Math121sol9 - Math 121: Linear Algebra and Applications...

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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 ² 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 k x + y k 2 = h x + y,x + y i . By linearity of the inner product, we get 1 4 k x + y k 2 - 1 4 k x - y k 2 = 1 4 ( h x + y,x + y i - h x - y,x - y i ) = 1 4 ( h x,x i + 2 h x,y i + h y,y i - h x,x i + 2 h x,y i - h y,y i ) = 1 4 (4 h x,y i ) = h x,y i An entirely analogous process, though a little longer, will prove the second identity. 1
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This note was uploaded on 10/23/2010 for the course MATH Math 121 taught by Professor Bieri during the Fall '07 term at Harvard.

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Math121sol9 - Math 121: Linear Algebra and Applications...

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