Math121sol9 - Math 121 Linear Algebra and Applications...

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Math 121: Linear Algebra and Applications Prof. Lydia Bieri Solution Set 9 Posted: January 9th Written by: Luca Candelori Exercise 1 (7.1/2b) . The eigenvalues are λ = 4 , - 1, each with multiplicity 1, with corresponding eigenvectors v 1 = (2 , 3) and v 2 = (1 , - 1) respectively. Each vector forms a basis for its generalized eigenspace. The matrix is diagonalizable with Jordan Form: J = 4 0 0 - 1 Exercise 2 (7.1/3c) . Answers in the back of the book. Exercise 3 (7.2/4b) . The eigenvalues are λ = 2 with multiplicity 2 and λ = 1 with multiplicity 1. A basis for the generalized eigenspace of λ = 2 is given by the vectors v 1 = ( - 1 , 0 , 2) and v 2 = (1 , 2 , 0). A basis for the generalized eigenspace of λ = 1 is given by the vector v 3 = (1 , 2 , 1). In both cases the geometric multiplicity equals the algebraic multiplicity, therefore the matrix is diagonalizable with Q = - 1 1 1 0 2 2 2 0 1 and J = 2 0 0 0 2 0 0 0 1 Exercise 4 (7.2/5a) . Answers in the back of the book Exercise 5 (6.1/20, 6.2/4) . (a) Note first that x + y 2 = x + y, x + y . By linearity of the inner product, we get 1 4 x + y 2 - 1 4 x - y 2 = 1 4 ( x + y, x + y - x - y, x - y ) = 1 4 ( x, x + 2 x, y + y, y - x, x + 2 x, y - y, y ) = 1 4 (4 x, y ) = x, y An entirely analogous process, though a little longer, will prove the second identity.
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