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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 3dimensional 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 skewsymmetric 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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