I am thoroughly confused at the moment. I have a result which shows that for compact and symmetric operators $K$ from the Hilbert space $H$ to itself, $\pm||K||$ is an eigenvalue.

But without requiring compactness on $K$, I can get the following assuming I have eigenvalues to begin with such that $Kx=\lambda x$.

$\lvert\langle Kx,x\rangle\rvert = \lvert\langle\lambda x,x\rangle\rvert = \lvert\lambda\rvert \lVert x\rVert^2$

Taking the supremum,

$\sup\{\lvert\langle Kx,x\rangle\rvert,\ \lVert x\rVert=1\} = \lvert\lambda\rvert = \lVert K\rVert$

What is wrong with my reasoning here?

  • 2
    $\begingroup$ A non-compact operator may have no eigenvalues at all $\endgroup$
    – ziggurism
    Jan 4, 2018 at 14:20
  • $\begingroup$ You need to take the sup over all $x$ and the eigenvectors might not form a basis if $K$ is not compact. $\endgroup$
    – Walton
    Jan 4, 2018 at 14:22
  • $\begingroup$ I understand that existence is not guaranteed. And I am taking the supremum over every element, the expression of the norm on the left is a result got by virtue of $K$ being symmetric. $\endgroup$ Jan 4, 2018 at 14:26
  • $\begingroup$ $\sup_{\|x\|=1} |\langle Kx,x\rangle|=\|K\|$, but this may not be true if you remove the absolute value. $\endgroup$ Jan 4, 2018 at 15:26
  • $\begingroup$ Sorry, that's a typo. Will fix $\endgroup$ Jan 4, 2018 at 15:51

1 Answer 1


Not every set of reals has a supremum, only bounded non-empty sets do. You are making two unstated assumptions: the set $\{\lvert\langle Kx,x\rangle\rvert:x\text{ an eigenvalue}\}$ is bounded, and that it is non-empty. Without the first assumption, $\sup\lvert\langle Kx,x\rangle\rvert$ may not exist, (or be $\infty$) if the set is unbounded. Assuming it exists is equivalent to assuming your operator $K$ is bounded.

So should we conclude that for all bounded $K$, $\lVert K\rVert$ is an eigenvalue? No, we're not there yet. We need also that $\{\lvert\langle Kx,x\rangle\rvert:x\text{ an eigenvalue}\}$ is nonempty.

For bounded symmetric operators it is true that $\lVert K\rVert=\sup_{\lambda\in\sigma(K)}\lvert \lambda\rvert.$ The norm of the operator is the largest value in the spectrum, the spectral radius. Recall that the spectrum $\sigma(K)$ of an operator is the set of points such that $K-\lambda$ is not invertible. In finite dimensions, an operator is singular iff it has a nontrivial nullspace. In other words, a point is in the spectrum iff it is an eigenvalue with an eigenvector. However in infinite dimensions, non-invertibility does not guarantee that $K-\lambda$ has a nullspace. There are other ways to fail to be invertible in infinite dimensions, for example it can fail to be surjective. Or $(K-\lambda)^{-1}$ may exist but fail to be continuous. So in infinite dimensions even a symmetric operator over an algebraically closed field can fail to have eigenvalues.

But compactness of the operator $K$ is a sufficient condition that a non-zero point of the spectrum be an eigenvalue with eigenvectors. Note that it is not a necessary condition. For example the identity operator is not compact, but has eigenvalues and eigenvectors.

For example, let $H=L^2([0,1])$ and $Kf=xf(x)$. Then the norm of the operator is $\lVert K\rVert = 1.$ And the spectrum of $K$ is $[0,1]$. The spectral radius is $\sup[0,1]=1.$ But $1$ is not an eigenvalue, because there are no non-zero $L^2$ functions satisfying $xf(x)=f(x).$ In fact this operator has no eigenvalues. How can it be that $1$ is in the spectrum but is not an eigenvalue? Because $K-1$ is injective (since $K$ has no eigenvectors) but not surjective (since $1/(x-1)$ is not in $L^2$, there is no $f$ such that $(x-1)f(x)=1.$) This operator is not compact. To see this directly, observe that $f_n=\sqrt{n}\cos(nx)$ is a bounded sequence, but $\lVert Kf_n\rVert\to \infty.$ Or consider the shift operator on $\ell^2(\mathbb{N}).$

To guarantee that a point in the spectrum is actually an eigenvalue with an eigenvector, at least for the nonzero spectrum, requires an additional assumption like compactness of the operator. This result, for compact operators, is called the Fredholm alternative: if $K$ is compact an $I$ is an isomorphism, then $K+I$ either has a finite dimensional kernel, or is invertible. Terry Tao has a nice blog post containing some proofs of this result, and discussion.

Note that zero may be in the spectrum of a compact operator without being an eigenvalue. For example $K(e_n)=\frac{1}{n}e_n$ on $\ell^2(\mathbb{N}).$ And zero must be in the spectrum, see Spectrum of a compact operator

  • 2
    $\begingroup$ For bounded operators it is true that $\lVert K\rVert=\sup_{\lambda\in\sigma(K)}\lvert \lambda\rvert.$ ... No, in general the norm need not equal the spectral radius. Even in 2 dimensions you can have nonzero operator with spectrum $\{0\}$. But your assertion is OK for symmetric (or Hermitian) operators, which may be all that you need to discuss the OP. $\endgroup$
    – GEdgar
    Jan 4, 2018 at 14:43
  • $\begingroup$ @GEdgar thanks for the correction. Yes, I meant to consider symmetric operators $\endgroup$
    – ziggurism
    Jan 4, 2018 at 14:46
  • $\begingroup$ @ziggurism I really appreciate the extra information you provided. But just to clarify whether I understood your answer correctly with regards to my original question - if I make the strong assumptions that eigenvalues and the supremum exist (and moreover confine $K$ to be bounded) then $||K||$ is an eigenvalue BUT nothing can be said about its associated eigenvectors unless $K$ is also compact? Lastly I believe you are missing a 'be' after 'may' in your last paragraph. $\endgroup$ Jan 5, 2018 at 14:20
  • 1
    $\begingroup$ @plebmatician Remember what it means for $\lambda$ to be an eigenvalue (rather than just a point of the spectrum): it means that there exist nonzero vector solutions to $Kv=\lambda v$ (rather than just the failure of $K-\lambda$ to be invertible). Thus, to assume you have eigenvalues is also to assume you have eigenvectors. So if you make the assumption that your operator is bounded and has eigenvalues (which is the same as to say it has eigenvectors), then you do not need the additional assumption of compactness. For example, the identity operator has eigenvalues, but is not compact. $\endgroup$
    – ziggurism
    Jan 5, 2018 at 14:24
  • 1
    $\begingroup$ Compactness is a sufficient condition for the existence of eigenvalues/eigenvectors. Not a necessary one. $\endgroup$
    – ziggurism
    Jan 5, 2018 at 14:24

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