Consider FT on $L^2$. I've seen many times the following argument:
"$\mathscr F^4=\textrm{id} \ \Rightarrow \sigma(\mathscr F)\subset \{\pm i, \pm 1\}$"
I don't really get it. It's clear that if $\lambda$ is an eigenvalue of $A$ and $A^n=\textrm{id}$, then $\lambda^n=1$. But spectrum does not consist of eigenvalues only, there's also situation when $A-\lambda I$ is not surjective. So my question is how should we deal with this case?
We know that $\mathscr F:L^2\rightarrow L^2$ is unitary, in particular - bijective and bounded. Why $\mathscr F-\lambda I$ is always surjective?
Question could be generalized: if $A^n=I$, what can I say about $\sigma(A)$ (perhaps with some additional restrictions on $A$).
Thanks!
This follows from the spectral mapping theorem. In the simplest case which is sufficient for your application, assume $p(z)$ is a polynomial, and $T:\mathcal H\rightarrow\mathcal H$ be an operator from a Hilbert space to itself, then the spectrum of $p(T)$ is exactly $\{p(\lambda) : \lambda\in\sigma(T)\}$. In particular, if $\lambda$ is in the spectrum of $A$ where $A^n=I$, then $\lambda^n$ is in the spectrum of $I$, hence $\lambda^n=1$.
Indeed, for any $\lambda\in\mathbb C$, consider the factorization of the polynomial $q(z) = p(z)-\lambda = \prod_i (z-z_i)$, we have $p(T)-\lambda I = \prod_i (T-\lambda z_i)$. If $\lambda$ is in the spectrum of $p(T)$, $\prod_i (T-\lambda z_i)$ is not invertible, hence at least one of its factors is not, that is $z_i\in\sigma(T)$ for some $i$, hence $p(z_i)-\lambda=0$, $\lambda=p(z_i)$. We have shown that $\sigma(p(T))\subset p(\sigma(T))$, and it's not hard to show the other direction.