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Let $X$ be Banach space. Is there a nonzero compact operator $A: X \to X$, whose spectrum consists of zero only?

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up vote 4 down vote accepted

This is wrong if $\dim X \le 1$, so let $\dim X \ge 2$ from now on.

Let $A\colon \mathbb R^2 \to \mathbb R^2$ given by $\hat A = \begin{pmatrix} 0 & 1\\\ 0 & 0\end{pmatrix}$. Let $Y \subseteq X$ a two-dimensional subspace and $P\colon X \to X$ a projection onto $Y$ (this exists as $Y$ is finite-dimensional). Let $S \colon \mathbb R^2 \to Y$ be an isomorphism and let $A = S\hat AS^{-1}P$. Then $A^2 = 0$ and $A$ has finite-dimensional range. So $A$ is compact and $0$ is its only spectral value.

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