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5.3(2,3,4,9h,15,20,21,22)

5.3(2,3,4,9h,15,20,21,22) - Math 431 Assignment 9 Solutions...

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Math 431 - Assignment 9 Solutions Section 5.3 2 .(b) Let A = - 1 . 4 0 . 8 - 2 . 4 1 . 8 . Then det ( A - tI 2 ) = det - 1 . 4 - t 0 . 8 - 2 . 4 1 . 8 - t = t 2 - 0 . 4 t - 0 . 6 = ( t - 1)( t + 0 . 6) . Hence, both eigenvalues λ 1 = 1 and λ 2 = - 0 . 6 of A are contained in S = { λ : | λ | < 1 or λ = 1 } . Since the eigenvalues of A are distinct, A is diagonalizable. From theorem 5.14, it follows that lim m →∞ A m exists. Next E λ 1 = n x y : - 1 . 4 x + 0 . 8 y = x and - 2 . 4 x + 1 . 8 y = y o = n x y : y = 3 x o = n x 1 3 : x R o , and E λ 2 = n x y : - 1 . 4 x + 0 . 8 y = - 0 . 6 x and - 2 . 4 x + 1 . 8 y = - 0 . 6 y o = n x y : y = x o = n x 1 1 : x R o . Let Q = 1 1 3 1 . Then Q - 1 = - 1 2 1 2 3 2 - 1 2 , because successive row operations yield 1 1 1 0 3 1 0 1 r 2: r 2 - 3 · r 1 -→ 1 1 1 0 0 - 2 - 3 1 r 1: r 1+ 1 2 · r 2 -→ 1 0 - 1 2 1 2 0 1 3 2 - 1 2 Thus, A m = Q 1 0 0 - 0 . 6 m Q - 1 = 1 1 3 1 1 0 0 ( - 0 . 6) m - 1 2 1 2 3 2 - 1 2 .
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Since | - 0 . 6 | < 1, lim m →∞ 1 0 0 ( - 0 . 6) m = 1 0 0 0 , and so lim m →∞ A m = 1 1 3 1 1 0 0 0 - 1 2 1 2 3 2 - 1 2 = 1 1 3 1 - 1 2 1 2 0 0 = - 1 2 1 2 - 3 2 3 2 . 3 . From the hypothesis, we have lim m →∞ ( A m ) ij = L ij , for all 1 i n and 1 j p . From this, it follows immediately, that lim m →∞ ( ( A m ) t ) ji = L ji , for all 1 j p and 1 i n . Therefore, lim m →∞ ( A m ) t = L t . 4 . Assume that A is similar to the n × n diagonal matrix D = λ 1 0 . . . 0 0 λ 2 . . . 0 . . . . . . . . . . . . 0 0 . . . λ n Then D m = λ m 1 0 . . . 0 0 λ m 2 . . . 0 . . . . . . . . . . . . 0 0 . . . λ m n for all m 1. From Corollary on p.285, lim m →∞ D m exists and is a n × n matrix, say M , similar to L . Thus, lim m →∞ λ m i exists for all 1 i n , which can happen only if | λ i | ≤ 1 with
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