TITLE: Classify the spectrum of $T$ is $T$ is a bounded linear operator on a complex Hilbert space $H$ and $-T = T^*, T^2 = -I,$ and $T \not= \pm iI$
QUESTION [0 upvotes]: I've been struggling with this question for some time. Here is what I have so far:
I guessed that the inverse of $T - \lambda I$ is of the form $c(T+\lambda I)$. Then, solving for $I = (T-\lambda I)c(T+\lambda I)$, we get the equation $c = \frac{-1}{1 + \lambda^2}$, which is well-defined for $\lambda \not= \pm i$. 
Thus, the spectrum of $T$ only contains $-i$ and $i$. (It has to contain these two since the spectrum is non-empty and closed). But I have no clue how to figure out if $i$ and $-i$ are in the point or continuous spectrum. (They're not in the residual spectrum since the residual spectrum of unitary operators is empty and $T$ is unitary). Any ideas would be helpful.

REPLY [2 votes]: Point spectrum.
Since $T$ is normal, the spectral theorem applies. Let $P_+$ and $P_-$ be the spectral projections on $\{+ i\}$ and $\{-i\}$ respectively.  These are both nontrivial since $T \ne \pm i I$.  Then $T P_+ = i P_+$ and $T P_- = -i P_-$, and nonzero vectors in the range of $P_\pm$ are eigenvectors for $\pm i$.
If you don't want to use the spectral theorem, you can do it explicitly: $P_+ = (I - i T)/2$ and 
$P_- = (I + iT)/2$.  Show that these are 
orthogonal projections with $T P_\pm = \pm i P_\pm$.
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