Measure, Integration & Real Analysis, 2021a

298 Chapter 10 Linear Maps on Hilbert Spaces
The next result provides a nice way to compute the spectrum of a polynomial
applied to an operator. For example, this result implies that if T is a bounded operator
on a complex Banach space, then the spectrum of T 2 consists of the squares of all
numbers in the spectrum of T.
As with the previous result, the next result fails on real Banach spaces. As you
can see, the proof below uses factorization of a polynomial with complex coefficients
as the product of polynomials with degree 1, which is not necessarily possible when
restricting to the field of real numbers.
10.40 Spectral Mapping Theorem
Suppose T is a bounded operator on a complex Banach space and p is a
polynomial with complex coefficients. Then
sp ( p(T) ) = p ( sp(T) ) .
Proof If p is a constant polynomial, then both sides of the equation above consist
of the set containing just that constant. Thus we can assume that p is a nonconstant
polynomial.
First suppose α ∈ sp ( p(T) ) . Thus p(T) − αI is not invertible. By the Fundamental
Theorem of Algebra, there exist c, β 1 ,...β n ∈ C with c ̸= 0 such that
10.41 p(z) − α = c(z − β 1 ) ···(z − β n )
for all z ∈ C. Thus
p(T) − αI = c(T − β 1 I) ···(T − β n I).
The left side of the equation above is not invertible. Hence T − β k I is not invertible
for some k ∈{1, . . . , n}. Thus β k ∈ sp(T). Now10.41 implies p(β k )=α. Hence
α ∈ p ( sp(T) ) , completing the proof that sp ( p(T) ) ⊂ p ( sp(T) ) .
To prove the inclusion in the other direction, now suppose β ∈ sp(T). The
polynomial z ↦→ p(z) − p(β) has a zero at β. Hence there exists a polynomial q
with degree 1 less than the degree of p such that
for all z ∈ C. Thus
p(z) − p(β) =(z − β)q(z)
10.42 p(T) − p(β)I =(T − βI)q(T)
and
10.43 p(T) − p(β)I = q(T)(T − βI).
Because T − βI is not invertible, T − βI is not surjective or T − βI is not injective.
If T − βI is not surjective, then 10.42 shows that p(T) − p(β)I is not surjective. If
T − βI is not injective, then 10.43 shows that p(T) − p(β)I is not injective. Either
way, we see that p(T) − p(β)I is not invertible. Thus p(β) ∈ sp ( p(T) ) , completing
the proof that sp ( p(T) ) ⊃ p ( sp(T) ) .
Measure, Integration & Real Analysis, by Sheldon Axler