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Unformatted text preview: 18.06 Problem Set 6 Due Wednesday, 9 April 2008 at 4 pm in 2106. Problem 1: a) Do problem 2 from section 6.1 (pg. 283) in the book. b) Do problem 9 from section 6.1 (pg. 284). Solution (5+5 points) a) To find the eigenvalues of A , we take the determinant of A I , where is a variable: det( A I ) = det 1 4 2 3 = (1 )(3 ) 8 This simplifies to 2 4  5, which has roots 5 , 1. (A shortcut to find this equation for a 2 by 2 matrix is to take 2 tr ( A ) + det( A ).) We then find the eigenvectors with eigenvalue 1 by taking the nullspace of A (5) I : the nullspace of A 5 I = 4 4 2 2 is given by the basis [1 , 1] T . Similarly, A ( 1) I = 2 4 2 4 has nullspace given by the basis [ 2 , 1] T . In sum, the eigenvalue 5 leads to eigen vector (1 , 1), and the eigenvalue 1 leads to the eigenvector ( 2 , 1). We now consider A + I . The equation giving the eigenvalues is 2 6 = 0. This has roots 0 , 6. Note that ( A + I ) I and ( A + I ) 6 I are exactly the matrices we had as before. So they have the same nullspaces. That is, A + I has the same eigenvectors as A , but the eigenvalues have increased by 1. b) Part 1: if Ax = x , then x is also an eigenvector of A 2 with eigenvalue 2 , because A 2 x = A ( x ) = ( Ax ) = ( x ) 1 Part 2: Supposing that A is invertible, x is an eigenvector of A 1 with eigenvalue  1 : we just take the equation Ax = x and multiply both sides by  1 A 1 to get  1 x =  1 A 1 ( x ) = A 1 x . Part 3: If Ax = x , then ( A + I ) x = x + x = ( + 1) x . Problem 2: Do problem 13 from section 6.1 (pg. 285) in the book. Solution (10 points) a) We have Pu = uu T u = u because u T u = u u = 1. So u is an eigenvector with eigenvalue 1. b) If v is perpendicular to u then 0 = u v = u T v . So Pv = uu T v = 0, showing that v has eigenvalue 0. c) By part b, it suffices to find three independent vectors perpendicular to u (and in fact this will be equivalent). Using the fact that the row space and null space are perpendicular, we need to find three independent vectors in the nullspace of [1 / 6 , 1 / 6 , 3 / 6 , 5 / 6]. We obtain special solutions [ 1 , 1 , , 0] T , [ 3 , , 1 , 0] T , and [ 5 , , , 1] T . Problem 3: Consider the matrix M = 2 2 1 1 14 6 9 7 2 1 2 1 8 1 7 4 a) One eigenvector is x 1 = (1 , 1 , , 3). What is the corresponding eigenvalue? b) Note that det( M ) = 0. Use this information to find another eigenvalue 2 how do you know this must be an eigenvalue? c) A third eigenvalue is 3 = 1. Write down (but dont solve) a linear system that can be solved to find x 3 . d) What is the fourth eigenvalue? (Hint: use the trace.) Solution (10 points) a) Since x 1 is an eigenvector, we have Mx 1 = 1 x 1 for the corresponding eigen value 1 . So we just calculate Mx 1 = [1 , 1 , , 3] T . Thus 1 = 1....
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 Spring '08
 CHANILLO

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