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pset8-s09-soln

pset8-s09-soln - 18.06 Problem Set 8 Solution Due Wednesday...

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18.06 Problem Set 8 Solution Due Wednesday, 22 April 2009 at 4 pm in 2-106. Total: 160 points. Problem 1: If A is real-symmetric, it has real eigenvalues. What can you say about the eigenvalues if A is real and anti-symmetric ( A = - A T )? Give both a general explanation for any n × n A (similar to what we did in class and in the book) and check by finding the eigenvalues a 2 × 2 anti-symmetric example matrix. Solution (15 points = 10(proof) + 5(example) ) If λ is an eigenvalue of A with a nonzero eigenvector v , that is Av = λv . Then, on one hand, we have v H Av = v H λv = λ k v k 2 , and on the other hand, v H Av = ( - v H A T ) v = - ( Av ) H v = - ( λv ) H v = - ¯ λ k v k 2 . Since v is nonzero, k v k 2 > 0. We conclude that λ = - ¯ λ . This implies that λ is purely imaginary, that is the real part of λ is zero. For example, we take A = 0 1 - 1 0 . (Since A is anti-symmetric, its diagonal entries must be zero.) We then solve det( A - λI ) = λ 2 +1 = 0 to get λ 1 = i, λ 2 = - i . They are purely imaginary numbers. Problem 2: Find an orthogonal matrix Q that diagonalizes A = - 2 6 6 7 , i.e. so that Q T AQ = Λ where Λ is diagonal. What is Λ? Solution (10 points) Since A is real-symmetric, we should be able to get orthonormal eigenvectors, and then Q is just the matrix whose columns are the eigenvectors (as in class and the textbook), and Λ is the diagonal matrix of eigenvalues. So, we just solve for the eigenvalues and eigenvectors of A . To get the eigenvalues, we solve det( A - λI ) = 0 = λ 2 - 5 λ - 50, obtaining λ 1 = 10 and λ 2 = - 5. Since the eigenvalues are distinct, the eigenvectors are automatically orthogonal, and we just need to normalize them to have length 1: λ 1 = 10 , A - λ 1 I = - 12 6 6 - 3 , v 1 = 1 2 , q 1 = 1 / 5 2 / 5 λ 2 = - 5 , A - λ 2 I = 3 6 6 12 , v 2 = - 2 1 , q 2 = - 2 / 5 1 / 5 . 1
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Hence, we have Q = 1 5 - 2 5 2 5 1 5 ! , Λ = 10 0 0 - 5 Problem 3: Even if the real matrix A is rectangular, the block matrix B = 0 A A T 0 is symmetric. An eigenvector ~x of B satisfies B~x = λ~x with: ~x = ~ y ~ z , 0 A A T 0 ~ y ~ z = λ ~ y ~ z , and thus A~ z = λ~ y and A T ~ y = λ~ z . (a) Show that - λ is also an eigenvalue of B , with the eigenvector ( ~ y, - ~ z ) T . (b) Show that A T A~ z = λ 2 ~ z , so that λ 2 is an eigenvalue of A T A . (c) Show that λ 2 is also an eigenvalue of AA T by finding a corresponding eigen- vector. (d) If A = I (2 × 2), find all four eigenvalues and eigenvectors of B . Solution (25 points = 5+5+5+10) (a) We check this by direct computation. B ~ y - ~ z = - A~ z A T ~ y = - λ~ y λ~ z = - λ ~ y - ~ z . Hence - λ is also an eigenvalue of B , with the eigenvector ~ y - ~ z . (b) Again, we check by direct computation. A T A~ z = A T ( λ~ y ) = λA T ~ y = λ ( λ~ z ) = λ 2 ~ z. Hence, λ 2 is an eigenvalue of A T A with eigenvector ~ z . (c) By “symmetry”, it is not hard to guess that ~ y may be an eigenvector of AA T .
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