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5.1 (2,3,4,10,11,14,15,17,19,22) 5.2(2,3,8,9,10,13)

5.1 (2,3,4,10,11,14,15,17,19,22) 5.2(2,3,8,9,10,13) -...

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Unformatted text preview: Selected answers to suggested problems: 5.1, 5.2 5.1 2. (b) T (3 + 4 x ) =- 2(3 + 4 x ), T (2 + 3 x ) =- 3(2 + 3 x ), so both are eigenvectors (and a basis) and [ T ] β =- 2- 3 . (d) T ( x- x 2 ) =- 4(- 1- x + x 2 ), T (- 1 + x 2 ) =- 2(- 1 + x 2 ), T (- 1- x + x 2 ) = 3( x- x 2 ). This is not a basis of eigenvectors; [ T ] β = 3- 2- 4 . (f) T 1 1 =- 3 1 1 , T- 1 2 =- 1 2 , T 1 2 = 1 2 , and T- 1 2 =- 1 2 , so this is a basis of eigenvectors, and [ T ] β = - 3 1 1 1 . 3. (b) (i) eigenvalues 1, 2, 3 (ii) E 1 = - z- z z : z ∈ R . E 2 = x- x : x ∈ R . E 3 = x- x : x ∈ R . (iii) One possible basis is { (1 , 1 ,- 1) , (1 ,- 1 , 0) , (1 , ,- 1) } . (iv) Q = 1 1 1- 1- 1- 1- 1 . D = 1 2 3 . (d) (i) eigenvalues 0, 1, 1 1 2 (ii) E = 1 2 z- 6 z z : z ∈ R . E 1 = y : y ∈ R . (iii) Not possible; E 1 is not of large enough dimension. 4. (c) Matrix representation of T in terms of the standard ordered basis is - 4 3- 6 6- 7 12 6- 6 11 . Determinant of [ T ]- tI is 2 + 3 t- t 3 ; eigenvalues 2 ,- 1 ,- 1. E 2 = 1 2 z z z : z ∈ R . E- 1 = y + 2 z y z : y, z ∈ R . One possible basis: {- 1 2 , 1 , 1) , (1 , 1 , 0) , (2 , , 1) } . (d) Matrix representation of T in terms of standard ordered basis ( { 1 , x } ) is 1- 6 2- 6 . Determinant of [ T ]- tI is 6 + 5 t + t 2 ; eigenvalues- 2 ,- 3. E- 2 = 2 y y : y ∈ R . E- 3 = 3 a 2 a : a ∈ R . One possible basis: { 2 + x, 3 + 2 x } . (e) Matrix representation of T in terms of standard ordered basis ( { 1 , x, x 2 } ) is 1 3 9 1 3 4 2 ....
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5.1 (2,3,4,10,11,14,15,17,19,22) 5.2(2,3,8,9,10,13) -...

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