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Just like the title: Assume $T$ is a skew-Hermitian but not a Hermitian operator an a finite dimensional complex inner product space V. Prove that the non-zero eigenvalues of $T$ are pure imaginary.

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up vote 2 down vote accepted

We need the following properties of the inner product

i) $<au,v> = a<u,v> \quad a \in \mathbb{C}$,

ii) $ \langle u, a v \rangle = \overline{\langle a v, u \rangle} = \overline{a} \overline{\langle v, u \rangle} = \overline{a} \langle u, v \rangle \quad a \in \mathbb{C}.$

Since T is skew Hermitian, then $T^{*}=-T$. Let $u$ be an eigenvector that corresponds to the eigenvalue $\lambda$ of $T$, then we have

$$ <Tu,u>=<u,T^{*}u> \Longleftrightarrow <Tu,u>=<u,-Tu>$$

$$ <\lambda u,u>=<u,-\lambda u> \Longleftrightarrow \lambda <u,u>= -\bar{\lambda} <u,u> $$

$$ \Longleftrightarrow \lambda = -\bar{\lambda} \Longleftrightarrow x+iy = -x+iy. $$

What can you conclude from the last equation?

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Hint $$\langle \mathbf{v},\ A\mathbf{v}\rangle = \mathbf{v}^*A\mathbf{v}= \lambda\|\mathbf{v}\|^2$$ $$\langle A\mathbf{v},\ \mathbf{v}\rangle=\mathbf{v}^*A^*\mathbf{v}=-\mathbf{v}^*A\mathbf{v}=-\lambda\|\mathbf{v}\|^2$$

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