An approximate eigenvalue that is not an eigenvalue

Question

Could you please help me understand why $\lambda$ in the example below is not an eigenvalue? It's easy to see that each $\lambda_n$ is an eigenvalue but I am having difficulty ascertaining that their limit is not.

Thank you in advance.

Answer 1

This answer is based on the comments of lulu :

Let $x$ be an eigenvector and $\mu$ the associated eigenvalue. If $x$ is in the span of the $e_i$s, we would have $\mu = \lambda_n$ for some $n$ and therefore $\mu\ne \lambda$.

Otherwise, since $$\mu x = \sum_{n=0}^\infty \lambda_n \langle x, e_n\rangle e_n$$ is in the closure of the span of the $e_n$, either :

• $\mu \ne0$ and $x$ is in the closure of the span of the $e_i$ in which case by continuity of $x \mapsto \langle x, e_n \rangle$, $\mu = \lambda_n$ for some $n$, therefore $\mu\ne \lambda$.

• or $\mu =0$ in which case $\langle x, e_n\rangle =0$ for all $n$, ie $x\in Span(e_n)^{\bot}$. Thus you also need $\lambda \ne0$ or $Span(e_n)^{\bot} = \{0\}$ (that is, the $(e_n)$ being an orthonormal basis) for the statement to hold (the reverse implication being rather straightforward).