Measure, Integration & Real Analysis, 2021a

Section 10B Spectrum 297
A major result in finite-dimensional
linear algebra states that every operator
on a nonzero finite-dimensional complex
vector space has an eigenvalue. We have
seen examples showing that this result
does not extend to bounded operators on
complex Hilbert spaces. However, the
next result is an excellent substitute. Although
a bounded operator on a nonzero
The spectrum of a bounded operator
on a nonzero real Hilbert space can
be the empty set. This can happen
even in finite dimensions, where an
operator on R 2 might have no
eigenvalues. Thus the restriction in
the next result to the complex case
cannot be removed.
complex Hilbert space need not have an eigenvalue, the next result shows that for
each such operator T, there exists α ∈ C such that T − αI is not invertible.
10.38 spectrum is nonempty
The spectrum of a bounded operator on a complex nonzero Hilbert space is a
nonempty subset of C.
Proof Suppose T ∈B(V), where V is a complex Hilbert space with V ̸= {0}, and
sp(T) =∅. Let f ∈ V with f ̸= 0. Take g = T −1 f in 10.37. Because sp(T) =∅,
10.37 implies that the function
α ↦→ 〈 (T − αI) −1 f , T −1 f 〉
is analytic on all of C. The value of the function above at α = 0 equals the average
value of the function on each circle in C centered at 0 (because analytic functions
satisfy the mean value property). But 10.34(c) implies that this function has limit 0
as |α| →∞. Thus taking the average over large circles, we see that the value of the
function above at α = 0 is 0. In other words,
〈
T −1 f , T −1 f 〉 = 0.
Hence T −1 f = 0. Applying T to both sides of the equation T −1 f = 0 shows that
f = 0, which contradicts our assumption that f ̸= 0. This contradiction means that
our assumption that sp(T) =∅ was false, completing the proof.
10.39 Definition p(T)
Suppose T is an operator on a vector space V and p is a polynomial with coefficients
in F:
p(z) =b 0 + b 1 z + ···+ b n z n .
Then p(T) is the operator on V defined by
p(T) =b 0 I + b 1 T + ···+ b n T n .
You should verify that if p and q are polynomials with coefficients in F and T is
an operator, then
(pq)(T) =p(T) q(T).
Measure, Integration & Real Analysis, by Sheldon Axler