Spectrum of a bounded operator $T$ satisfying $T^n=I$ Let $\mathcal{H}$ be an infinite dimensional Hilbert space, suppose $T\in \mathcal{B}(\mathcal{H})$ is a bounded operator and suppose that $n$ is the smallest natural number so that $T^n=I$.
Let $f(z) = z^n$, then $f$ is entire and so by the spectral mapping theorem, $f(\sigma(T)) = \sigma(f(T))$. $f(T) = I$ by our above supposition and so $\sigma(f(T)) = \{1\}$.
If $\sigma(T) = \{\lambda_i:i\in I\}$ (where $I$ is some index set), then we have that $f(\sigma(T)) = \{f(\lambda_i):i\in I\} = \{\lambda_i^n:i\in I\}$. Thus $\lambda_i^n = 1$.
So what we see is that $\sigma(T)$ is comprised of $n$th roots of unity. This however does not tell us which of the $n$th roots comprise the spectrum of $T$. Do all $n$th roots of unity comprise the spectrum of $T$ or can it only be some of them? If only some, what condition can we place on $T$ to force all $n$th roots to be in the spectrum?
 A: Suppose $T^{n}=I$. Let $p_{k}$ be the Lagrange polynomials
$$
                   p_{k}=\prod_{j=0,j\ne k}^{n-1}(\lambda-e^{2\pi ji/n})\left/\prod_{j=0,j\ne k}^{n-1}(e^{2\pi ki/n}-e^{2\pi ji/n})\right.\;.
$$
Notice that $p_{0}+\cdots+p_{n-1}=1$ because it is an (n-1)-st degree polynomial that equals $1$ at all n-th roots of unity. Define $P_{k}=p_{k}(T)$. Then $P_{0}+\cdots+P_{n-1}=I$. Furthermore $P_{j}P_{k}=P_{k}P_{j}=0$ for $j \ne k$ because such a product can be represented as $p(T)$ where $(\lambda^{n}-1)$ divides $p$. So,
$$
         I = (P_{0}+\cdots+P_{n-1})^{2}=P_{0}^{2}+\cdots+P_{n-1}^{2}.
$$
Using this, one obtains
$$
\begin{align}
    P_{k}^{2}= (I-\sum_{j=0,j\ne k}^{n-1}P_{j})^{2} & =I-2\sum_{j=0,j\ne k}^{n-1}P_{j}
             + \sum_{j=0,j\ne k}^{n-1}P_{j}^{2} \\
             & = I-2(I-P_{k})+(I-P_{k}^{2})
\end{align}
$$
Therefore, $P_{k}^{2}=P_{k}$ is a projection. Let $\lambda =e^{2\pi i/k}$. Then $(T-\lambda^{k}I)P_{k}=0$, and
$$
           T=T(P_{0}+\cdots+P_{n-1}) = \lambda^{0}P_{0}+\cdots+\lambda^{k-1}P_{k-1}.
$$
In other words, $\mathcal{H}$ is a direct sum of closed subspaces on which $T$ is $\lambda^{k}$ times the identity. A particular $k$ can be missing in such a sum if $P_{k}=0$, which is certainly possible because there is no assumption that $\lambda^{n}-1$ is a minimal annihilating polynomial for $T$.
Conversely, given any finite direct sum decomposition of $\mathcal{H}$ into non-trivial closed subspaces $M_{j}$, it is possible to define $T$ as a scalar multiple of the identity on each $M_{j}$ with scalars taken to be any $n$-th root of unity. Then $T^{n}=I$. So the spectrum of such a $T$ can be any non-empty subset of the n-th roots of unity.
These arguments work the same for any annihilating polynomial with distinct roots.
A: This is all I can think of.  "There exists an invertible operator $S$ such that $ST = e^{2\pi i/n} TS$."  This will imply what you want, because the spectrum of $STS^{-1}$ is equal to the spectrum of $T$.
My intuition was when $T$ is diagonal, and $S$ is some kind of operator that permutes the basis elements.
