I know that for a compact self-adjoint operator (I assume a separable Hilbert space), the eigenfunctions form a Schauder basis for the entire space. But if the operator is not self-adjoint, does this property still hold at least for the range of the operator? I cannot find a reference of it, which looks suspicious to me.
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$\begingroup$ Let $(\Omega, \mathcal{F}, \mu)$ be a positive measure space and, for any $p \in [1,+\infty[,$ let $L^{p}((\Omega, \mathcal{F}, \mu);\mathbb{R})$ be the space of (equivalence classes of) Borel measurable functions $x: \Omega \rightarrow \mathbb{R}$ such that $\int_{\Omega} | x(\omega) |^{p} \mu(d \omega) < +\infty.$ Set $\Omega = [0,1]$ and let $\mathcal{X} = L^{2}((\Omega, \mathcal{B}(\Omega), \mu);\mathbb{R})$ where $\mu$ is the Lebesgue measure. Now for any $t \in \Omega,$ consider the operator $B : \mathcal{X} \rightarrow \mathcal{X} : x(t) \mapsto t x(t).$ What are the eigenvalues of $B?$ $\endgroup$– Trouble Is BackCommented May 23 at 16:10
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$\begingroup$ This operator is know to have no eigenvalues right? But I do not think that it is compact, so it cannot stand for a counterexample. $\endgroup$– Cyril SolerCommented May 23 at 16:50
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$\begingroup$ Ah, yes, my apologies. I did not see that specification in your statement. For compact operators, there is a theorem that says the spectrum of the operator, say $K,$ either coincides with the point $0$ or has the form $\{ 0 \} \cup \{ k_n \},$ where all numbers $k_n$ are eigenvalues of $K$ of finite multiplicity. That is, $\textrm{dim}\,\textrm{ker}(K - k_n \textrm{Id}) < +\infty$ and the collection $\{ k_n \}$ is either finite or is a sequence converging to zero. This is directly from "Real and Functional Analysis" by Bogachev (7.3.2. Theorem). $\endgroup$– Trouble Is BackCommented May 23 at 17:01
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$\begingroup$ As an explicit example (from the same book - 7.3.3. Example), it cites the Volterra operator on $L^{2}([0,1])$ or on $C([0,1]).$ The book can be found here: link.springer.com/book/10.1007/978-3-030-38219-3 If you would like for me to explicitly show that this example is true, please let me know, and I'll provide the details in a solution. $\endgroup$– Trouble Is BackCommented May 23 at 17:04
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This is not even true in finite dimensions. Say, the matrix $\begin{pmatrix} 1 & 1\\0 & 1 \end{pmatrix}$ has range the entire $\mathbb{C}^2$, but the span of eigenvectors is $\text{span}\{\begin{pmatrix} 1\\0 \end{pmatrix}\}$.
