Assignment 3 solution

# Assignment 3 solution - MATH 223 Linear Algebra Winter 2008...

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MATH 223, Linear Algebra Winter, 2008 Solutions to Assignment 3 1. Let V = P 2 ( x ) be the vector space of polynomials with real coeﬃcients of degree at most two. For each of the following subsets of V , decide if it is independent, if it is a spanning set for V , and if it is a basis for V . Justify your answers. (a) S 1 = { 1 + x, 1 + x 2 ,x + x 2 } Solution: This is independent, and a spanning set, and hence a basis for V . For independence, suppose that a (1 + x ) + b (1 + x 2 ) + c ( x + x 2 ) = 0. Then a + b = a + c = b + c = 0; this easily implies that a = b = c = 0. Since P 2 ( x ) is 3-dimensional, this implies it is a basis, but let’s check from the deﬁnition that it’s also a spanning set for V . Given any polynomial a 0 + a 1 x + a 2 x 2 V , let α = a 0 + a 1 - a 2 2 , β = a 0 + a 2 - a 1 2 and γ = a 1 + a 2 - a 0 2 ; it is easily seen that α (1+ x )+ β (1+ x 2 )+ γ ( x + x 2 ) = a 0 + a 1 x + a 2 x 2 . (b) S 2 = { 1 + x - 2 x 2 , 3 - 2 x + x 2 , 5 - 3 x 2 , 13 - 2 x - 5 x 2 } . Solution: This set cannot possibly be independent, as it has four polynomials (vectors) in it. Also, 2(1 + x - 2 x 2 ) + 1(3 - 2 x + x 2 ) = 5 - 3 x 2 and 4(1 + x - 2 x 2 ) + 3(3 - 2 x + x 2 ) = 13 - 2 x - 5 x 2 . So S 2 spans the same subspace as { 1 + x - 2 x 2 , 3 - 2 x + x 2 } ; it cannot be a spanning set for all of V either. But let’s just see directly that the constant polynomial 1 is not in Span ( S 2 ). If it were, we could solve x 1 (1 + x - 2 x 2 ) + x 2 (3 - 2 x + x 2 )+ x 3 (5 - 3 x 2 )+ x 4 (13 - 2 x - 5 x 2 ) = 1, or x 1 +3 x 2 +5 x 3 +13 x 4 = 1, x 1 - 2 x 2 + 0 x 3 - 2 x 4 = 0 - 2 x 1 + x 2 - 3 x 3 - 5 x 4 = 0. The corresponding augamented matrix is 1 3 5 13 | 1 1 - 2 0 - 2 | 0 - 2 1 - 3 - 5 | 0 . It row-reduces to 1 0 2 4 | 2 5 0 1 1 3 | 1 5 0 0 0 0 | 2 5 , so there is no solution. (c) S 3 = { 1 - x 2 , 2 + x + 3 x 2 , 5 - x + 4 x 2 , 2 x - 7 x 2 } . Solution: Again, this set cannot be independent. Speciﬁcally, if we try to solve x 1 (1 - x 2 )+ x 2 (2+ x +3 x 2 )+ x 3 (5 - x +4 x 2 ) = 2 x - 7 x 2 , we ﬁnd we do have a solution x 1 = 63 14 , x 2 = 11 14 , x 3 = - 17 14 . In fact for any

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## This note was uploaded on 04/29/2008 for the course MATH 223 taught by Professor Loveys during the Fall '07 term at McGill.

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Assignment 3 solution - MATH 223 Linear Algebra Winter 2008...

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