assign8_sol

# assign8_sol - MATH 235 Assignment 8 Solutions Textbook...

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Unformatted text preview: MATH 235: Assignment 8 Solutions Textbook Questions 1. [ 2 marks ] (p.454 #34) Give the spectral decomposition of A from Example 3. The example gives us the diagonalization A = PDP- 1 where P = 1 √ 2- 1 √ 18- 2 3 4 √ 18- 1 3 1 √ 2 1 √ 18 2 3 , D = 7 0 0 7 0 0- 2 . From this, we can simply read off the eigenvalues and orthonormal eigen- vectors: A = 7 u 1 u T 1 + 7 u 2 u T 2- 2 u 3 u T 3 where u 1 = 1 √ 2 1 √ 2 , u 2 = - 1 √ 18 4 √ 18 1 √ 18 , u 3 = - 2 3- 1 3 2 3 . 2. [ 6 marks total ] (p.454 #35) Let u ∈ R n be a unit vector and let B = uu T . (a) [ 2 marks ] Show that for any x ∈ R n , B x is the orthogonal projection of x onto u. We have B x = uu T x = u ( u · x ) = u · x u · u u , the last step being true since u is a unit vector. This is exactly the definition of the orthogonal projection of x onto u . (b) [ 2 marks ] Show that B is symmetric and B 2 = B . We have B T = ( uu T ) T = ( u T ) T u T = uu T = B, B 2 = ( uu T )( uu T ) = u ( u T u ) u T = uu T = B. (c) [ 2 marks ] Show that u is an eigenvector of B and find its eigenvalue. Since B u = ( uu T ) u = u ( u T u ) = u , u is an eigenvector of B with eigenvalue 1. 1 3. [ 4 marks ] (p.462, #8) Let A be the matrix of the quadratic form 9 x 2 1 + 7 x 2 2 + 11 x 2 3- 8 x 1 x 2 + 8 x 1 x 3 , which has the eigenvalues 3, 9, and 15. Find an orthogonal ma- trix P such that the transformation x = P y turns x T A x into a quadratic form with no cross-product terms; state P , the new quadratic form, and the principal axes. We have A = 9- 4 4- 4 7 4 11 . We need a unit eigenvector for each eigenvalue; they will be orthogonal since the eigenvalues are all different. For the eigenvalue 3, we have = ( A- 3 I ) x = 6- 4 4- 4 4 4 8 x 1 x 2 x 3 = 6 x 1- 4 x 2 + 4 x 3- 4 x 1 + 4 x 2 4 x 1 + 8 x 3 ∴ x 1 = x 2 , x 1 =- 2 x 3-→ x 1 x 2 x 3 =- x 3 2 2- 1 . This is the eigenvector u 1 . In the same way, we find u 2 = - 1 2 2 , u 3 = 2- 1 2 . Normalizing these, we construct the matrices P and D such that A = PDP- 1 : P = 2 3- 1 3 2 3 2 3 2 3- 1 3- 1 3 2 3 2 3 . , D = 3 0 0 9 0 0 15 If x = P y , we have x T A x = y T P T AP y = y T D y (recall that P is orthog- onal). Thus the new quadratic form is 3 y 2 1 + 9 y 2 2 + 15 y 2 3 ....
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assign8_sol - MATH 235 Assignment 8 Solutions Textbook...

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