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This is another exercise from Golan's book.

Problem: Let $V$ be an inner product space over $\mathbb{C}$ and let $\alpha$ be an endomorphism of $V$ satisfying $\alpha^*=-\alpha$, where $\alpha^*$ denotes the adjoint. Show that every eigenvalue of $\alpha$ is purely imaginary.

My proposed solution is below.

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Isn't there a geometric interpretation for the notion of adjoint? –  math-visitor Jun 15 '12 at 3:18
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2 Answers

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Let me show another argument which applies to a more general setting: if $\alpha$ is a linear operator on a Hilbert space satisfying $\alpha^*=-\alpha$, then the spectrum of $\alpha$ is purely imaginary (i.e. real part equal zero).

Indeed, one simply needs to notice that $\alpha-\lambda\,\text{id}$ is invertible if and only if $(\alpha-\lambda\,\text{id})^*$ is invertible. As $$(\alpha-\lambda\,\text{id})^*=\alpha^*-\overline\lambda\,\text{id}=-\alpha-\overline\lambda\,\text{id}=-(\alpha+\overline\lambda\,\text{id}),$$ we conclude that any $\lambda$ in the spectrum of $\alpha$ satisfies $\overline\lambda=-\lambda$.

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Suppose $c$ is an eigenvalue of $\alpha$ and $v$ is a corresponding eigenvector. We have

$$c \langle v,v\rangle= \langle \alpha(v),v\rangle =\langle v, \alpha^*(v)\rangle =\langle v,-\alpha(v)\rangle=\langle v, -cv\rangle=\overline{\langle -cv, v\rangle} = \overline{-c\langle v,v\rangle}$$

Because $\langle v,v\rangle$ is real, as it is the norm of a vector, we have $c=\overline{-c}$, which immediately shows $c$ is imaginary (write $c$ out in terms of imaginary and real parts to see this).

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Looks fine. Why though couldn't $\,c=0\,$ ? A line about this could enhance the solution. –  DonAntonio Jun 15 '12 at 0:10
    
@Potato: $0$ is purely imaginary (in that it has zero real part). –  Qiaochu Yuan Jun 15 '12 at 1:04
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