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M235A4aSoln

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

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MATH 235 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 ( 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 be an eigenvector of A . 1

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2. Recall that the trace of an n × n matrix A = [ a ij ], denoted by tr( A ), is the sum of the diagonal elements, that is, tr( A ) = n i =1 a ii .
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M235A4aSoln - MATH 235 Solutions Assignment 4a 1 Let A be...

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