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

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter7: Distance And Approximation
Section7.4: The Singular Value Decomposition
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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
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
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:

Advanced Math homework question answer, step 1, image 1

 

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.

Advanced Math homework question answer, step 2, image 1

 

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