= Solution
Suppose the <spectrum> were empty. The resolvent
$$
R(\lambda)=(\lambda I-T)^{-1}
$$
would then be an entire operator-valued <function>. For $|\lambda|>\|T\|$,
$$
R(\lambda)=\frac1\lambda\sum_{n=0}^\infty\left(\frac T\lambda\right)^n,
$$
because the <series> converges in operator norm and multiplication by $\lambda I-T$ telescopes to $I$. Hence $\|R(\lambda)\|\leq(|\lambda|-\|T\|)^{-1}$.
For fixed $x,y\in\ell^2$, the <scalar> <function> $\langle R(\lambda)x,y\rangle$ is entire, bounded outside a disc by the estimate and bounded inside by compactness. Liouville's theorem makes it constant, and its <limit> at infinity makes that constant zero. This for all $x,y$ would imply $R(\lambda)=0$, impossible for an inverse. Thus the <spectrum> is nonempty.
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