Suppose that $ T\in \mathcal L (V) $ is normal. Prove that $T^{*} = -T $ iff every eigenvalue of T is purely imaginary.

Question

Suppose that $ T\in \mathcal L (V) $ is normal. Prove that $T^{*} = -T $ iff every eigenvalue of T is purely imaginary.

$(\Rightarrow)$ assume $T^{*} = -T $ Suppose $\lambda $ is an eigenvalue with an eigenvector $v\neq 0 $

$$\lambda \| v\|^2 = \lambda \langle v,v \rangle $$ $$ = \langle \lambda v,v \rangle $$ $$ = \langle T v,v \rangle $$ $$ = \langle v,T^{*} v \rangle $$ $$ = \langle v,-T v \rangle $$ $$ = \langle v,-\lambda v \rangle $$ $$ = -\bar{\lambda} \langle v, v \rangle $$ $$ = -\bar{\lambda} \| v\|^2 $$ this shows us for any $ \lambda = - \bar \lambda $ so all eigenvalues are imaginary.

$(\Leftarrow ) $ Assume all the eigenvalues are imaginary WTS $T^*= -T$ can someone help me with this direction?

Answer 1

$\Longleftarrow$: Assume that $T$ is normal with imaginary eigenvalues. By the spectral theorem, there exists a unitary $U$ such that $$ T = U\pmatrix{\lambda_1\\ & \ddots \\ && \lambda_n}U^* $$ Since the eigenvalues $\lambda_i$ are imaginary, we compute $$ T^* = \left[U\pmatrix{\lambda_1\\ & \ddots \\ && \lambda_n}U^*\right]^* = U \pmatrix{\bar \lambda_1\\ & \ddots \\ && \bar \lambda_n} U^* = - U \pmatrix{\lambda_1\\ & \ddots \\ && \lambda_n} U^* = -T $$ as desired.