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Theorem 3. Let A be a bounded linear operator acting between two real Hilbert spaces. Then 1. [R(A)]+ = N(A*). 2. R(A) = [N(A*)]*. 3. [R(A*)]* = N (A). 4. R(A*) = [N(A)]+. %3D %3D

Theorem 3. Let A be a bounded linear operator acting between two real Hilbert spaces. Then 1. [R(A)]+ = N(A*). 2. R(A) = [N(A*)]*. 3. [R(A*)]* = N (A). 4. R(A*) = [N(A)]+. %3D %3D

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Theorem 3. Let A be a bounded linear operator acting between two real Hilbert spaces. Then 1. [R(A)] = Nr(A*). 2. R(A) = [N(A*)]*. 3. [R(A*)]+ = N(A). 4. R(A*) = [N(A)]+. Proof. Part 1 is just Theorem 1. To prove part 2, take the orthogonal complement of both sides of 1 obtaining [R(A)] = [N(A*)]*. Since R(A) is a subspace, the result follows. Parts 3 and 4 are obtained from 1 and 2 by use of the relation A** = A. I
Transcribed Image Text:Theorem 3. Let A be a bounded linear operator acting between two real Hilbert spaces. Then 1. [R(A)] = Nr(A*). 2. R(A) = [N(A*)]*. 3. [R(A*)]+ = N(A). 4. R(A*) = [N(A)]+. Proof. Part 1 is just Theorem 1. To prove part 2, take the orthogonal complement of both sides of 1 obtaining [R(A)] = [N(A*)]*. Since R(A) is a subspace, the result follows. Parts 3 and 4 are obtained from 1 and 2 by use of the relation A** = A. I
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