M235A4aSoln

M235A4aSoln - MATH 235 Solutions Assignment 4a 1. Let A be...

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Solutions Assignment 4a 1. Let A be an n × n matrix. Also, let α, β, γ be three distinct eigenvalues of A having corresponding eigenvectors x , y , z , respectively. Consider the vector v = x + y + z . Can v be an eigenvector of A corresponding to an eigenvalue λ (possibly different from α, β, γ ? Explain. Solution : Given A * x = α * x, A * y = β * y , A * z = γ * z and * v = * x + * y + * z , then A * v = A * x + A * y + A * z = α * x + β * y γ * z . (1) Assume * v is an eigenvector of A corresponding to the eigenvalue λ . Then A * v = λ * v . From (1), it follows that α * x + β * y + γ * z = λ ± * x + * y + * z ² ( α - λ ) * x +( β - λ ) * y +( γ - λ ) * z = * 0 . (2) Since the eigenvectors * x, * y , * z belong to distinct eigenvalues, the eigenvectors * x, * y , * z are linearly independent. Thus, (2) gives that α - λ = β - λ = γ - λ = 0 α = β = γ = λ which is a contradiction (since α, β and γ are distinct). Hence, it follows that * v cannot
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This note was uploaded on 01/05/2012 for the course MATH 235 taught by Professor Celmin during the Fall '08 term at Waterloo.

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M235A4aSoln - MATH 235 Solutions Assignment 4a 1. Let A be...

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