17.3. INVARIANT SUBSPACES 427

and let 0 < δ < min{δ i, i = 1, · · · ,n}. Then letting Ui ≡ Ki +B(0,δ ) , it follows that theUi are disjoint open sets. Let Γ j be a oriented cycle such that for z ∈ K j,n(Γ j,z) = 1 andfor z ∈ Ki, i ̸= j,n(Γ j,z) = 0, and if z ∈ Γ∗j , then n(Γi,z) = 0. Let Γ be the sum of theseoriented cycles. Thus n(Γ,z) = 0 if z /∈Ω≡ ∪n

i=1Ui and Ui ⊇ Γ∗i . Define

fi (λ )≡{

1 on Ui0 on U j for j ̸= i (17.8)

Thus fi is analytic on Ω. Then fi (A) ≡ Pi ≡ 12πi∫

Γfi (λ )(λ I−A)−1 dλ . By the spectral

mapping theorem, Theorem 17.2.6,

σ ( fi (A)) = fi (σ (A)) = {0,1} (17.9)

Note that for λ ∈ ρ (A) ,A(λ I−A)−1 = (λ I−A)−1 A as can be seen by multiplyingboth sides by (λ I−A) and observing that the result is A on both sides. Then since (λ I−A)is one to one, the identity follows. Now let Pk ∈ L (X ,X) be the linear transformationgiven by Pk =

12πi∫

Γk(λ I−A)−1 dλ .

From Lemma 17.1.1,

APk = A1

2πi

∫Γk

(λ I−A)−1 dλ =1

2πi

∫Γk

A(λ I−A)−1 dλ

=1

2πi

∫Γk

(λ I−A)−1 Adλ =1

2πi

(∫Γk

(λ I−A)−1 dλ

)A = PkA (17.10)

With these introductory observations, the following is the main result about invariantsubspaces. First is some notation.

Definition 17.3.1 Let X be a vector space and let Xk be a subspace. Then X =

∑nk=1 Xk means that every x ∈ X can be written in the form x = ∑

nk=1 xk,xk ∈ Xk. We write

X =n⊕

k=1

Xk if whenever 0 = ∑k xk, it follows that each xk = 0. In other words, we use the

new notation when there is a unique way to write each vector in X as a sum of vectors inthe Xk. When this uniqueness holds, the sum is called a direct sum. In case AXk ⊆ Xk, wesay that Xk is A invariant and Xk is an invariant subspace.

Theorem 17.3.2 Let σ (A) = ∪nk=1Kk where K j ∩Ki = /0, each K j being compact.

There exist Pk ∈L (X ,X) for each k = 1, · · · ,n such that

1. I = ∑nk=1 Pk

2. PiPj = 0 if i ̸= j

3. P2i = Pi for each i

4. X =n⊕

k=1

Xk where Xk = PkX and each Xk is a Banach space.

5. AXk ⊆ Xk which says that Xk is A invariant.

6. Pkx = x if x ∈ Xk. If x ∈ X j, then Pkx = 0 if k ̸= j.

17.3. INVARIANT SUBSPACES 427and let 0 < 6 < min{6;,i=1,--- ,n}. Then letting U; = K; +B (0,5), it follows that theU; are disjoint open sets. Let I; be a oriented cycle such that for z € Kj,n(Tj,z) = 1 andfor z € Ki,i A j,n(Tj,z) =0, and if z € T, then n(Ij,z) = 0. Let I’ be the sum of theseoriented cycles. Thus n(I’,z) = 0 if z ¢ Q = U*_,U; and U; DT. Define— f lonU;ont 0 on U; for j Ai (17.8)Thus f; is analytic on Q. Then f;(A) =P = sa Ir fila )(AI—A) ‘dA. By the spectralmapping theorem, Theorem 17.2.6,o(fi(A)) = fi(o(A)) = {0,1} (17.9)Note that for A € p(A),A(AI—A) | = (AI—A)~'A as can be seen by multiplyingboth sides by (AJ — A) and observing that the result is A on both sides. Then since (AJ — A)is one to one, the identity follows. Now let Rh, € &(X,X) be the linear transformationgiven by P, = oo Jr, AL —A)'da.From Lemma 17.1.1,1 yet 1AP = Ax (AL A) da = = ACAI A) 'da1 1 _ yt 7= ami I, (AI—A)~ ‘Add = = (fa A) ah) A=RAl (17.10)With these introductory observations, the following is the main result about invariantsubspaces. First is some notation.Definition 17.3.1 Let xX be a vector space and let X, be a subspace. Then X =Yee Xk means that every x € X can be written in the form x = Vj, Xk, Xn © Xp. We writenX= Dx: if whenever 0 = YX, it follows that each x, =0. In other words, we use thek=1new notation when there is a unique way to write each vector in X as a sum of vectors inthe Xx. When this uniqueness holds, the sum is called a direct sum. In case AX; © Xx, wesay that X; is A invariant and X; is an invariant subspace.Theorem 17.3.2 Let o(A) = Up Ky where Kj K; =, each K; being compact.There exist P, € & (X,X) for each k = 1,-++ ,n such that1. T=YVh_, Pk2. PP; =OifiF j3. P? =P, for eachin4,.X= Dx where X; = P,X and each X; is a Banach space.k=15. AX, © X; which says that X; is A invariant.6. Pox =x if x © X,. If x € Xj, then Px = 0 ifk F j.