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
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
Problem 26EQ
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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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