# Math Help - Spectrum

1. ## Spectrum

If $\cal{X}$ is a Banach space and $T: \cal{X} \rightarrow \cal{X}$ is an isometry , then either $\sigma(T) \subseteq \partial \cal{D}$ or $\sigma(T) = cl \cal{D}$

where
$\partial\cal{D}$ = {z in C: |z| = 1}

Case 1: isometry is bijective
When $\sigma(T) \subseteq \partial \cal{D}$ all points of the specrum are on the boundary

Case 2: the isometry is not surjective.
$\sigma(T) = cl \cal{D}$

the spectrum is the whole disk.

2. Originally Posted by Nusc
If $\cal{X}$ is a Banach space and $T: \cal{X} \rightarrow \cal{X}$ is an isometry , then either $\sigma(T) \subseteq \partial \cal{D}$ or $\sigma(T) = cl \cal{D}$

where
$\partial\cal{D}$ = {z in C: |z| = 1}

Case 1: isometry is bijective
When $\sigma(T) \subseteq \partial \cal{D}$ all points of the specrum are on the boundary

Case 2: the isometry is not surjective.
$\sigma(T) = cl \cal{D}$
the spectrum is the whole disk.
LaTeX note: use \mathcal, not \cal, in formulas (\cal works globally, and converts the rest of the formula to gibberish).

Use the theorem which says that the boundary of the spectrum is contained in the approximate point spectrum. Suppose that $\lambda$ is an approximate eigenvalue for T, with approximate eigenvector x. Then x is a unit vector, and $Tx\approx\lambda x$. But $\|Tx\| =1$, and it follows that $|\lambda|=1$.

Thus $\partial\sigma(T)\subseteq\partial\mathcal{D}$. It then follows on purely geometrical grounds that either $\sigma(T) \subseteq\partial\mathcal{D}$ or $\sigma(T) = \overline{\mathcal{D}}$. The first alternative occurs when $0\notin\sigma(T)$ (in other words, when T is unitary), and the second alternative in the case when $0\in\sigma(T)$ (in other words, when T is a proper isometry).