I am currently dealing with the spectral theorem for compact, self-adjoint operators. From this theorem, we know that for a linear, bounded, compact and self-adjoint operator the seuqence of eigenvalues is real and the only possible accumulation point is $0$. Hence, we can analyze e.g. the convergence of the sum of all eigenvalues of such operators.

I was wondering, if we there are statements for 'the other way around'. Assume, we have a sequence $(\lambda_n)_{n \in\mathbb{N}}$ of real values with accumulation point $0$. Let's assume that $(\lambda_n)_{n \in\mathbb{N}} \in \ell^1(\mathbb{R})$. Is there any linear, bounded, compact, self-adjoint operator $T$, such that this operator has the eigenvalues $(\lambda_n)_{\mathbb{N}}$. In other words, is there an one-to-one correspondence between the set of $\ell^1$ sequences and such operators $T$, or at least some relation?

I haven't seen anything in my lecture notes or in books on this topic. Is this, because it is only an 'unconventional' question or is it because there is no such statement?

Thank you in advance for your help!

EDIT: Of course, we need an operator, acting on an infinite-dimensional space

  • $\begingroup$ you need at least infinite dimension $\endgroup$
    – Exodd
    Commented Jul 2, 2020 at 18:12
  • $\begingroup$ You can define the operator which has "matrix form" $\text{diag}(\lambda_1,\lambda_2,\dots)$. I think this is bounded, self-adjoint, and compact because $\lambda_n\to 0$? $\endgroup$
    – Keshav
    Commented Jul 2, 2020 at 18:14
  • $\begingroup$ Just make sure that the eigenspace associated with the non-zero eigenvalue is finite-dimensional. Then all you need is $\lambda_n\rightarrow 0$ as $n\rightarrow\infty$. $\endgroup$ Commented Jul 2, 2020 at 19:00
  • $\begingroup$ Thank you for your comments. Of course, we need at least infinite dimensions - I've added this in my text. I suppose that the compact, bounded, self-adjoint and compact operators are kind of the 'most natural extension' of matrices to infinite dimensional vector spaces, so the diagonal-form might work. @DisintegratingByParts if I get you right you suggest to take a sequence with only finitely many non-zero entries? This would be the easiest way to force finite-dimensional eigenspaces corresponding to the non-zero eigenvalues. So e.g. $\lambda_n = 1/n^2$ won't work? $\endgroup$
    – pcalc
    Commented Jul 2, 2020 at 21:07
  • $\begingroup$ Why not take $\ell^2$ with standard basis $e_k$, and define the operator $T$ by $Te_k = \lambda_ke_k$? $\endgroup$
    – Neal
    Commented Jul 2, 2020 at 21:47

1 Answer 1


Let $\sigma = (\lambda_n)_{n=0}^\infty\subset \mathbb{R}$ be a sequence of real numbers accumulating at zero, such that $\sum|\lambda_n| < \infty$. (The summability condition implies that $|\lambda_n|$ has finite multiplicity and is bounded above.)

Let $H$ be a separable Hilbert space with inner product $\langle \cdot,\cdot\rangle$ and let $\{e_n\}_{n=0}^\infty$ be an orthonormal basis of $H$. Define an operator $T$ by setting $Te_n = \lambda_ne_n$, with domain $$D(T) = \bigg\{ v \in H\ \bigg|\ \sum_n (\lambda_n\langle v, e_n\rangle)^2 < \infty \bigg\}$$ Then:

  1. Because $\sigma$ is absolutely summable, the operator $T$ is bounded and has domain $D(T) = H$.
  2. As $\langle e_j, T^*e_k \rangle = \lambda_k$, the operator $T$ is manifestly self-adjoint.
  3. For each $k \geq 0$, define the finite rank operator $T_k$ by setting $$T_ke_n = \begin{cases}\lambda_ne_n, & n \leq k \\ 0, & n > k\end{cases}$$ Denoting by $\|\cdot \|$ the operator norm, observe that $\|T_k - T\| \leq \sup_{i > k} |\lambda_k| \to 0$, so $T_k\to T$ in the operator norm. As $T$ is the operator norm limit of finite rank operators, it is compact.

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