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Let X be a normed space and A € BL(X) be of finite rank. Then σe (A) = σ₁(A) = o(A).

Let X be a normed space and A € BL(X) be of finite rank. Then σe (A) = σ₁(A) = o(A).

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 58EQ
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12.2 Theorem Let X be a normed space and A € BL(X) be of finite rank. Then σ₂(A) = σa (A) = o(A). Proof: It is enough to show that o(A) C oe(A). Let koe(A), that is, let A kI be injective. We show that A - kI is invertible. Case 2: X is infinite dimensional. In this case k 0, since oth- erwise A would be an injective operator on X and if {x₁,x2,...} is an infinite linearly independent subset of X, then {A(x₁), A(x₂),…} would be an infinite linearly independent subset of R(A), contradict- ing its finite dimensionality. Let B denote the restriction (A-kI)|R(A) : R(A) → R(A). Since A kI is injective, so is B. As we have seen in Case 1, B is then surjective as well. Now consider y E X. Then A(y) = R(A). Hence there is some u € R(A) with B(u) = A(y), that is, A(u) - ku = A(y), or A(u - y) = ku. Letting r = (u - y)/k, we find that A(z) = u = kx + y, that is, (A − kI)(x) = y. Thus A - kI is surjective. Next, we claim that A - kI is bounded below. Otherwise, there is a sequence (Tn) in X such that ||A(zn) - kxn|| → 0 as n → ∞ and ||₁|| = 1 for each n. Then (A(n)) is a bounded sequence in the finite dimensional normed space R(A). By 5.5, there is a convergent subsequence (A(xn;)). If A(xn,) →y, then kæn, → y as well. Since ||y|| = |k| lim,→∞ ||n, || = |k| #0, we see that y # 0. But A(y) = A(lim kän,) = k lim A(xn,) = ky, and since A-kI is injective, we must have y = 0. This contradiction justifies our claim. Now A - kI is invertible by 12.1.
Transcribed Image Text:12.2 Theorem Let X be a normed space and A € BL(X) be of finite rank. Then σ₂(A) = σa (A) = o(A). Proof: It is enough to show that o(A) C oe(A). Let koe(A), that is, let A kI be injective. We show that A - kI is invertible. Case 2: X is infinite dimensional. In this case k 0, since oth- erwise A would be an injective operator on X and if {x₁,x2,...} is an infinite linearly independent subset of X, then {A(x₁), A(x₂),…} would be an infinite linearly independent subset of R(A), contradict- ing its finite dimensionality. Let B denote the restriction (A-kI)|R(A) : R(A) → R(A). Since A kI is injective, so is B. As we have seen in Case 1, B is then surjective as well. Now consider y E X. Then A(y) = R(A). Hence there is some u € R(A) with B(u) = A(y), that is, A(u) - ku = A(y), or A(u - y) = ku. Letting r = (u - y)/k, we find that A(z) = u = kx + y, that is, (A − kI)(x) = y. Thus A - kI is surjective. Next, we claim that A - kI is bounded below. Otherwise, there is a sequence (Tn) in X such that ||A(zn) - kxn|| → 0 as n → ∞ and ||₁|| = 1 for each n. Then (A(n)) is a bounded sequence in the finite dimensional normed space R(A). By 5.5, there is a convergent subsequence (A(xn;)). If A(xn,) →y, then kæn, → y as well. Since ||y|| = |k| lim,→∞ ||n, || = |k| #0, we see that y # 0. But A(y) = A(lim kän,) = k lim A(xn,) = ky, and since A-kI is injective, we must have y = 0. This contradiction justifies our claim. Now A - kI is invertible by 12.1.
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