235_tt2_s10_soln

# 235_tt2_s10_soln - Math 235 1. Short Answer Problems Term...

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Term Test 2 Solutions 1. Short Answer Problems [2] a) Let B = { ~v 1 ,...,~v k } be an orthonormal basis for a subspace S of an inner product space V . Deﬁne proj S and perp S . Solution: Let ~v V , then proj S ( ~v ) and perp S ( ~v ) are the unique vectors such that ~v = proj S ( ~v ) + perp S ( ~v ) where proj S ( ~v ) S and perp S ( ~v ) S . In particular, proj S ( ~v ) = < ~v,~v 1 > ~v 1 + ··· + < ~v,~v k > ~v k and perp S = ~v - proj S ( ~v ). [2] b) Consider the quadratic form Q ( ~x ) = 4 x 2 1 - 5 x 2 2 + 2 x 2 3 - 3 x 1 x 2 + 4 x 1 x 3 + 2 x 2 x 3 . Write down the symmetric matrix A such that Q ( ~x ) = ~x T A~x . Solution: A = 4 - 3 / 2 2 - 3 / 2 - 5 1 2 1 2 [2] c) Find a basis for the orthogonal compliment of the subspace S = Span ±² 1 0 0 1 ³ , ² 0 0 1 - 1 ³´ of M (2 , 2) under the standard inner product. Solution: Observe that dim S = dim M (2 , 2) - dim S = 4 - 2 = 2 and that ~v 1 = ² 0 1 0 0 ³ and ~v 2 = ² - 1 0 1 1 ³ are orthogonal to the basis vectors of S . Hence, S = Span { ~v 1 ,~v 2 } . [2] d) Let Q ( x 1 ,x 2 ) = ax 2 1 + 2 bx 1 x 2 + cx 2 2 where ac - b 2 < 0. Prove that Q is indeﬁnite. Solution: Let A = ² a b b c ³ have eigenvalues λ 1 and λ 2 . Then we have λ 1 λ 2 = det A = ac - b 2 < 0 and so one eigenvalue is positive and the other is negative. Hence Q is indeﬁnite. [2] e) Use the Triangularization theorem to prove that every symmetric matrix is orthogonally diagonalizable. Solution: Let A be a symmetric matrix. Then A has all real eigenvalues, so by the triangu- larization theorem, we have that there exists an orthogonal matrix P such that P T AP = T where T is upper triangular. But, then T T = ( P T AP ) T = P T A T P = P T AP = T. Hence,

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## This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.

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235_tt2_s10_soln - Math 235 1. Short Answer Problems Term...

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