Let V be a complex vector space of dimension n and let T E L(V). Suppose that A is the only eigenvalue of T and its geometric multiplicity is 1. Prove that there do not
Let V be a complex vector space of dimension n and let T E L(V). Suppose that A is the only eigenvalue of T and its geometric multiplicity is 1. Prove that there do not
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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![Let V be a complex vector space of dimension n and let T E L(V). Suppose that A is
the only eigenvalue of T and its geometric multiplicity is 1. Prove that there do not](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b0185de-0645-4c2a-aea4-e046d61ab5cb%2F59a2b426-f06e-48b3-b452-4caafe1233f8%2Fodgerjb.jpeg&w=3840&q=75)
Transcribed Image Text:Let V be a complex vector space of dimension n and let T E L(V). Suppose that A is
the only eigenvalue of T and its geometric multiplicity is 1. Prove that there do not
Expert Solution
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Step 1
Suppose that there exists a basis of V with respect to the matrix of T is block diagonal with two square blocks, and the square blocks is smaller than n * n.
It is given that the lambda is the only Eigen value of T and its geometric multiplicity is 1. And a geometric multiplicity of an Eigen value can-not be exceeding by it algebraic multiplicity.
Let there are two square blocks be A1 and A2 of order k and order l respectively:
Step 2
Let’s take two Eigenvectors. In v1, first k entries are one, and rest are 0 and vector v2 first k entries are 0, and rest are 1.
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