Chapter 2

Chapter 2 - Chapter 2 Finite-Dimensional Vector Spaces 2.1...

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Chapter 2: Finite-Dimensional Vector Spaces 2.1 Let v V ; since ( v 1 ,...,v n ) spans v , there exist scalars a 1 ,...,a n F such that v = n i =1 a n v n . We wish to ﬁnd b 1 ,...,b n F such that v = b 1 ( v 1 - v 2 ) + ··· + b n - 1 ( v n - 1 - v n ) + b n v n (since v is an arbitrary element of V , doing so ensures that the new list also spans V ). To do so, set b 1 = a 1 , b n = a n , and b i - b i - 1 = a i otherwise. Then we have b 1 ( v 1 - v 2 ) + ··· + b n - 1 ( v n - 1 - v n ) + b n v n = b 1 v 1 + ( b 2 - b 1 ) v 2 + ··· + ( b n - 1 - b n - 2 ) v n - 1 + b n v n = a 1 v 1 + a 2 v 2 + ··· + a n v n = v Since a i ,...,a n are determined for each v V , we have shown that these deﬁnitions for b 1 , ··· ,b n determine a linear combination expressing v in terms of the new list. Hence, that list spans V . 2.2 Suppose that 0 = a 1 ( v 1 - v 2 ) + ··· + a n v n ; we wish to show that a 1 = ··· = a n = 0. To do so, we will use the fact that ( v 1 ,...,v n ) is a linearly independent in list in V . To wit, we expand out this expression: 0 = a 1 ( v 1 - v 2 ) + ··· + a n v n = a 1 v 1 + ( a 2 - a 1 ) v 2 + ··· + a n v n The independence of ( v 1 ,...,v n ) implies that a 1 = 0 ,a 2 - a 1 = 0 ,...,a n = 0. Starting from the second expression we have that a 2 = a 1 = 0; continuing as such down the list shows that a 1 = a 2 = ··· = a n = 0, just as desired. Hence, the list ( v 1 - v 2 ,...,v n - 1 - v n ,v n ) is linearly independent in V . 2.3 Once again, let 0 = a 1 ( v 1 + w ) + ··· + a n ( v n + w ). If we expand this expression, we obtain 0 = a 1 ( v 1 + w ) + ··· + a n ( v n + w ) = a 1 v 1 + ··· + a n v n + ( a 1 + ··· + a n ) w = ⇒ - ( a 1 + ··· + a n ) w = a 1 v 1 + ··· + a n v n Now, since ( v 1 , ··· ,v n ) is a linearly independent list in V , we have that α = n i =1 a i 6 = 0. Hence, we can divide both sides of our equality by this sum to obtain w = - a 1 α v 1 - ··· - a n α v n Since we have expressed w as a linear combination of ( v 1 ,...,v n ), we must have w span ( v 1 ,...,v n ). 2.4 The answer is no. Consider the following polynomials, both of which are in the described subset P m ( F ): p + ( z ) = a 0 + a 1 z + ··· + a m z m and p - ( z ) = a 0 + a 1 z + ··· - a m z m If we take the sum, we have p + ( z ) + p - ( z ) = a 0 + a 1 z + ··· + a m - 1 z m - 1 6 = P m ( F ), for the degree is not

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Chapter 2 - Chapter 2 Finite-Dimensional Vector Spaces 2.1...

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