Given a sequences of eigenvalues could we find a corresponding operator

Question

Given $a_n \in \mathbb{C}$ for n=1,2,... Could one expect to find a a self-adjoint operator $A$ on a Hilbert Space $\mathcal{H}$, such that $\lambda_n(A)=a_n$ for all n=1,2,... Just this characterize the spectrum of such an operator that would have a cardinality of $\mathbb{N}$.

What kind of assumptions on the coefficients $a_n$ one needs in order to have such an existence? Is there something on the literature about this problem?

Thank you in advance

Answer 1

As pointed out in the comments, we need $a_i\in\mathbb{R}$ if you want something self-adjoint. If you just want normal operators this condition is not necessary anymore.

Think about a matrix: $$A=\begin{bmatrix}a_1&0&0&\cdots\\ 0&a_2&0&\cdots\\ 0&0&a_3&\cdots\\ \vdots&\vdots&\vdots&\ddots\end{bmatrix}$$

If we were in finite dimension $k$, and there were only values $a_1,\ldots,a_k$, this would work. What is the associated operator in this case? You can then figure out what the infinite-dimensional analogue is.

In the infinite-dimensional case,

• The operator associated to $A$ is continuous if $\sup_i|a_i|<\infty$, but in this case you may have extra eigenvalues (e.g. if $a_i=\sqrt{2}^{-i}$ then $0$ is also an eigenvalue)
• The operator associated to $A$ is only densely defined otherwise.