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Assume that $A$ is an $n\times n$ skew-symmetric real matrix, i.e. $$A^T=-A.$$

Since $\det(A-\lambda I)=\det(A^T-\lambda I)$, $A$ and $A^T$ have the same eigenvalues. On the other hand, $A^T$ and $-A$ also have the same eigenvalues. Thus if $\lambda$ is an eigenvalue of $A$, so is $-\lambda$. If $n$ is odd, $\lambda = 0 $ is an eigenvalue.

A curious search in Google returns that the nonzero eigenvalues of $A$ are all pure imaginary and thus are of the form $iλ_1, −iλ_1, iλ_2, −iλ_2,$ … where each of the $λ_k$ are real.

Here is my question:

How can I prove the fact that "the nonzero eigenvalues of $A$ are all pure imaginary"?

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If you are given that the eigenvalues of a Hermitian operator are real, just consider $H=iA$. –  bobobinks Aug 12 '11 at 17:06
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Zero eigenvalues are purely imaginary too :-) –  TonyK Aug 12 '11 at 17:25
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2 Answers

up vote 10 down vote accepted

Consider $A$ as a matrix over $\mathbb{C}$. Then we have that for all $\mathbf{x},\mathbf{y}\in\mathbb{C}^n$, $$\langle A\mathbf{x},\mathbf{y} \rangle = \langle \mathbf{x},A^*\mathbf{y}\rangle,$$ where $\langle-,-\rangle$ is the standard complex inner product, and $A^*$ is the adjoint (which relative to the standard complex inner product is given by the conjugate transpose of $A$). Since $A$ is a real matrix, the adjoint is equal to the transpose, so for every $\mathbf{x},\mathbf{y}\in\mathbb{C}^n$, you have $$\langle A\mathbf{x},\mathbf{y}\rangle = \langle \mathbf{x},A^T\mathbf{y}\rangle = \langle \mathbf{x},-A\mathbf{y}\rangle = -\langle \mathbf{x},A\mathbf{y}\rangle.$$

Now suppose that $\mathbf{x}$ is an eigenvector with eigenvalue $\lambda$. Setting $\mathbf{y}=\mathbf{x}$, we have $$\langle A\mathbf{x},\mathbf{x}\rangle = \langle \lambda\mathbf{x},\mathbf{x}\rangle = \lambda \lVert\mathbf{x}\rVert^2.$$ On the other hand, $$-\langle \mathbf{x},A\mathbf{x}\rangle = -\langle\mathbf{x},\lambda\mathbf{x}\rangle = -\overline{\lambda}\langle\mathbf{x},\mathbf{x}\rangle = -\overline{\lambda}\lVert\mathbf{x}\rVert^2.$$ These two are equal, and since $\mathbf{x}$ is an eigenvector, then $\lVert\mathbf{x}\rVert\neq 0$. Therefore, we have that $\lambda=-\overline{\lambda}$, and hence $\lambda$ is either $0$ or a pure imaginary number.

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You can see the proof here:

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Hmm, this is also a standard trick to prove that all the eigenvalues of a Hermitian matrix are real. –  Jack Aug 12 '11 at 17:15
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