Let $A$ be a compact self-adjoint linear operator on a $\mathbb R$-Hilbert space $H$ and $(\lambda_n)_{n\in\mathbb N}\subseteq\mathbb R$ denote an enumeration of the spectrum $\sigma(A)$ with$^1$ $$|\lambda_n|\ge|\lambda_{n+1}|\tag1\;\;\;\text{for all }n\in\mathbb N.$$
Let $p>0$. Considering the spectral decomposition of $A$, it is easy to see that $\lambda^p_n$ is an eigenvalue$^2$ of $A^p$ for all $n\in\mathbb N$. How can we show that $(\lambda_n^p)_{n\in\mathbb N}$ is in fact an enumeration, nonincreasing in absolute value, of $\sigma(A^p)$, i.e. there are no other eigenvalues of $A^p$?
If $A$ has finite rank, the claim is easy to prove by a dimension argument, since eigenspaces with respect to different eigenvalues are known to be orthogonal. But I'm unsure what to do in the infinite-dimensional case.
$^1$ If $N:=|\sigma(A)\setminus\{0\}|\in\mathbb N$, then $\lambda_n=0$ for all $n>N$.
$^2$ $0$ is terminologically treated as an eigenvalue here.
Best Answer
This has nothing to do specifically with compact, nor with selfadjoint. If $A^p$ makes sense (say via Holomorphic, Continuous, or Borel Functional Calculus), the Spectral Mapping Theorem gives you that $$ \sigma(A^p)=\{\lambda^p:\ \lambda\in\sigma(A)\}. $$