I am working on the following question, and would very much appreciate a proof-check/some final guidance on the questionable part below!
Let $V$ be a finite-dimensional complex inner-product space and let $T \in $ End$(V)$. Suppose that $T$ satisfies $T^* = -T$.
(a) Prove that all eigenvalues of $T$ are purely imaginary.
(b) Explain why $V$ has an orthonormal basis consisting of eigenvectors for $T$.
(c) The operator $I - T$ is invertible. Show that the operator $S = (I + T)(I - T)^{-1}$ is an isometry.
Part (A) My Proof: Take $v \neq 0$ such that $Tv = \lambda v$. Then we have $\langle Tv, v \rangle = \langle \lambda v, v \rangle = \lambda\langle v, v\rangle$. Further, we can say that $\langle Tv, v \rangle = \langle v, T^*v\rangle = \langle v, -Tv\rangle = \langle v, -\lambda v\rangle = \overline{-\lambda}\langle v, v\rangle$.
Thus, we must have that $\lambda = \overline{-\lambda}$. If we write $\lambda = a + bi$ then $a + bi = -(a - bi) = -a + bi \implies a = -a \implies a = 0$.
And so with $a = 0$ we have that every eigenvalue must be purely imaginary.
Part (B) By the 'Complex Spectral Theorem' we know that an operator $T$ has an orthonormal eigenbasis if it is a normal operator (namely that $TT^* = T^*T$). Now we have that $TT^* = T(-T) = -T^2$ and we have that $T^*T = (-T)T = -T^2$ and so $T$ is normal and the Spectral Theorem can be invoked.
Part (C) My thought for part C was to show that $S^*S = I$ which would demonstrate that $S$ is an isometry. However, my attempt at that does not quite work out as I was hoping.
$$ S = (I + T)(I - T)^{-1} \\ \implies S^* = ((I - T)^{-1})^*(I + T)^* \\ = (I - T^*)^{-1}(I + T^*)\\ = (I + T)^{-1}(I - T) $$
Then $S^*S = (I + T)^{-1}(I - T)(I + T)(I - T)^{-1}$.
Would it be valid to do something like: $(I - T)(I + T) = (I - T^2) = (I + T)(I - T)$ which would then give $S^*S = I$ as needed?