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Homework question.

Let $V$ be a finite dimensional inner-product space over $\mathbb{C}$. Let $T \in L(V,V)$ satisfy $T^*=-T$. Show that all eigenvalues of $T$ are purely imaginary, i.e., if $\lambda$ is an eigenvalue of $T$, then $\lambda = ia$ with $a \in \mathbb{R}$.

I can recall a proof we did in class that if $T$ is self-adjoint then all eigenvalues are real, and the basic gist of it was let $\lambda$ be an eigenvalue, then lambda is equal to its complex conjugate. I understand this; it implies that if $\lambda = a+bi = a-bi$ then $b = 0$.

It follows (in my mind, at least) that I need to show that given that $T^* = -T$ that $\lambda = -\lambda$.

I'm kind of stuck here, though. I've tried adapting the proof I have for self-adjoint operators, but I'm unable to do it without having the conjugate in the result, which is not what I want. (that seems to imply that all eigenvalues are 0, I don't think that's true).

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3 Answers 3

up vote 1 down vote accepted

Showing that $\lambda=-\lambda$ would only let you say that $\lambda=0$, not that $\lambda$ was imaginary, so that is not what you need to show.

Instead, look at what exactly was originally done for $S$ symmetric and $\lambda,v$ an eigenpair:

$$\lambda \langle v,v\rangle =\langle Sv,v\rangle=\langle v,Sv\rangle = \overline{\langle Sv,v\rangle} =\overline{\lambda \langle v,v\rangle}=\overline{\lambda}\langle v,v\rangle.$$

The first equality is by linearity, the second by $S=S^*$, the third by conjugate symmetry of the inner product, the fourth by linearity again, and the last by $\overline{ab}=\bar{a}\,\bar{b}$ along with the fact $\langle v,v\rangle$ is purely real and so is invariant under complex conjugation. From this we deduce $\lambda = \lambda^*$.

So what happens if instead $T=-T^*$?

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An alternative proof of this, probably not as enlightening but I like it: if $T^* = -T$, then $iT$ is self-adjoint (why?) and so the eigenvalues of $iT$ are real. It then follows that the eigenvalues of $T$ are purely imaginary. (again, why?)

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Suppose $\lambda$ is an eigenvalue, and $v$ a corresponding eigenvector of unit length. Then $$<v,Tv> = <T^*v,v> = -<Tv,v> = -\overline{<v,Tv>}.$$ Since $<v,Tv> = \lambda$, we have $\lambda = -\overline{\lambda}$, from which it follows that $\lambda$ is purely imaginary.

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Yes, I've found this result. But doesn't this imply that $\lambda = 0$? It seems that if $-\lambda = \overline\lambda$, then that's like saying that $-a+bi = a-bi$, which would only be true if $a = b = 0$. Am I missing something? –  Daniel Harms May 6 '12 at 7:37
@Precision: $-\lambda=-(a+bi)=-a\color{Red}-bi,$ not $-a+bi$. –  anon May 6 '12 at 7:40
You are missing something. It implies $a+ib = -a+ib$ from which you can conclude $a=0$, hence imaginary. –  copper.hat May 6 '12 at 7:40
Ah... yeah. Well, that's awkward. Makes so much sense now, thank you. –  Daniel Harms May 6 '12 at 7:43

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