MIT18_06S10_exam3_s10_soln

MIT18_06S10_exam3_s10_soln - 18.06 Quiz 3 Solutions May 8...

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Unformatted text preview: 18.06 Quiz 3 Solutions May 8, 2010 Profes- sor Strang Your PRINTED name is: 1. Your recitation number is 2. 3. 1. (40 points) Suppose u is a unit vector in R n , so u T u = 1 . This problem is about the n by n symmetric matrix H = I − 2 uu T . (a) Show directly that H 2 = I. Since H = H T , we now know that H is not only symmetric but also . Solution Explicitly, we find H 2 = ( I − 2 uu T ) 2 = I 2 − 4 uu T + 4 uuu T uu T (2 points): since u T u = 1, H 2 = I (3 points). Since H = H T , we also have H T H = 1, implying that H is an orthogonal (or unitary) matrix. (b) One eigenvector of H is u itself. Find the corresponding eigenvalue. Solution Since Hu = ( I − 2 uu T ) u = u − 2 uu T u = u − 2 u = − u , λ = − 1. (c) If v is any vector perpendicular to u, show that v is an eigenvector of H and find the eigenvalue. With all these eigenvectors v, that eigenvalue must be repeated how many times? Is H diagonalizable ? Why or why not? Solution For any vector v orthogonal to u (i.e. u T v = 0), we have Hv = ( I − 2 uu T ) v = v − 2 uu T v = v , so the associated λ is 1. The orthogonal complement to the space spanned by u has dimension...
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MIT18_06S10_exam3_s10_soln - 18.06 Quiz 3 Solutions May 8...

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