sol3 - SOLUTIONS FOR HOMEWORK 3 Section 1.5 18 This is a...

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Section 1.5 18. This is a bonus problem , with very little partial credit given. Suppose S consists of polynomials of p 1 , p 2 , . . . , of degrees 0 d 1 < d 2 < . . . (we can relabel the polynomials to enumerate them in the order of decreasing degree). That is, we can write p k ( x ) = c 0 k + c 1 k x + . . .c d k k x d k , with c d k k n = 0. Suppose, for the sake of contradiction, that the set S is linearly dependent. Then there exists N N and scalars a 1 , . . ., a N , not all of them equal 0, such that a 1 p 1 + . . . + a N p N = 0. Let n be the highest number for which a n n = 0. Then the coeFcient of x d n in a 1 p 1 + . . . + a N p N equals a n c d n n n = 0, contradicting the assumption that a 1 p 1 + . . . + a N p N = 0. Section 1.6 2. By Corollary 2 (pp. 47-48), a set of n elements in an n -dimensional space is a basis i± it is linearly independent. In the examples below, we deal with sets in the 3-dimensional space R 3 , containing 3 elements each. To determine whether such a set is a basis, we only need to test the sets for linear independence. The sets in the examples below contain 3 elements each, which equals to the dimension of R 3 . (b) No , the set { (2 , 4 , 1) , (0 , 3 , 1) , (6 , 0 , 1) } is not a basis for R 3 . Indeed, { (2 , 4 , 1) , (0 , 3 , 1) , (6 , 0 , 1) } is linearly independent i± (0 , 0 , 0) is the only so- lution for the equation (1) t 1 (2 , 4 , 1) + t 2 (0 , 3 , 1) + t 3 (6 , 0 , 1) = (0 , 0 , 0) , or equivalently, the system 2 t 1 + 6 t 3 = 0 4 t 1 + 3 t 2 = 0 t 1 t 2 t 3 = 0 has only the trivial solution. By placing the last equation ²rst, and eliminating t 1 from the other two equations, we obtain the system t 1 t 2 t 3 = 0
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

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sol3 - SOLUTIONS FOR HOMEWORK 3 Section 1.5 18 This is a...

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