hmw10 - Solutions to Homework 10. Math 110, Fall 2006. Prob...

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Solutions to Homework 10. Math 110, Fall 2006. Prob 5.1.3. (a) The eigenvalues are - 1 and 4, with the eigenspaces E - 1 = span { [1 - 1] t } , E 4 = span { [2 3] t } . The vectors [1 - 1] t , [2 3] t form a basis for IR 2 . The matrix Q diagonalizes A , where Q = ± 1 2 - 1 3 ² , Q - 1 AQ = ± - 1 0 0 4 ² . (b) The eigenvalues are 3, 2 and 1, with the eigenspaces E 1 = span { [1 1 - 1] t } , E 2 = span { [1 - 1 0] t } , E 3 = span { [1 0 - 1] t } . These three vectors vectors form a basis for IR 3 . The matrix Q diagonalizes A , where Q = 1 1 1 1 - 1 0 - 1 0 - 1 , Q - 1 AQ = 1 0 0 0 2 0 0 0 3 . (c) The eigenvalues are 1 and - 1, with the eigenspaces E 1 = span { [1+ i 2] t } , E - 1 = span { [ - 1+ i 2] t } . These two vectors form a basis for C 2 . The matrix Q diagonalizes A , where Q = ± 1 + i - 1 + i 2 2 ² , Q - 1 AQ = ± 1 0 0 - 1 ² . (d) The eigenvalues are 1 (with multiplicity two) and 0, with the eigenspaces E 1 = span { [0 1 0] t , [1 0 1] t } , E 0 = span { [1 4 2] t } . These three vectors vectors form a basis for IR 3 . The matrix Q diagonalizes A , where Q = 0 1 1 1 0 4 0 1 2 , Q - 1 AQ = 1 0 0 0 1 0 0 0 0 . Prob 5.1.22. (a) and (b) If Tx = λx , then T j x = λ j x , hence, for any polynomial g ( t ) = a 0 + a 1 t + ··· + a m t m , we get ( a 0 I + a 1 T + ··· + a m T m ) x = a 0 x + a 1 λx + ··· + a m λ m x = ( a 0 + a 1 λ + ··· + a m λ m ) x = g ( λ ) x. (c) ³ 2 ± 1 2 3 2 ² 2 - ± 1 2 3 2 ² + ± 1 0 0 1 ² ! ± 2 3 ² = ± 14 10 15 19 ²± 2 3 ² = 29 ± 2 3 ² = (2 · 4 2 - 4 + 1) ± 2 3 ² . Prob 5.2.1. (a) False: take the identity map. (b) False: they can be multiples of each other. (c) False: the zero vector is not an eigenvector. (d) True (by Theorem 5.5). (e) True (by the Corollary to Theorem 5.5). (f) False: the characteristic polynomial must also split into linear factors. (g) True due to the complete splitting into linear factors. (h) True (follows from the defintion of a direct sum). (i) False:
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hmw10 - Solutions to Homework 10. Math 110, Fall 2006. Prob...

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