{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

pset6-soln

# pset6-soln - 18.06 Problem Set 6 Due Wednesday 9 April 2008...

This preview shows pages 1–3. Sign up to view the full content.

18.06 Problem Set 6 Due Wednesday, 9 April 2008 at 4 pm in 2-106. 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 - λI , where λ is a variable: det( A - λI ) = det 1 - λ 4 2 3 - λ = (1 - λ )(3 - λ ) - 8 This simplifies to λ 2 - 4 λ - 5, which has roots 5 , - 1. (A shortcut to find this equation for a 2 by 2 matrix is to take λ 2 - tr ( A ) λ + 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 λ 2 - 6 λ = 0. This has roots 0 , 6. Note that ( A + I ) - 0 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 = λx , then x is also an eigenvector of A 2 with eigenvalue λ 2 , because A 2 x = A ( λx ) = λ ( Ax ) = λ ( λx ) 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Part 2: Supposing that A is invertible, x is an eigenvector of A - 1 with eigenvalue λ - 1 : we just take the equation Ax = λx and multiply both sides by λ - 1 A - 1 to get λ - 1 x = λ - 1 A - 1 ( λx ) = A - 1 x . Part 3: If Ax = λx , then ( A + I ) x = λx + x = ( λ + 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 , 0] T , [ - 3 , 0 , 1 , 0] T , and [ - 5 , 0 , 0 , 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 , 0 , - 3). What is the corresponding eigenvalue? b) Note that det( M ) = 0. Use this information to find another eigenvalue λ 2 - how do you know this must be an eigenvalue? c) A third eigenvalue is λ 3 = - 1. Write down (but don’t 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 = λ 1 x 1 for the corresponding eigen- value λ 1 . So we just calculate Mx 1 = [1 , 1 , 0 , - 3] T . Thus λ 1 = 1.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}