HW10_soln - Homework#10 Solutions p 348#10 The keys to this...

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Homework #10 Solutions p 348, #10 The keys to this exercise are the following. Lemma 1. Let V be a vector space over a field F . If { v 1 , . . . , v n } is a linearly independent set in V and w v 1 , . . . , v n then { v 1 , . . . , v n , w } is linearly independent as well. Proof. Let a 1 , . . . , a n , b F so that a 1 v 1 + · · · + a n v n + bw = 0. If b = 0 then we have w = ( - b - 1 a 1 ) v 1 + · · · + ( - b - 1 a n ) v n v 1 , . . . , v n which is a contradiction. It follows that b = 0 and so 0 = a 1 v 1 + · · · + a n v n + bw = a 1 v 1 + · · · + a n v n , which implies, via the linear independence of v 1 , . . . , v n , that a 1 = · · · = a n = 0. That is, the only linear combination of v 1 , . . . , v n , w that equals 0 is the trivial combination. Hence, { v 1 , . . . , v n , w } is linearly independent. Lemma 2. Let V be a vector space over a field F . If { v 1 , . . . , v n } is a basis for V and { w 1 , . . . , w m } is a linearly independent set in V then m n . Proof. The proof of Theorem 19.1 can be used, word for word. Now let S = { v 1 , v 2 , . . . , v n } be a set of linearly independent vectors in a finite dimensional vector space V . If S = V then S is a basis for V and we are finished. Otherwise we can find a vector w 1 V , w 1 S
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