Finite order operators has pure point spectrum I have to prove that the spectrum of an operator $T:H\rightarrow H$, $H$ Hilbert, s.t. $T^k=id_H$ has pure point spectrum. Now, it is obvious that $\sigma_0(T)\subseteq\{e^{i\mu2\pi/k}\}_{\mu=0}^{k-1}$ is the point spectrum of $T$ and $\sigma_0(T)\subseteq\sigma(T)$. However, I cannot find out how to prove that the two spectra coincide. I guess one has to prove that $T-\lambda{id_H}$ is invertible whenever $\lambda\notin\sigma_0(T)$. I thought I can do it in this way: $T-\lambda id_H$ is clearly injective under the assumptions on $\lambda$ (otherwise $\lambda\in\sigma_0(T)$). To prove that it's surjective, one may use the fact that $H=ker(T-\lambda id_H)\oplus Im(T-\lambda id_H)=\{0\}\oplus Im(T-\lambda id_H)=Im(T-\lambda id_H)$ (the equalities are isometric equivalences). Then, the open mapping theorem would imply the continuity of $(T-\lambda id_H)^{-1}$. Is it correct?
However, I'd like to avoid to invoke the open mapping theorem. Is there an easy way to do so?
 A: Hint: You can directly write down the inverse of $T -\lambda I$: Compute $(I+S+S^{2}+\cdots +S^{k-1})) (T-\lambda I)$ where $S=\frac T {\lambda}$. You will see that $T-\lambda I$ is invertible whenever $\lambda ^{k} \neq 1$ and $\lambda \neq 0$.
A: Using the hint I was given, I prove that $T$ is a pure point spectrum operator. Clearly, $0\notin\sigma(T)$, since $T^k=id_H \ \Longrightarrow \ T^{k-1}=T^{-1}$, that is $T$ is invertible and its inverse is obviously a continuous mapping (as a composition of bounded operators from $H$ into itself). Let $\lambda\in\mathbb{C}\setminus(\{0\}\cup\sigma_0(T))$. Set $T^0=id_H$.
\begin{equation}
\underbrace{\sum_{j=0}^{k-1}\frac{T^j}{\lambda^j}}_{$=:S$}(Tx-\lambda x)=\sum_{j=0}^{k-1}\frac{T^{j+1}x}{\lambda^j}-\frac{T^j(\lambda x)}{\lambda^j}=\frac{1-\lambda^{k}}{\lambda^{k-1}}x=x.
\end{equation}
That is, $\frac{\lambda^{k-1}}{1-\lambda^{k}}S=T^{-1}$ and the assumptions on $\lambda$ imply the well-posedness of the factor $\frac{\lambda^{k-1}}{1-\lambda^{k}}$ and the fact that it does not vanish.
A: Your argument to show that $\sigma(T)\subset\{\lambda:\ \lambda^k=1\}$, minus the typos, is fine. It is part of the more general Spectral Mapping Theorem (the "polynomial version"s says that if $p(T)=0$ for some polynomial, then $\sigma(p(T))=p(\sigma(T))$).
But, it has nothing to do with the point spectrum. Your argument for the point spectrum is flawed, as far as I can tell, because you have not shown that the image of $T-\lambda I$ is closed.
I might be wrong, of course, but I don't think there is an elementary argument to show this.
One way to prove the result by using some machinery, is to use the Holomorphic Functional Calculus. Once you know that the spectrum is discrete, given $\lambda_0\in\sigma(T)$ you can construct a holomorphic function (a polynomial, actually) $p$ with $p(\lambda_0)=1$ and $p(\lambda)=0$ for $\lambda\in\sigma(T)\setminus\{\lambda_0\}$. The Functional Calculus then gives you an operator $p(T)$. As $p^2=p$ on $\sigma(T)$, we get $p(T)^2=p(T)$. As this is nonzero, take $x$ such that $p(T)x\ne0$; this is an eigenvector for $\lambda_0$, since
$$
Tp(T)x=\lambda_0\,p(T)x.
$$
This equality occurs due to the equality of functions $tp(t)=\lambda_0p(t)$ on $\sigma(T)$.
