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Unformatted text preview: Winter 2011 • Math 67 • Linear Algebra Homework 3 Problem 2.3 Suppose that ( v 1 ,...,v n ) is linearly independent and that ( v 1 + w,...,v n + w ) is linearl dependent. Then there are scalars a i , not all zero, such that ∑ n i =1 a i ( v i + w ) = 0. Expanding this and moving the terms with w ’s to the right side of the equation yields a 1 v 1 + ··· + a n v n = ( a 1 + ··· + a n ) w The right side is not zero, since it was then we would see that ( v 1 ,...,v n ) was linearly dependent, which it is not. This means ( a 1 + ··· + a n ) 6 = 0 so we can divide it away and get 1 a 1 + ··· + a n ( a 1 v 1 + ··· + a n v n ) = w. Thus, w is in the span of v 1 ,...,v n . Problem 2.4 The set of polynomials in P ( F ) of degree m is not a subspace of P ( F ), since it is not closed under additions. To wit, ( z + z 2 ) z 2 = z is the sum of two polynomials of degree 2. However, z is not a polynomial of degree 2....
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 Winter '07
 Schilling
 Linear Algebra, Algebra, Vectors, Scalar, Vector Space, basis

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