What does it mean that an eigenvalue is "isolated"? My intuitive understanding says it is when one can find an open ball around it such that there is no other eigenvalue in that open ball.

However, I am reading a book ("Perturbation of Spectra in Hilbert Space" by Friedrichs) that says:

...we assume that the undisturbed operator $H_0$ is Hermitian, possesses a spectral resolution, and has a single eigenvalue $\omega_0$ with an eigenvector $X_0$ [such that $X_0\neq 0$]. We assume this eigenvalue to be isolated, so that the equation $$ (H_0-\omega_0)X=\Psi $$ has a solution $X$ whenever the given right member $\Psi$ is orthogonal to $X_0$.

My question is: how does it follow from $\omega_0$ being isolated that the equation has a solution only if $<X_0,\Psi>=0$?

No matter if $\omega_0$ is isolated or not, we have that $$ <X_0, (H_0-\omega_0)X> = <(H_0-\omega_0)X_0, X> = <0,X> = 0$$so that if the equation $$ (H_0-\omega_0)X=\Psi $$ is satisfied by $X$ then $<X_0, \Psi>=0$.

What does the isolatedness of $\omega_0$ have to do with it?

  • 1
    $\begingroup$ The paragraph you quoted is saying something different than what you stated. It says if $\Psi$ is orthogonal to $X_0$ then there is a solution. You are saying the converse, if there is a solution then $\Psi$ is orthogonal to $X_0$. $\endgroup$
    – muaddib
    May 22, 2015 at 8:58
  • $\begingroup$ I can see that, and I was wondering if perhaps that has something to do with assumption of $\omega_0$ being isolated? At any rate, I was able to show that if there is a solution, $<\Psi, X_0>=0$, but how do you show the converse? $\endgroup$
    – PPR
    May 22, 2015 at 9:16

2 Answers 2


What does it mean that an eigenvalue is "isolated"?

An eigenvalue is isolated if it is an isolated point of the spectrum, i.e., it has a neighborhood in which there are no other points of the spectrum. This is a stronger property than not having other eigenvalues around.

The claim is that the equation $ (H_0-\omega_0)X=\Psi $ has a solution for all $\Psi$ such that $\Psi\perp X_0$. Simply put, the range of $H_0-\omega_0I$ is $X_0^\perp$.

You may recall that the closure of the range of a Hermitian operator is the orthogonal complement of the kernel. Since the kernel is the span of $X_0$, it follows that the range of $H_0-\omega_0I$ is dense in $X_0^\perp$.

It remains to show that the range of $H_0-\omega_0I$ is closed. This is a consequence of the spectral theorem for normal operators: e.g., see Proposition IX.4.5 in Conway's Functional Analysis which says that for a normal operator $T$ the range of $T-\lambda I$ is closed if and only if $\lambda$ is not a limit point of $\sigma(T)$.

  • $\begingroup$ > This is a stronger property than not having other eigenvalues around. \\ How is it any different than the neighborhood definition? $\endgroup$ May 22, 2015 at 22:16
  • $\begingroup$ Because not every point in the spectrum is an eigenvalue. $\endgroup$
    – user147263
    May 22, 2015 at 22:18
  • $\begingroup$ What is the difference between "neighborhood" and "around"? Maybe I don't know what "spectrum" means in this context. $\endgroup$ May 22, 2015 at 22:27
  • 1
    $\begingroup$ No difference. See Spectrum (functional analysis) $\endgroup$
    – user147263
    May 22, 2015 at 22:29
  • $\begingroup$ @Yes thanks for your answer, it helps a lot. I am wondering why is $ker(H_0-\omega_0) = span(X_0)$? Couldn't it be that there are other eigenvectors for the same $\omega_0$? $\endgroup$
    – PPR
    May 23, 2015 at 12:36

I have good reasons to believe that the author of the book is treating a non-degenerate case but forgot to mention it. That is why he has treated the degenerate case separately in appendix I. It is evident because if there were another linearly independent eigenvector $X_1$ to the same eigenvalue, the statement must be modified to "the equation $(H_0−ω_0)X=Ψ$ has a solution $X$ whenever the given right member $\Psi$ is orthogonal to $X_0$ and $X_1$" which automatically satisfies "$\Psi$ is orthogonal to $X_0$", no problem there. But then you can apply Gram-Schmidt process and obtain a vector $X_1'$ that is orthogonal to $X_0$ and does not belong to Ran$(H_0 -\omega_0)$ which contradicts the original statement.


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