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I have a question regarding operator theory and would be glad if someone could help. I have a linear operator $A$ that is non-self-adjoint, unbounded and is densely defined in a Hilbert space $H$. I know how the spectrum of this operator looks like. However, not all points of the spectrum are eigenvalues. Given a point in the spectrum $\lambda_j$ (in my case $\lambda_j \in \mathbb{R}$), is there a way to check, if this point is an eigenvalue in the mathematical meaning of this word, i.e., that there exist a function $\varphi_j \in H$ such that $A\varphi_j = \lambda_j\varphi_j$?

Of course, the operator $A$ is given explicitly, i.e., I know how the corresponding linear prescription looks like.

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If the only data you have available is the spectrum, then the answer is no. You can have two operators with the same spectrum, where a certain point is an eigenvalue for one operator and not an eigenvalue for the other.

For an easy example, let $H$ be a separable Hilbert space and fix an orthonormal basis $\{e_n\}$. Let $\{\alpha_n\}$ be an enumeration of $\mathbb Q\cap=[0,1]$. Consider the two diagonal operators $S$ and $T$, where $ Se_n=\alpha_ne_n,\ \ Te_1=\frac{\sqrt2}2\,e_1,\ \ Te_n=\alpha_{n-1}e_n,\ \ n\geqq2. $ The both the spectrum of $S$ and $T$ is $[0,1]$, but $\sqrt2/2$ is an eigenvalue of $T$ and not an eigenvalue of $S$.