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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?

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    $\begingroup$ I don't see why in the finite dimensional case you can say that $z^n-1$ is the minimum polynomial. For example, if $n=4$, then $z^2+1$ might be the minimum polynomial, say with $T = \begin{bmatrix}0&1\\-1&0\end{bmatrix}$. $\endgroup$ – Stephen Montgomery-Smith Jun 23 '14 at 19:27
  • $\begingroup$ Thanks @StephenMontgomery-Smith. I edited the post accordingly. $\endgroup$ – Cameron Williams Jun 23 '14 at 19:32
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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.

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  • $\begingroup$ I completely forgot about this post until now. So what you're saying is that if $T^n=I$ is the minimum polynomial, all of the $n$th roots are in fact eigenvalues? $\endgroup$ – Cameron Williams Jul 13 '14 at 20:48
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    $\begingroup$ If $\mathcal{H}$ is over $\mathbb{C}$, then some or any of the roots can be. Some of the $P_{n}$ may be $0$. In any event $\mathcal{H}$ splits into a direct sum of closed subspaces on which $T$ is $\lambda I$, where $\lambda$ is a root of $\lambda^{n}-1=0$. $\endgroup$ – DisintegratingByParts Jul 13 '14 at 21:02
  • $\begingroup$ Oh I see. I think I misinterpreted. Thanks! $\endgroup$ – Cameron Williams Jul 13 '14 at 22:58
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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.

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