ass3-sol - Math 235 Assignment 3 Due 9:15am Wednesday 1...

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Unformatted text preview: Math 235 Assignment 3 Due 9:15am, Wednesday Jan 31, 2007. 1. From the Text § 5.1 #16. Find a basis for the eigenspace corresponding to the listed eigenvalue. 3 2 1 3 1 1 1 4 , λ = 4 Solution. We need to find a basis for null( A- 4 I ) . For that purpose we carry out some row operations: A- 4 I = - 1 2 1- 1 1 1- 3 R 1 + R 2 + R 3 → R 1-→ 1- 1 1 1- 3 -→ 1- 1 1 1- 3 . So we want to find vectors x = x 1 x 2 x 3 x 4 such that 1- 1 1 1- 3 x 1 x 2 x 3 x 4 = . The solution is given by x 1 x 2 x 3 x 4 = 2 3 1 s + 1 t, where s and t are free. Therefore a basis for the eigenspace corresponding to the eigenvalue 4 is 2 3 1 , 1 . #22. In the following exercise A is an n × n matrix. Mark each statement True or False. Justify your answer. Solution. a. If A x = λ x for some scalar λ, then x is an eigenvector of A. False. A zero vector satisfies the above property but it is not an eigenvector. b. If v 1 and v 2 are linearly independent eigenvectors, then they correspond to distinct eigenvalues....
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ass3-sol - Math 235 Assignment 3 Due 9:15am Wednesday 1...

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