5.1(1, 2d, 3a, 4bj, 11, 17, 21) 5.2 (1,2ce, 3be, 7, 8, 11, 12, 14a)

5.1(1, 2d, 3a, 4bj, 11, 17, 21) 5.2 (1,2ce, 3be, 7, 8, 11, 12, 14a)

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Unformatted text preview: MATH321 HOMEWORK SOLUTIONS HOMEWORK #8 Section 5.1: 1, 2(d), 3(a), 4(b)(j), 11, 17, 21 Section 5.2: 1, 2(c)(e), 3(b)(e), 7, 8, 11, 12, 14(a) Krzysztof Galicki Problem 5.1.1 (See Answers to Selected Exercises ). Problem 5.1.2(d) We compute T ( x- x 2 ) = 4 + 4 x- 4 x 2 , T (- 1 + x 2 ) = 2- 2 x 2 , T (- 1- x + x 2 ) = 3 x- 3 x 2 . Hence, [ T ] = 3- 2 0- 4 and the basis does not consist of eigenvectors. Problem 5.1.3(a) We compute p ( t ) = (1- t )(2- t )- 6 = 2- 3 t + t 2- 6 = t 2- 3 t- 4 = ( t- 4)( t + 1) giving the two distinct eigenvalues { t 1 ,t 2 } = {- 1 , 4 } . The corresponding eigenvectors are x 1 =- 1 1 , x 2 = 2 3 . Hence, Q =- 1 2 1 3 . Problem 5.1.4 (b) Here we need to diagonalize the matrix A = 7- 4 10 4- 3 8- 2 1- 2 . The characteristic polynomial p ( t ) = det 7- t- 4 10 4- 3- t 8- 2 1- 2- t = det - 1- t 2- 4 t 4- 3- t 8- 2 1- 2- t = = (- 1- t )[(3+ t )(2+ t )- 8]+(2- 4 t )[4- 2(3+ t )] =- (1+ t )[ t 2 +5 t +2]- (1+ t )[4- 8 t ] = =- (1 + t )[ t 2 + 5 t- 2 + 4- 8 t ] =- (1 + t )( t 2- 3 t + 2) =- (1 + t )( t- 1)( t- 2) . The spectrum { t 1 ,t 2 ,t 3 } = {- 1 , 1 , 2 } consists of 3 distinct eigenvalues so the matrix is diagonalizbale. The corresponding eigenvectors are x 1 = 1 2 , x 2 = 1- 1- 1 , x 2 = 2- 1 . For example, when t 1 =- 1 we must reduce 8- 4 10 4- 2 8- 2 1- 1 1- 1 / 2 0 1 which gives x 1 . Similarly, when t 1 = 1 we must reduce 6- 4 10 4- 4 8- 2 1- 3 1 0 1 0 1 1 0 0 0 which gives x 2 . Finally, when t 3 = 2 we must reduce 5- 4 10 4- 5 8- 2 1- 4 1 0 2 0 1 0 0 0 0 which gives x 3 . (j) In general, one could simply first write the operator T relative to any basis (say the standard basis ) and then diagonalize the corresponding 4 4 matrix. However, there is a way of solving this problem without doing any computations. Let us consider a subspace V M 2 2 ( R ) of traceless matrices, i.e., V = { A M 2 2 ( R ) | tr( A ) = 0 } . V is clearly a 3-dimensional vector subspace consisting of matrices of the from A = a b c- a . Note, now that on V the operator T is simply T ( A ) = A t and the eigenvalues and eigenvectors of this operator were analyzed in problem 17 (see below). Here the eigenvalues are 1 = 1 (with eigenvectors traceless symmetric matrices) and 2 =- 1 (with eigenvalues traceless skew-symmetric matrices) and we get the following basis...
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

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5.1(1, 2d, 3a, 4bj, 11, 17, 21) 5.2 (1,2ce, 3be, 7, 8, 11, 12, 14a)

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