TITLE: $T: \ell^2 \rightarrow \ell^2$ defined by $T(x_1,x_2,x_3, \dots) = (\alpha_1x_1,\alpha_2x_2,\alpha_3x_3\dots)$
QUESTION [2 upvotes]: Kreyszig Section 8.4 Q7. Let, $T: \ell^2 \rightarrow \ell^2$ defined by $T(x_1,x_2,x_3, \dots) = (\alpha_1x_1,\alpha_2x_2,\alpha_3x_3\dots)$ where $(\alpha_i)_{i \in \mathbb{N}}$ are dense in $[0,1]$. Show that $T$ is not compact.
We have this theorem:

The set of all eigenvalues of a compact operator on a normed linear spaces is countable.

Let, us consider the orthonormal basis $(e_n)_{n \in \mathbb{N}}$ of $\ell^2$ Then $T(e_n) = \alpha_n e_n$. So $\alpha_n$ is a eigenvalue of $T$.

REPLY [3 votes]: As you already proved, all $\alpha_n$ are eigenvalues, hence the spectrum $\sigma(T)$ contains the set $\{\alpha_n, n \in \mathbb{N}\}$.
The spectrum is a closed set and the set $\{\alpha_n, n \in \mathbb{N}\}$ is dense in $[0,1]$, hence the spectrum $\sigma(T)$ contains $[0,1]$.
Any nonzero element of the spectrum of a compact operator is also an eigenvalue, hence elements of the interval $(0,1]$ are all eigenvalues of $T$.
The interval $(0,1]$ is uncountable, which contradicts the theorem you mentioned.
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