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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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    $\begingroup$ If you are given that the eigenvalues of a Hermitian operator are real, just consider $H=iA$. $\endgroup$ – Benji Aug 12 '11 at 17:06
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    $\begingroup$ Zero eigenvalues are purely imaginary too :-) $\endgroup$ – TonyK Aug 12 '11 at 17:25
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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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  • $\begingroup$ Why can we say that $-<x,\lambda x>=-\bar{\lambda}<x,x>?$ I understand why the lamda can be pulled out but why can it be conjugated? $\endgroup$ – MathIsHard Aug 31 '17 at 21:28
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    $\begingroup$ @Math4Life: By definition. Because complex inner products are sesquilinear, not bilinear. They are, by definition, homogeneous in the first coordinate and conjugate homogeneous in the second. Remember that $\langle x,y\rangle = \overline{\langle y,x\rangle}$ is a property required of complex inner products, so if $\langle \lambda\mathbf{x},\mathbf{y}\rangle = \lambda\langle\mathbf{x},\mathbf{y}\rangle$, then scalars pulled out of the second argument must "come out" conjugated. $\endgroup$ – Arturo Magidin Aug 31 '17 at 22:24
  • $\begingroup$ Oh thank you very much. I appreciate it $\endgroup$ – MathIsHard Sep 1 '17 at 17:11

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