By row reduction we find that a basis for the

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By row reduction, we find that a basis for the nullspace of this matrix is 1 0 0 - 1 , 0 1 0 - 1 , 0 0 1 - 1 . Next, A - 3 I = - 1 1 1 1 0 - 2 0 0 0 0 - 2 0 1 1 1 - 1 . The vector 1 0 0 1 spans the nullspace of this matrix, so we have altogether found a basis of R 4 consisting of eigenvectors for A . Converting back to polynomials, we have the basis β = { 1 - x 3 , x - x 3 , x 2 - x 3 , 1 + x 3 } . The corresponding diagonal matrix is D = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 3 . Problem 6. Consider the matrix A = 0 1 + i 2 - 2 i 0 1 - i 3 4 i 0 ! . 6
4 Written Problems for Review (a) Calculate det A . Make sure you simplify your final answer. Answer: We use the “triangular rule” for 3 × 3 determinants: det( A ) = 0 · 0 · 0+3(1+ i )(1 - i )+2( - 2 i )(4 i ) - 2 · 0 · 3 - 0(1 - i )(4 i ) - 0(1+ i )( - 2 i ) = 22 . (b) If char A ( t ) = at 3 + bt 2 + ct + d , what are the coefficients a, b, and d equal to? Show all relevant calculations and explain. Answer: The constant term d is det( A ) = 22. The leading coefficient a is ( - 1) 3 = - 1. The quadratic coefficient is b = ( - 1) 2 tr( A ) = 0 . Problem 7. We would like to use Linear Algebra to find a direct formula for the Fibonacci sequence a n = a n - 1 + a n - 2 for n 2 with a 0 = 0 , and a 1 = 1. (a) Let ~v n = a n a n +1 for all n 0. Find a 2 × 2 matrix A such that ~v n +1 = A~v n for all n 0. Justify your answer! Answer: We have ~v n +1 = a n +1 a n +2 = a n +1 a n +1 + a n = 0 1 1 1 a n a n +1 = 0 1 1 1 ~v n , so we have A = 0 1 1 1 . (b) For any n 0, express ~v n in terms of the matrix A and the vector ~v 0 . Justify your answer Answer: We have ~v n = A~v n 1 = A 2 ~v n - 2 = . . . = A n ~v 0 . (c) How does the direct formula for a n depend on the eigenvalues λ 1 and λ 2 of A ? You do not have to diagonalize A or show the full direct formula for the Fibonacci sequences. It suffices to show where/how the eigenvalues appear in this direct formula. Answer: The Fibonacci numbers will be of the form a n = αλ n 1 + βλ n 2 for some fixed α, β R . (d) To find the direct formula for the Fibonacci sequence, do we need to just find a diagonal matrix D similar to A , or do we also need to find an eigenbasis for A ? Explain briefly. Answer: If we use the above shortcut, we just need the eigenvalues λ 1 , λ 1 and do not need to find the change of coordinates matrix Q since we can solve for α and β by substituting n = 0 and n = 1. Problem 8. Let T : R 3 lin. R 3 with matrix A = 0 1 6 - 1 0 5 0 0 4 ! in the standard basis E = { ~e 1 ,~e 2 ,~e 3 } . (a) Find a 2-dimensional T -invariant subspace W of R 3 . Give its basis and explain why it is T -invariant. Answer: Let W = Span { ~e 1 ,~e 2 } , which has { ~e 1 ,~e 2 } as a basis. Now T ( ~e 1 ) = - ~e 2 W and T ( ~e 2 ) = ~e 1 W , so W is T -invariant. (b) Is the subspace W you found in part (a) also T -cyclic? Explain why it is or why it isn’t. Answer: It is because the vectors ~e 1 and T ( ~e 1 ) = - ~e 2 span W and T k ( ~e 1 ) W for all k . So W is the T -cyclic subspace generated by ~ e 1 . (c) Does W have a 1-dimensional T -invariant subspace U ? Why or why not?

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