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Answer by David Gao for Is the span of the eigenvectors of a non self-adjoint...
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...
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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...
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