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Unformatted text preview: Math 423, Answers for homework 1 1.2.1. (a) True, by definition. (b) False (homework). (c) False: take x = 0 . Comment: true if x 6 = 0 . (d) False: take a = 0 . Comment: true if a 6 = 0 . (e) True. (f) False: m rows, n columns. (g) False. (h) False: highest terms may cancel. (i) True. (j) True. (k) True. 1.2.2. 0 0 0 0 0 0 0 0 0 0 0 0 . 1.2.7. We observe that f (0) = 1 , f (1) = 3 , g (0) = 1 , g (1) = 3 , h (0) = 2 , h (1) = 6 . 1.2.9. Corollary 1. The vector described in (VS 3) is unique. Proof. We need to show that if there is another element z V such that for all x V , x + z = x , then z = 0 . Assume that z has the above property. Consider z + 0 . On the one hand, it is equal to z by (VS 3). On the other hand, it is equal to 0 + z by (VS 1), which is equal to by the defining property of z . We conclude that z = z + 0 = 0 . Another proof: since x + 0 = x = x + z , by (VS 1) we conclude that 0 + x = z + x , and by cancellation it follows that z = 0 . Corollary 2. The vector y described in (VS 4) is unique. Proof. We fix x V . The vector y V is defined by the property that x + y = 0 . Suppose there is another vector z V , also with the property x + z = 0 . Consider....
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This note was uploaded on 01/19/2011 for the course MATH 121a taught by Professor Staff during the Winter '08 term at UC Irvine.
 Winter '08
 staff
 Math, Linear Algebra, Algebra

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