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MTL 411:

Lecture D:

April 18, 2020

Let T : V → V be a linear on a vector V over C. We say that a scalar λ ∈ C is said to be an eigenvalue of T if there exists a non-zero vector x ∈ V such that T x = λx. If V is finite dimensional space, then T has atleast one eigenvalue (over C). But in the infinite dimensional case, this may not be true. For example, consider the right shift operator 2 T (x1, x2,...) = (0, x1, x2,...) on ` . It is easily verified that if T x = λx, then x = 0. Hence T has no eigenvalues. So the notion of eigenvalue should be treated in more general way which is called spectrum. Exercise. Show that the T : C[0, 1] → C[0, 1] is defined by

x Z (T f)(x) = f(t)dt, x ∈ [0, 1] 0 has no eigenvalues. Remark. A scalar λ is an eigenvalue of an operator T iff N (T ) 6= {0}. Recall, if λ is an eigenvalue of an operator T on a finite dimensional space V . Then λI −T is not invertible on V . Alternatively, if λ is not an eigenvalue of T , then λI − T is invertible on V since that 1-1 implies onto in the finite dimensional space. Moreover, (λI − T )−1 is bounded linear operator because of every linear operator is bounded in the finite dimensional space. Of course, this may not be true in the infinite dimensional space. Throughout this section we assume that X is a and T ∈ B[X].

Definition 0.1 (Resolvent). The resolvent ρ(T ) of T is defined by

−1 ρ(T ) = {λ ∈ C :(λI − T ) exists and }.

Here I denotes the identity operator on X.

Definition 0.2 (Spectrum). The spectrum of T (denoted by σ(T )) is defined by

σ(T ) = C \ ρ(T ) = {λ ∈ C : N (λI − T ) 6= {0} or R(λI − T ) 6= X}.

Lemma 0.3. If ||T || < 1, then I − T is invertible, and

∞ X (I − T )−1 = T n. (Neumann series) n=0

1 Pn k Proof. Consider the partial sum Sn = k=0 T . Then we have

n X k ||Sn − Sm|| = || T || (n > m) k=m+1 n X ≤ ||T ||k. k=m+1

Pn k Since ||T || < 1, we get the sequence of operators (Sn) is Cauchy in B[X]. Hence Sn = k=0 T P∞ k converges in B[X], say S = k=0 T . It is easy to verify that (I − T )S = S(I − T ) = I. Theorem 0.4 (Properties).

(a) If |λ| > ||T ||, then λ ∈ ρ(T ).

1 (b) If T is invertible and ||S|| < ||T −1|| , then T − S has the bounded inverse. (c) The resolvent set ρ(T ) is open, and the spectrum σ(T ) is compact.

T T Proof. (a) If |λ| > ||T ||, then || λ || < 1. Hence by above lemma, we get I − λ has the bounded inverse =⇒ (λI − T )−1 ∈ B[X], i.e., λ ∈ ρ(T ). (b) Since ||ST −1|| ≤ ||S|| ||T −1|| < 1, we get I − ST −1 has the bounded inverse. Then T − S = (I − ST −1)T . So T − S has the bounded inverse. (c) Let λ ∈ ρ(T ). We have to show λ is an interior point of ρ(T ), i.e., there exists a δ > 0 such that λ − µ ∈ ρ(T ) whenever |µ| < δ. 1 Choose δ = ||(λI−T )−1|| . Then by above result (b), we get (λ−µ)I −T has the bounded inverse whenever |µ| < δ. Thus λ − µ ∈ ρ(T ) whenever |µ| < δ. Hence ρ(T ) is open in C and it’s complement σ(T ) is closed and also bounded (why?). Therefore, σ(T ) is a compact subset of C.

Remark. The spectrum σ(T ) is nonempty. The proof involves complex analysis techniques, we omit the proof here. Examples. 2 (1) Right shift operator: T (x1, x2,...) = (0, x1, x2,...), x ∈ ` . Since ||T || = 1, the spectrum σ(T ) ⊆ {λ ∈ C : |λ| ≤ 1}. For |λ| ≤ 1,

(λI − T )x = (λx1, λx2 − x1, λx3 − x2,...).

1 1 1 2 Observe that if w = ( λ , λ2 , λ3 ,...) then (λI − T )w = (1, 0, 0,...) and w∈ / ` . So λI − T is not onto for |λ| ≤ 1. Hence σ(T ) = {λ ∈ C : |λ| ≤ 1}.

x2 x3 2 (2) Diagonal operator: T (x1, x2,...) = (x1, 2 , 3 ,...), x ∈ ` . 1 1 1 Since ( n I −T )en = 0, we get N (( n I −T )) 6= {0} for every n ∈ N. Hence n ∈ σ(T ). For λ = 0, 1 T does not have bounded inverse (why?). On the other hand, for λ∈ / { n : n ∈ N} ∪ {0}, we 1 can show that λ∈ / σ(T ). Therefore, σ(T ) = {0} ∪ { n : n ∈ N}. Exercise. Consider the Multiplication operator: T : f 7→ mf from the space (C[a, b], || · ||∞) to itself, where m ∈ C[a, b]. Show that the spectrum σ(T ) = m([a, b]).

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