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

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

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

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