F08_M235_Assn4_sols

# F08_M235_Assn4_sols - Math 235 Assignment 4 Solutions 1 We...

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Unformatted text preview: Math 235 Assignment 4 Solutions 1. We have that A x = α x , A y = β y , A z = γ z . We then get that A v = A x + A y + A z = α x + β y + γ z . Now, if v is an eigenvector of A corresponding to the eigenvalue λ , then A v = λ v . From the above equation, it follows that α x + β y + γ z = λ ( x + y + z ) , or equivalently, ( α- λ ) x + ( β- λ ) y + ( γ- λ ) z = 0 . Since x , y , z are eigenvectors of A , corresponding to distinct eigenvalues, they are linearly independent. So, each coefficient must equal zero. That is α- λ = β- λ = γ- λ = 0 , or equivalently α = β = γ = λ. Since α,β,γ are distinct, we have a contradiction, and conclude that v is not an eigenvector of A . 2. a) Let A = CD and B = DC . Then, tr( A ) = n X i =1 a ii = n X i =1 n X j =1 c ij d ji = n X i =1 n X j =1 d ji c ij = n X j =1 ( n X i =1 d ji c ij ) = n X j =1 b jj = tr( B ) . Hence, tr( CD ) = tr( DC ), as required. b) If A is similar to B , then there exists an invertible matrix P such that P- 1 AP = B . Now, tr( B ) = tr( P- 1 AP ) = tr( P- 1 ( AP )) = tr(( AP ) P- 1 ) from(a) = tr( APP- 1 ) = tr( A ) Hence, tr( A ) = tr( B ), as required. 3. We wish to show that B = P- 1 AP for some invertible matrix P . Since M is invertible, we get that N = M- 1 A and N = BM- 1 . So, BM- 1 = M...
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F08_M235_Assn4_sols - Math 235 Assignment 4 Solutions 1 We...

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