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About LinkBacksIfis a Banach space and
is an isometry , then either
or
where
= {z in C: |z| = 1}
Case 1: isometry is bijective
Whenall points of the specrum are on the boundary
Case 2: the isometry is not surjective.
the spectrum is the whole disk.
Spectrum
If
where
Case 1: isometry is bijective
When
Case 2: the isometry is not surjective.
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).Originally Posted by Nusc
If
where
Case 1: isometry is bijective
When
Case 2: the isometry is not surjective.
the spectrum is the whole disk.
Use the theorem which says that the boundary of the spectrum is contained in the approximate point spectrum. Suppose thatis an approximate eigenvalue for T, with approximate eigenvector x. Then x is a unit vector, and
. But
, and it follows that
.
Thus. It then follows on purely geometrical grounds that either
or
. The first alternative occurs when
(in other words, when T is unitary), and the second alternative in the case when
(in other words, when T is a proper isometry).
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