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Unformatted text preview: SOLUTION TO ASSIGNMENT 1  MATH 251, WINTER 2007 1. Consider the following vector spaces (you do not need to prove those are vector spaces, unless specified): (1) The vector space V 1 of continuous functions f : [0 , 1] → R , where we define f + g to be the function ( f + g )( x ) = f ( x )+ g ( x ) and αf to be the function ( αf )( x ) = αf ( x ). (2) The vector space V 2 of polynomials over a field F with the usual addition of poly nomials and multiplication by a scalar (which is a special case of multiplying two polynomials). (3) The vector space V 3 = { ( x 1 ,x 2 ,x 3 ,... ) : x i ∈ F } , where we define ( x 1 ,x 2 ,x 3 ,... ) + ( y 1 ,y 2 ,y 3 ,... ) = ( x 1 + y 1 ,x 2 + y 2 ,x 3 + y 3 ,... ) , and α ( x 1 ,x 2 ,x 3 ,... ) = ( αx 1 ,αx 2 ,αx 3 ,... ) . (4) Fix scalars A,B ∈ F . Prove that the vectors ( x 1 ,x 2 ,x 3 ,... ) of V 3 that satisfy: x n = Ax n 1 + Bx n 2 , n ≥ 3 form a subspace V 4 . (5) The vector space of functions f : { 1 , 2 ,...,n } → R , where again ( f + g )( x ) = f ( x ) + g ( x ) and ( αf )( x ) = αf ( x ). In each case determine whether the vector space is finite dimensional or infinite dimen sional. In case it is finite dimensional, give a basis. In case it is infinite dimensional, prove that by providing an explicit infinite set of linearly independent vectors. Proof. (1) We claim that the functions 1 ,x,x 2 ,... are a linearly independent set. In deed, suppose that for some N and some scalars a 1 ,...,a N , the function a + a 1 x + ··· + a N x N is the zero function on [0 , 1]. This means that the polynomial a + a 1 x + ··· + a N x N has infinitely many zeros and so must be the zero polynomial. That is, every a i = 0. Since we have arbitrarily large independent sets the dimension must be infinite. (2) This is even easier. The polynomials 1 ,x,x 2 ,... are a linearly independent set, because by definition if a + a 1 x + ··· + a N x N is the zero polynomial then each a i = 0. (Note that in (1) we must consider when is such an expression zero as a function , while here it is easier: we just need to consider when it is not zero as a formal expression .) 1 2 SOLUTION TO ASSIGNMENT 1  MATH 251, WINTER 2007 (3) Suppose we have two series ( x 1 ,x 2 ,x 3 ,... ) , ( y 1 ,y 2 ,y 3 ,... ) satisfying the recursion relation. Then α ( x 1 ,x 2 ,x 3 ,... ) = ( αx 1 ,αx 2 ,αx 3 ,... ) satisfies αx n = α ( Ax n 1 + B x n 2 ) = A · αx n 1 + B · αx n 2 and so is an element of V 4 . Similarly ( x 1 ,x 2 ,x 3 ,... )+ ( y 1 ,y 2 ,y 3 ,... ) = ( x 1 + y 1 ,x 2 + y 2 ,x 3 + y 3 ,... ) satisfies x n + y n = Ax n 1 + Bx n 2 + Ay n 1 + Ay n 2 = A ( x n 1 + y n 1 )+ B ( x n 2 + y n 2 ). Finally, V 4 is not empty, because the sequence (0 , , ,... ) satisfies the recursion relation. Thus, V 4 is a subspace....
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 Spring '08
 Goren
 Vector Space, Z/pZ

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