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Unformatted text preview: 18.06 Problem Set 8 Solution Due Wednesday, 22 April 2009 at 4 pm in 2106. Total: 160 points. Problem 1: If A is realsymmetric, it has real eigenvalues. What can you say about the eigenvalues if A is real and antisymmetric ( 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 antisymmetric 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 = 1 1 0 . (Since A is antisymmetric, 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 realsymmetric, 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 Hence, we have Q = 1 √ 5 2 √ 5 2 √ 5 1 √ 5 ! , Λ = 10 5 Problem 3: Even if the real matrix A is rectangular, the block matrix B = A A T is symmetric. An eigenvector ~x of B satisfies B~x = λ~x with: ~x = ~ y ~ z , A A T ~ 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 ....
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This note was uploaded on 05/06/2010 for the course 18 18.06 taught by Professor Strang during the Spring '08 term at MIT.
 Spring '08
 Strang

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