1

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.

0xbadf00d
  • 13,422
  • How are you defining $A^p$ for noninteger $p$? (One definition of $A^p$ makes this question trivial.) – Neal May 06 '20 at 19:21
  • The spectral theorem for a compact self-adjoint $A$ is essentially the same as for the finite-dimensional case. $A=\sum_{n=1}^{\infty}\lambda_n E_n$, where $E_n$ is the orthogonal projection onto the eigenspace with eigenvalue $\lambda_n$, which must be finite-dimensional (and the eigenvalues have $0$ as the only possible point of accumulation.) – Disintegrating By Parts May 06 '20 at 20:09

1 Answers1

1

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 $$ \sigma(A^p)=\{\lambda^p:\ \lambda\in\sigma(A)\}. $$

Martin Argerami
  • 205,756
  • Simply my the identity $A=\sum_{n\in\mathbb N}\lambda_ne_n\otimes e_n$, right? – 0xbadf00d May 07 '20 at 04:06
  • Sorry, I don't know what you mean. – Martin Argerami May 07 '20 at 04:07
  • I mean that, by the spectral theorem (for compact operators here), $A$ has the expansion above for some ONB $(e_n)$ of $H$. – 0xbadf00d May 07 '20 at 04:12
  • Yes. $\ \ \ \ \ $ – Martin Argerami May 07 '20 at 04:28
  • Thanks. If $\mu_1,\ldots,$ are the distinct eigenvalues (this sequence may be end and filled with $0$'s) and $\pi_i$ is the orthogonal projection onto $E_i:=\mathcal N(\mu_i-A)$, can we infer that $A^p=\sum_{n\in\mathbb N}\mu_n^p\pi_n$? I'm not sure. I mean $\pi_n$ is an orthogonal projection, but is $\pi_n^p=\pi_n$ for a fractional $p$? – 0xbadf00d May 07 '20 at 04:47
  • 1
    Yes, $p(\pi_n)=\pi_n,p(1)$ for any polynomial with $p(0)=0$. Then take limits. – Martin Argerami May 07 '20 at 04:54
  • Okay, last question: Can we also conclude that $\sigma(|A|)={|\lambda|:\lambda\in\sigma(A)}$? If so, is there a relation between the eigenspaces, i.e. if $\lambda\in\sigma(A)$, are $\mathcal N(\lambda-A)$ and $\mathcal N(|\lambda|-|A|)$ related? I've asked for that here https://math.stackexchange.com/q/3659689/47771. – 0xbadf00d May 07 '20 at 09:50