Quiz 4 - / ∈ span( S 1 ), and consider the set S 2 = S 1...

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Math 3A (44351) Quiz 4 Problem 1. Let n > 0 and suppose that S R n contains less than n linearly independent vectors. That is, S = { v 1 ,...,v k } for some k < n . Prove that S can be extended to be a basis for R n . If S spanned R n , then since it has k linearly independent vectors, the dimension of R n = k , which we know to be false since k < n and the dimension of R n is n (we can clearly show this by looking at the standard basis.) Thus, S cannot span R n . Because of this, we must be able to find at least one new vector v k +1 R n s.t. v k +1 / span(S). If we add this to our set, and consider S 1 = S S { v k +1 } , then S 1 is still linearly independent. If k + 1 = n , then S 1 is a set of n linearly independent vectors in R n , and thus S 1 is a basis for R n . If k +1 < n , then we repeat the above reasoning to find a new vector v k +2
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Unformatted text preview: / ∈ span( S 1 ), and consider the set S 2 = S 1 S { v k +2 } . This process terminates at S n-k , at which point we have a set of n linearly independent vectors in R n , and since we only added vectors to S, S ⊂ S n-k and we have extended S to be a basis for R n . Problem 2. Let u 1 = ± 2 1 ² ,u 2 = ± 4 3 ² ,v 1 = ±-3 1 ² ,v 2 = ± 2-5 ² a) Find the transition matrix from [ v 1 ,v 2 ] to [ u 1 ,u 2 ]. S = U-1 V = 1 2 ± 3-4-1 2 ²±-3 2 1-5 ² = 1 2 ±-13 26 5-12 ² b) If x = 3 v 1-4 v 2 = ± 3-4 ² V , determine the coordinates of x with respect to [ u 1 ,u 2 ]. x U = Sx V = 1 2 ±-13 26 5-12 ²± 3-4 ² = 1 2 ±-143 63 ² 1...
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This note was uploaded on 05/08/2010 for the course MAT MATH taught by Professor Pantano during the Spring '10 term at UC Irvine.

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