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Unformatted text preview: Math 116 Problem Set 2 October 10, 2007 (Thanks to Ameya Velingker for his writeup.) 1. (a) This modi cation does not make our system a vector space. Notice that Observe that if the distributive law were to hold, we would have θ = 0 S = (1 + ( 1)) S = 1 · S + ( 1) · S = S + S = S , a contradiction. (b) Recall from lecture that for each x ∈ X , ∃ x such that x +( x ) = θ . Therefore, given z + z = z , we can add z to both sides: z + z + ( z ) = z + z z + ( z + ( z )) = θ z + θ = θ z = θ. Thus, z = θ is the unique solution. 2. Notice that θ = (( 1) + 1)( a + b ) = ( a + b ) + ( a + b ) = a b + a + b so a + θ = b + a + b , and a + θ = a , so we have a = b + a + b , which implies b + a = a + b , as desired. (Notice that by using 0 = ( 1) + 1 instead of 0 = 1 + ( 1) , we avoid the trouble of proving that θ is a left identity.) 3. We shall establish by induction on k , that any linear combination of the form α 1 x j 1 + α 2 x j 2 + ··· + α k x j k , with α i ≥ , ∑ k i =1 α k = 1 and is in co ( S ) . For the base case, k = 1 , the result is obvious, since x 1 ∈ co ( S ) . Now, suppose the result is true for k = m (where m < n ) and we wish to prove the result for k = m +1 . Assume we are given α 1 ,α 2 ,...,α m +1 with α i ≥ , ∑ m +1 i =1 α i = 1 . Without loss of generality, assume that α 1 6 = 0 (otherwise, at least one of the α i is nonzero, and we can reorder the α i so that α 1 = 0 ). Then, α 1 x j 1 + α 2 x j 2 + ··· + α m +1 x j m +1 = α 1 x 1 + m +1 X i =2 α i ! α 2 ∑ m +1 i =2 α i x j 2 + ··· + α m +1 ∑ m +1 i =2 α i x j m +1 ! . Because α 2 P m +1 i =2 α i + ··· + α m +1 P m +1 i =2 α i = 1 , we have that P = α 2 P m +1 i =2 α i x j 2 + ··· + α m +1 P m +1 i =2 α i x j m +1 ∈ co ( S ) (by the induction hypothesis). But, note that α 1 x j 1 + α 2 x j 2 + ··· + α m +1 x j m +1 = α 1 x j 1 +( α 2 + ··· + α m +1 ) P , i.e. it is on the straight line between x j 1 and P , two points already established to be in co ( S ) . Therefore, α 1 x j 1 + α 2 x j 2 + ··· + α m +1 x j m +1 ∈ co ( S ) , as desired. Thus, we have shown that K ⊆ co ( S ) . To show that K = co ( S ) , it su ces to show that K is convex. Thus, suppose we are given two points α 1 x 1 + ··· + α n x n and β 1 x 1 + ··· + β n x n in K (with α i ,β i ≥ and ∑ n i =1 α i = ∑ n i =1 β i = 1 ). Then, consider any point on the line segment connecting these two points. Such a point must be of the form Q = c ( α 1 x 1 + ··· + α n x n ) + d ( β 1 x 1 + ··· + β n x n ) , 1 where c,d ≥ and c + d = 1 . But the above point can be reexpressed as ( cα 1 + dβ 1 ) x 1 + ··· + ( cα n + dβ n ) x n , and note that ∑ n i =1 ( cα i + dβ i ) = c ∑ n i =1 α i + d ∑ n i =1 β i = c + d = 1 . Therefore, clearly, Q ∈ K ....
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 Fall '09
 Linear Algebra, Vector Space, WI, Linear combination

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