ProblemSet2

ProblemSet2 - Math 116 Problem Set 2 October 10, 2007...

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Unformatted text preview: Math 116 Problem Set 2 October 10, 2007 (Thanks to Ameya Velingker for his write-up.) 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 = 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 ) = . Therefore, given z + z = z , we can add- z to both sides: z + z + (- z ) = z +- z z + ( z + (- z )) = z + = z = . Thus, z = is the unique solution. 2. Notice that = ((- 1) + 1)( a + b ) =- ( a + b ) + ( a + b ) =- a- b + a + b so a + =- b + a + b , and a + = 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 is a left identity.) 3. We shall establish by induction on k , that any linear combination of the form 1 x j 1 + 2 x j 2 + + k x j k , with i , k i =1 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 1 , 2 ,..., m +1 with i , m +1 i =1 i = 1 . Without loss of generality, assume that 1 6 = 0 (otherwise, at least one of the i is nonzero, and we can reorder the i so that 1 = 0 ). Then, 1 x j 1 + 2 x j 2 + + m +1 x j m +1 = 1 x 1 + m +1 X i =2 i ! 2 m +1 i =2 i x j 2 + + m +1 m +1 i =2 i x j m +1 ! . Because 2 P m +1 i =2 i + + m +1 P m +1 i =2 i = 1 , we have that P = 2 P m +1 i =2 i x j 2 + + m +1 P m +1 i =2 i x j m +1 co ( S ) (by the induction hypothesis). But, note that 1 x j 1 + 2 x j 2 + + m +1 x j m +1 = 1 x j 1 +( 2 + + 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, 1 x j 1 + 2 x j 2 + + 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 1 x 1 + + n x n and 1 x 1 + + n x n in K (with i , i and n i =1 i = n i =1 i = 1 ). Then, consider any point on the line segment connecting these two points. Such a point must be of the form Q = c ( 1 x 1 + + n x n ) + d ( 1 x 1 + + n x n ) , 1 where c,d and c + d = 1 . But the above point can be reexpressed as ( c 1 + d 1 ) x 1 + + ( c n + d n ) x n , and note that n i =1 ( c i + d i ) = c n i =1 i + d n i =1 i = c + d = 1 . Therefore, clearly, Q K ....
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ProblemSet2 - Math 116 Problem Set 2 October 10, 2007...

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