Prove non-zero eigenvalues of skew-Hermitian operator are pure imaginary

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.

2 Answers

We need the following properties of the inner product

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

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

$$\langle Tu,u \rangle= \langle u,T^{*}u \rangle \Longleftrightarrow \langle Tu,u \rangle = \langle u,-Tu \rangle$$

$$\langle \lambda u,u \rangle = \langle u,-\lambda u \rangle \Longleftrightarrow \lambda \langle u,u \rangle = -\bar{\lambda} \langle u,u\rangle$$

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

What can you conclude from the last equation?

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