235_tt2_w10_soln

235_tt2_w10_soln - 1. Short Answer Problems [1] a) State...

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Unformatted text preview: 1. Short Answer Problems [1] a) State the Principal Axis Theorem. Solution: Every symmetric matrix is orthogonally diagonalizable. [1] b) Let A be an m × n matrix. Prove that AT A is symmetric. Solution: (AT A)T = AT AT T = AT A, hence AT A is symmetric. [2] c) State the definition of a quadratic form Q(x) on Rn being negative definite. Solution: Q(x) is negative definite if Q(x) < 0 for all x = 0. [2] d) Consider the quadratic form Q(x) = 3x2 + 5y 2 + 3z 2 − 2xy − 2xz + 2yz . Write down the symmetric matrix A such that Q(x) = xT Ax. 3 −1 −1 1 . Solution: A = −1 5 −1 1 3 −7 −2 −3 −5 5 −1 1 0 2 −5 −5 0 4 2 7 −10 −4 6 [3] e) For the matrix A = , 0 −5 10 10 0 −4 −6 −2 −3 −5 7 2 6 2 3 5 −5 0 −1 is an eigenvalue with multiplicity three, and 2 is an eigenvalue with multiplicity two. Determine the determinant of A. Solution: We know that the trace of A is the sum of the eigenvalues. Hence, we have We also know that the determinant of A is equal to the product of the eigenvalues. Hence det A = (−1)(−1)(−1)(2)(2)(−3) = 12. (−7) + 1 + 2 + (−5) + 7 + 0 = (−1) + (−1) + (−1) + 2 + 2 + λ ⇒ λ = −3. 1 2 422 [6] 2. Let A = 2 4 2. Find an orthogonal matrix P that diagonalizes A and 224 the corresponding diagonal matrix. Solution: We have characteristic polynomial 4−λ 0 2 4−λ 0 2 4−λ 2 2 2 2−λ 2 2 2−λ 2 2 4−λ 2 = = C (λ ) = 4 0 6−λ 2 −(2 − λ) 4 − λ 2 2 4−λ Hence, the eigenvalues are λ = 2 and λ = 8. For λ = 2 we get 222 111 A − λI = 2 2 2 ∼ 0 0 0 . 222 000 −1 −1 1 and v2 = 0 . Since we need orthogonal vectors, Thus, we get eigenvectors v1 = 0 1 we apply the Gram-Schmidt procedure. We take w1 = v1 and −1 −1/2 −1 v2 · w1 1 w2 = v2 − w = 0 − 1 = −1/2 . 21 w1 20 1 1 For λ = 8 we get 1 0 −1 −4 2 2 A − λI = 2 −4 2 ∼ 0 1 −1 . 00 0 2 2 −4 1 Thus, we get eigenvector v3 = 1. 1 After normalizing the three orthogonal eigenvectors √ √ −1/ 2 −1/√6 √ P = 1/ 2 −1/ 6 √ 0 2/ 6 = −(λ − 2)(λ2 − 10λ + 16) = −(λ − 2)2 (λ − 8). and we get the orthogonal matrix √ 1 / √3 1 / √3 , 1/ 3 200 P T AP = 0 2 0 . 008 3 3. On M (2, 2) define the inner product < A, B >= tr(AT B ). [4] a) Use the Gram-Schmidt procedure to produce an orthonormal basis for the subspace S = span 10 and 01 are already orthogonal. Then 2 1 −1 v3 = − 01 2 Normalize each matrix and we get 1 10 √ 2 01 Solution: We let v1 = 10 01 1 −1 , , 01 10 01 v2 = 10 01 . 01 . We observe that v1 , v2 = 0, hence they 10 − . −1 0 1 0 −1/2 = 10 1/2 0 2 an orthonormal basis 1 1 01 0 −1 ,√ ,√ . 10 2 210 [2] b) Consider A = Solution: This matrix B is the projection of A onto S . Using the orthonormal basis from part (a), we have 11 −3 1 21 10 01 +√ √ +√ √ B=√ √ 22 01 22 10 22 10 0 1/ 2 0 3/ 2 = + + 01 1/2 0 −3/2 0 = 12 −1 1 . 0 −1 10 02 . Find the matrix B in S such that A − B is minimized. −1 2 4 [4] 4. Find b and c to obtain the best fitting equation of the form y = bx + cx2 for the following data: x −1 0 1 y 4 11 4 −1 0 1 T 1. Then we have Solution: Let X = and y = 1 01 1 a = (X T X )−1 X T y = = = 3 So y = − 2 x + 5 x2 . 2 20 02 −1 −3 5 −3 5 1/2 0 0 1/ 2 −3/2 5/2 [4] 5. Let A = 11 . Find an orthogonal matrix P and upper triangular matrix T −1 3 such that P T AP = T . 1−λ 1 Solution: Consider A − λI = . Then C (λ) = det(A − λI ) = λ2 − 4λ + 4 = −1 3 − λ −1 1 (λ − 2)2 , so we have one eigenvalue λ = 2. This gives A − λI = so a corresponding −1 1 1 eigenvalue is v1 = (so observe that A is not diagonalizable). 1 We now extend {v1 } to a orthonormal basis for R2 . So, we normalize v1 and pick v2 = −1 1 −1 1 1 √ . Thus we have P = √2 and then 2 1 11 1 11 P T AP = √ −1 1 2 1 1 1 1 −1 22 √ = , −1 3 11 02 2 as required. 5 6. Let A = 21 and Q(x) = xT Ax. 12 [4] a) Find an orthogonal matrix and diagonal matrix D such that P T AP = D . Solution: The eigenvalues are λ = 1 and λ = 3. For λ = 3 we get A − λI = For λ = 1 we get A − λI = Hence P = 1 √ 2 −1 1 1 −1 ∼ . Hence, v1 = 1 −1 00 11 11 ∼ . Hence, v1 = 11 00 1 √ 2 1 √ 2 1 . 1 −1 . 1 1 −1 30 ,D= . 11 01 [1] b) Classify Q(x) as positive definite, negative definite or indefinite. Solution: Since the eigenvalues of A are both positive, Q(x) is positive definite. [2] c) Write Q so that it has no cross terms and give the change of variables which brings it into this form. 2 Solution: Taking y = P T x we get Q = 3x2 + y1 . 1 [2] d) Sketch the graph of Q(x) = 1 showing both the original and new axes. Solution: Sketch we get 6 [3] 7. Show that if A is a symmetric matrix, then eigenvectors corresponding to distinct eigenvalues are orthogonal to each other. Solution: Let v1 , v2 be eigenvectors of A with corresponding eigenvalues λ1 and λ2 with λ1 = λ2 . Then Av1 · v2 = (λ1 v1 ) · v2 = λ1 (v1 · v2 ), and v1 · Av2 = v1 · λ2 v2 = λ2 (v1 · v2 ). But, since A is symmetric we have Av1 · v2 = v1 · Av2 , thus since λ1 = λ2 . λ1 (v1 · v2 ) = λ2 (v1 · v2 ) ⇒ v1 · v2 = 0, 8. Let A be an n × n matrix with eigenvector v and let u be orthogonal to v . [2] a) Prove that if A is symmetric, then Au is orthogonal to v . Solution: We have v · Au = Av · u since A is symmetric. Hence v · Au = Av · u = (λv ) · u = λ(v · u) = 0. [1] b) Give an example of a 2 × 2 matrix A with an eigenvector v , and a vector u which is orthogonal to v such that Au is not orthogonal to v . Solution: One example is A = 11 1 0 and v = and u = . 01 0 1 1 which is not orthogonal to 1 Observe that u is orthogonal to v , but Av = v and Au = Av . 7 [3] 9. Let V be an inner product space, and let v1 , . . . , vk be (not necessarily linearly independent) vectors in V . Suppose that S = span{v1 , . . . , vk }. For a vector x ∈ V , prove that x ∈ S ⊥ if and only if x, vi = 0 for each i = 1, . . . , k . Solution: (⇒) Let x ∈ S ⊥ . Then x, y = 0 for every y ∈ S . In particular, v1 , . . . , vk are in S , hence x, vi = 0 for each i. (⇐) Suppose that x, vi = 0 for each i. For every y ∈ S , y = a1 v1 + · · · + ak vk for some ai ’s. So x, y = x, a1 v1 + · · · + ak vk = a1 x, v1 + · · · + ak x, vk = 0. Since x, y = 0 for every y ∈ S , x ∈ S ⊥ . [3] 10. Let A be an invertible symmetric matrix. Prove that if the quadratic form xT Ax is positive definite, then so is the quadratic form xT A−1 x. Solution: If xT Ax is positive definite, then the eigenvalues of A are all positive. Let λ be any eigenvalue of A and let v be the corresponding eigenvector. Then we have Av = λv A Av = A−1 (λv ) v = λA−1 v 1 A−1 v = v λ 1 −1 Hence, λ is an eigenvalue of A . Thus, all the eigenvalues of A−1 are also positive, hence A−1 is also positive definite. −1 ...
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This note was uploaded on 04/18/2010 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.

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