Definition 8.5.1.
Let \(E\) be a Banach space and \(A \in \L(E)\text{.}\) The spectrum of \(A\text{,}\) denoted \(\sigma(A)\) is the set
\begin{equation*}
\sigma(A) = \{\la \in \C : \la I - A \text{ is not invertible.}\}
\end{equation*}
Section 8.5 The spectrum
If \(A\) is an \(n \times n\) complex matrix, then \(A\) has at least one eigenvalue. That is, there exists \(\la \in \C\) such that \(Av = \la v\) for some non-zero vector \(v \in \C^n\text{.}\) The spectrum of a matrix, denoted \(\sigma(A)\) is the set consisting of its eigenvalues. The spectrum of a matrix is used to study its structure and behavior as a linear map. We’d like to extend the notion of spectrum to the infinite dimensional setting, so we’ll take a moment to look at how we find it for matrices. In elementary linear algebra, we learn that the eigenvalues of \(A\) are the solutions to the equation \(\det(A - \la I) = 0\text{.}\) Of course what we’re actually measuring with this equation is the (lack of) invertibility of the operator \(A - \la I\text{,}\) which we could also encode geometrically with the condition that \(\ker (A - \la I) \neq \{0\}\text{.}\) Since determinants aren’t all that useful or easy to define in the infinite dimensional setting, we’ll use these to define the spectum of an operator.
Definition 8.5.2.
The resolvent set of \(A \in \L(\HH)\text{,}\) denoted \(\rho(A)\text{,}\) is the set of complex numbers \(\la\) so that \(\la I - A\) is invertible.
Given \(A \in \L(E)\text{,}\) we have that \(\sigma(A)\) is non-empty. (Else, the map \(\la \mapsto (\la I - A)\inv\) would be a bounded, non-constant, entire \(\L(E)\)-valued map, which would violate a version of Liouville’s theorem). If \(E\) is finite dimensional, then \(\sigma(A)\) consists of the eigenvalues of \(A\text{.}\) We can still use the term eigenvalue in the infinite dimensional setting in the exact same way as the finite case. Notice though that operators need not possess eigenvalues. For example, the shift operator \(S\) acting on \(\ell^2\) has no eigenvalues. (Exercise). On the other hand, since \(S\) isn’t invertible, \(0 \in \sigma(A)\text{.}\) This illustrates the significantly more complex situation that occurs in infinite dimensions. Notice that if \(\la\) is an eigenvalue of \(A\text{,}\) then \(\la \in \sigma(A)\text{.}\)
1
en.wikipedia.org/wiki/Liouville%27s_theorem_(complex_analysis)The proof of the following result, analogous to the finite case, is left as an easy exercise.
Proposition 8.5.3.
If \(A \in \L(E)\) is a Hermitian operator on a Hilbert space \(\hilbert\text{,}\) and \(\la\) is an eigenvalue of \(A\text{,}\) then \(\la \in \R\text{.}\)
A useful topological property of the spectrum is that it forms a compact subset of \(\C\text{.}\) (In the finite dimensional case, this is apparent.)
Theorem 8.5.4.
Let \(E\) be a Banach space and \(A \in \L(E)\text{.}\) Then \(\sigma(A)\) is a compact subset of \(\C\) and
\begin{equation*}
\sigma(A) \subseteq \{z \in \C: \abs{z} \leq \norm{A}\}.
\end{equation*}
Proof.
Define a function \(F: \C \to \L(E)\) by \(F(\la) = \la I - A\text{.}\) For any \(\la, \mu \in \C\text{,}\)
\begin{align*}
\norm{F(\la) - F(\mu)} \amp= \norm{(\la I - A) - (\mu I - A)}\\
\amp = \abs{\la - \mu}
\end{align*}
and so \(F\) is Lipschitz and therefore continuous in the operator norm. Now let \(GL(E)\) denote the invertible elements of \(E\text{,}\) which is an open subset of \(\L(E)\) by Theorem 8.2.2. Then \(\L(E)-GL(E)\) is closed, and by continuity so is \(F\inv(\L(E)-GL(E)) = \sigma(A)\text{.}\)
Now let \(\la \in \C\) with \(\abs{\la} \gt \norm{A}\text{.}\) Then \(\norm{\la\inv A} \lt 1\text{,}\) and so \(I - \la\inv A\) is invertible by Theorem 8.2.1. But then \(\la\inv(\la I - A)\) is invertible and so must be \(\la I - A\text{,}\) and so \(\la \notin \sigma(A)\text{.}\) Then \(\sigma(A)\) is in the closed disk of radius \(\norm{A}\) and hence is bounded. As closed and bounded subsets of \(\C\) are compact, we have established the result.
