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10w_math115a_hw3_solutions

10w_math115a_hw3_solutions - Math 115A Homework 3 Solutions...

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Math 115A Homework # 3 Solutions Prof. Yehuda Shalom TA: Darren Creutz Date Due: 28 January 2010 1.6.1 (a) F: the empty set is the basis for { 0 } (b) T: Thm 1.9 (c) F: the space of all polynomials over is infinite-dimensional (d) F: { (1 , 0) , (0 , 1) } and { (1 , 1) , (0 , 1) } are distinct bases for F 2 (e) T: Cor 1 (f) F: dim ( P n ( F )) = n + 1 (g) F: dim ( M n × m ( F ) = mn (h) T: Thm 1.10 (i) F: only true when S is a basis (j) T: Thm 1.11 (k) T: Thm 1.11 (l) T: Thm 1.10 1.6.5 Is { (1 , 4 , - 6) , (1 , 5 , 8) , (2 , 1 , 1) , (0 , 1 , 0) } a linearly independent subset of 3 ? Proof. No. dim ( 3 ) = 3 but there are 4 vectors in the set. Let W = span ( { (1 , 4 , - 6) , (1 , 5 , 8) , (2 , 1 , 1) , (0 , 1 , 0) } ) Then W is a subspace so dim ( W ) dim ( 3 ) = 3 (Theorem 1.11). By Theorem 1.9 there is then a set of 3 vectors in { (1 , 4 , - 6) , (1 , 5 , 8) , (2 , 1 , 1) , (0 , 1 , 0) } which generates W . Pick a vector not in that set. It must be a linear combination of the ones in that set. 1.6.11 Let u and v be distinct vectors of a vector space V . Show that if { u, v } is a basis for V and a and b are nonzero scalars then both { u + v, au } and { au, bv } are also bases for V . Proof. Assume that { u, v } is a basis for V . Then u and v are linearly independent and together generate V . Call this set G = { u, v } . Let L = { u + v, au } . If u + v and au are not linearly independent then u + v = cau for some c or au = c ( u + v ) 1
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for some b . So then v = ( ba - 1) u or cv = ( a - c ) u which would make v and u linearly dependent. This contradiction means that L is linearly independent. By the Replacement Theorem (1.10), there is a set H containing exactly 2 - 2 = 0 vectors such that L H generates V . Now H = since it contains nothing and therefore L generates V . Being linearly independent and generating V this means
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