do not. I guess this was actually two questions: one was if someone could show me a proof of this general fact, using 'adjoint $\endgroup$
    – FGerard
    Commented Jul 2, 2012 at 16:36
  • $\begingroup$ - ' properties. The second question was if somebody could give a geometric or topological idea of what was going on. I read that skew-symmetric matrices for the Lie Algebra for O(n) (in the real case ?), so perhaps this property for odd-dimensions might be explained from that perspective. $\endgroup$
    – FGerard
    Commented Jul 2, 2012 at 16:39
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The proof you wrote down works in any field of characteristic not equal to $2$. Here is a similar such proof which avoids determinants due to Ian Frenkel.

Pass to the algebraic closure. Find $P$ such that $A = PDP^{-1}$ with $D$ upper triangular (for example using Jordan normal form, but you don't need to work this hard; it follows from the existence of eigenvectors). Then the diagonal entries of $D$ are the eigenvalues of $A$. Moreover, $$A^T = (P^T)^{-1} D^T P^T$$

where $D^T$ is lower triangular, so the diagonal entries of $D$ are also the eigenvalues of $A^T$. Since $A^T = -A$ it follows that the multiset of eigenvalues of $A$ is closed under negation. Since there are an odd number of them there must be an eigenvalue which is its own negative, so it must be zero.

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  • $\begingroup$ I am not completely sure whether I need determinants to show that the multiset of eigenvalues is well-defined in the case that $A$ has nontrivial Jordan blocks, so run this proof for $A$ having distinct eigenvalues and then observe that such matrices are Zariski dense so the conclusion holds in general. $\endgroup$ Commented Jul 8, 2012 at 6:11

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