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math102HW7 - Math 102 Homework 7(selected problems David...

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Math 102 - Homework 7 (selected problems) David Lipshutz Problem 1. (Strang, 5.2: #4) If a 3 by 3 upper triangular matrix has diagonal entries 1, 2, 7, how do you know it can be diagonalized? What is Λ? Proof. Its eigenvalues must be 1, 2 and 7 since A - λI where λ = 1 , 2 or 7 is upper triangular with a 0 along the diagonal. So A - λI is singular, thus λ is an eigenvalue. Since the eigenvalues are all distinct, A can be diagonalized. Λ = 1 0 0 0 2 0 0 0 7 Problem 2. (Strang, 5.2: #8) Suppose A = uv T is a column times a row (a rank-1 matrix). (a) By multiplying A times u , show that u is an eigenvector. What is λ ? (b) What are the other eigenvalues of A ? (c) Compute trace( A ) from the sum on the diagonal and the sum of λ ’s. Proof. (a) Au = uv T u = ( u T vu T ) T = (( u T v ) u T ) T = λu where λ = u T v . (b) The other eigenvalues are zero since A is a rank-1 matrix. (c) trace( A ) = n i =1 u i v i = u T v = λ Problem 3. (Strang, 5.2: #30) Find Λ and S to diagonalize A in Problem 29. What is the limit of Λ k as k → ∞ ? What is the limit of S Λ S - 1 ? In the columns of this limiting matrix what do you see? Proof. det( A - λI ) = det " . 6 - λ . 4 . 4 . 6 - λ # = λ 2 - 1 . 2 λ + . 2 = ( λ - . 2)( λ - 1) We get A (1 , 1) T = (1 , 1) T and A (1 , - 1) T = . 2(1 , - 1) T . So A = " 1 1 1 - 1 # " 1 0 0 . 2 # " 1 / 2 1 / 2 1 / 2 - 1 / 2 # 1
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lim k →∞ Λ k = lim k →∞ " 1 k 0 0 . 2 k # = " 1 0 0 0 # So lim k →∞ SA k S = " 1 1 1 - 1 # " 1 0 0 0 # " 1 / 2 1 / 2 1 / 2 - 1 / 2 # = " 1 / 2 1 / 2 1 / 2 1 / 2 # The columns give you the equilibrium distribution of A . Problem 4. (Strang, 5.2: #34) Suppose that A = S Λ S - 1 . Take determinants to prove that det A = λ 1 λ 2 · · · λ n = product of λ ’s. Proof. det A = det S det Λ(det S ) - 1 = det Λ = λ 1 λ 2 · · · λ n This quick proof only works when A is diagonalizable. Problem 5. (Strang, 5.2: #40) Substitue A = S Λ S - 1 into the product ( A - λ 1 I )( A - λ 2 I ) · · · ( A - λ n I ) and explain why this produces the zero matrix. Proof. ( A - λ 1 I ) · · · ( A - λ n I ) = ( S Λ S - 1 - λ 1 I )( S Λ S - 1 - λ 2 I ) · · · ( S Λ S - 1 - λ n I ) = S - λ 1 I ) S - 1 S - λ 2 I ) S - 1 · · · S - λ n I ) S - 1 = S - λ 1 I )(Λ - λ 2 I ) · · · - λ n I ) S - 1 Even if A is not diagonalizable, it can be written as
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