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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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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 ] β = 0 0 3 0 - 2 0 - 4 0 0 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 .

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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 0 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 0 , x 2 = 1 - 1 - 1 , x 2 = 2 0 - 1 . For example, when t 1 = - 1 we must reduce 8 - 4 10 4 - 2 8 - 2 1 - 1 1 - 1 / 2 0 0 0 1 0 0 0 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 β ) 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 λ 1 = 1 (with eigenvectors traceless symmetric matrices) and λ 2 = - 1 (with eigenvalues traceless skew-symmetric matrices) and we get the following basis of eigenvectors for V β 0 = ( 1 0 0 - 1 , 0 1 1 0 , 0 - 1 1 0 ) .

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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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