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**Unformatted text preview: **Math 51 Exam 2 Solutions May 22, 2007 1. (12 points) (a) Find the eigenvalues of the following matrix. A = 1- 1 1 1- 1 1- 1 1 1 (6 points) To find the eigenvalues, we must solve the equation det( A- I ) = 0 for . det( A- I ) = 1- - 1 1 1- 1- 1- 1 1 1- To evaluate the determinant, we expand it along a row or column. Here, we show it expanded along the top row: (1- )- 1- 1 1 1- - (- 1) 1 1- 1- 1- + 1- 1- - 1 1 = (1- )( 2- 2) + (2- )- =- 3 + 2 + 2 - 2 + 2- - =- 3 + 2 Setting this equal to 0 and factoring yields 2 (1- ) = 0 from which we conclude that = 0 or 1. So the eigenvalues are 0 or 1. (b) Compute the eigenspace of the largest eigenvalue you found in part (a). (3 points) The largest eigenvalue is = 1. The eigenspace E 1 is precisely the nullspace of the matrix A- I . Lets compute N ( A- I ) by row-reducing: A- I = - 1 1 1- 2 1- 1 1 ; 1- 2 1- 1 1- 1 1 ; 1- 2 1- 1 1- 1 1 ; 1- 2 1- 1 1 ; 1- 2 1 1- 1 ; 1- 1 1- 1 which is row-reduced. There are two pivots, and their equations are x 1- x 3 = 0 x 1 = x 3 and x 2- x 3 = 0 x 2 = x 3 So N ( A- I ) = - x = x 1 x 2 x 3 = x 3 x 3 x 3 = span 1 1 1 . (c) Is there a nonzero vector- v R 3 such that A- v =- v ? Briefly explain. (3 points) Yes. There is a nonzero vector- v such that A- v =- v exactly when and only when 1 is an eigenvalue of A . In part a), we found that 1 is an eigenvalue of A . Math 51, Spring 2007 Exam 2 Solutions May 22, 2007 Page 2 of 9 2. (12 points) Let A = 1 3- 3 4- 2 16 . (a) Find a basis for the orthogonal complement of the null space N ( A ). (5 points) Since N ( A ) = C ( A T ), we will find a basis for C ( A T ). A T = 1 4 3- 2- 3 16 By definition, the two columns of A T span C ( A T ). These vectors are linearly independent, since they are clearly not scalar multiples of each other. Thus a basis for N ( A ) = C ( A T ) is 1 3- 3 , 4- 2 16 ....

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