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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 , = 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 = x for some scalar , 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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This note was uploaded on 01/17/2011 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.
 Spring '08
 CELMIN
 Math

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