426 CHAPTER 17. SPECTRAL THEORY OF LINEAR MAPS

We may not be able to explicitly evaluate f (A) in case f (λ ) is not analytic off σ (A) butat least the definition is reasonable and seems to work as it should in the special case whenf (A) is known by another method, say a power series. These details about a power seriesare left to the reader.

With Theorem 17.2.5, one can prove the spectral mapping theorem.

Theorem 17.2.6 Suppose Ω ⊇ σ (A) where Ω is an open set and let f be analyticon Ω. Let Γ be an oriented cycle, the union of oriented simple closed curves such thatn(Γ,z) = 1 if z ∈ σ (A) ,n(Γ,z) is an integer if z ∈Ω, and n(Γ,z) = 0 if z /∈Ω. Then for

f (A)≡ 12πi

∫Γ

f (λ )(λ I−A)−1 dλ ,

it follows that σ ( f (A)) = f (σ (A))

Proof: Let µ ∈ Ω. µ → f (λ )− f (µ)λ−µ

has a removable singularity at λ so it is equal to ananalytic function of µ called g(µ). Recall why this is. Since f is analytic at λ , we knowthat for µ near λ

f (µ) =∞

∑k=0

ak (µ−λ )k

and so, when this is substituted into the difference quotient, one obtains the power seriesfor an analytic function called g(µ).

Then f (λ )− f (µ) = g(µ)(λ −µ). From Theorem 17.2.5,

f (A)− f (λ ) I =1

2πi

∫Γ

( f (µ)− f (λ ))(µI−A)−1 dµ

=1

2πi

∫Γ

g(µ)(µ−λ )(µI−A)−1 dµ

= g(A)(A−λ I) = (A−λ I)g(A)

If λ ∈ σ (A) then either (A−λ I) fails to be one to one or it fails to be onto. If (A−λ I)fails to be one to one, then f (A)− f (λ ) I also fails to be one to one and so f (λ )∈σ ( f (A)).If (A−λ I) fails to be onto, then f (A)− f (λ ) I also fails to be onto and so f (λ )∈σ ( f (A)).In other words, f (σ (A))⊆ σ ( f (A)).

Now suppose ν ∈σ ( f (A)) . Is ν = f (λ ) for some λ ∈σ (A)? If not, then ( f (λ )−ν)−1

is analytic function of λ ∈ σ (A) and by continuity, this must hold in an open set Ω̂ whichcontains σ (A). Therefore, using this open set in the above considerations, it follows fromTheorem 17.2.5 that for Γ pertaining to this new open set as above,

( f (A)−ν)−1 ( f (A)−ν) =1

2πi

∫Γ

( f (µ)−ν)−1 ( f (µ)−ν)(µI−A)−1 dµ = I

and so ν was not really in σ ( f (A)) after all. Hence the equality holds. ■

17.3 Invariant SubspacesThis is where we need the machinery of Theorem 17.2.1. Up till now, we could havedone most things by simply considering large circles containing σ (A). Here the idea is toconsider pieces of σ (A). Let σ (A) = ∪n

i=1Ki where the Ki are disjoint and compact. Let

δ i = min{

dist(z,∪ j ̸=iK j

): z ∈ Ki

}> 0