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
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.2: Linear Independence, Basis, And Dimension
Problem 15EQ
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2F8ad465aa-19b2-49dd-a8d7-ecea433923bf%2F3y0w5wc_processed.png&w=3840&q=75)
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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