Solution02_121

Solution02_121 - Problem Set 2 Solutions J Scholtz Question...

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Problem Set 2 - Solutions J. Scholtz October 15, 2006 Question 1: 1.5/4 In order to show that the set { e 1 , . . . , e n } is linearly independent we need to show that the following equation is satisfied only if c 1 = . . . = c n = 0: c 1 e 1 + c 2 e 2 + . . . + c n e n = 0 But since e 1 = (1 , 0 , . . . , 0) and so on, then the LHS is equal to ( c 1 , c 2 , . . . , c n ). Hence the last equation reads: ( c 1 , c 2 , . . . , c n ) = (0 , 0 , . . . , 0) Which can only be satisfied if all c i s are zero. Question 2: 1.5/13 a) Two vectors Claim: If u and v are linearly independent then so are u + v and u - v . Proof: The equation a ( u + v )+ b ( u - v ) = 0 expands as ( a + b ) u +( a - b ) v = 0, but since u, v are linearly independent by our assumption, then a + b = 0 and a - b = 0, as F is not of charcteristic 2 then a = - b and a = b imply a = b = 0 and so our vectors are linearly independent. Claim: If u + v and u - v are linearly independent then so are u and u . Proof: If u + v and u - v are linearly independent then so are x = u + v 2 and y = u - v 2 . (Dividing by 2 is posssible since Char( F ) 6 = 2). Then observe that u = x + y and v = x - y and use the previous part. b) Three vectors This is almost the same as the previous part. Please see me if you need more explanation. 1
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Question 3: 1.5/17 An upper triangular matrix is of form: A = c 11 c 21 . . . c n 1 0 . . . . . . c n 2 . . . 0 . . . . . . 0 0 0 c nn Then the collumns form vectors of form v i = (0 , . . . , 0 , k ii , . . . , k in ) where the first i - 1 elements are zeroes. This means that their linear combination gives: a 1 v 1 + . . . + a n v n = ( a 1 k 11 , a 1 k 12 + a 2 k 22 , . . . , X a j k jn ) = 0 But k 11 6 = 0 since the diagonal entries are non-zero, hence a 1 = 0. Then the second entry reads a 2 k 22 = 0, and agains since k 22 6 = 0,then a 2 = 0. This process can be repeated finitely many times (n-times) to show that a 1 = . . . = a n = 0 and hence the set is linearly independent. Question 4: 1.6/3(d-e) part d) We can check, the only solution to: a ( - 1 + 2 x + 4 x 2 ) + b (3 - 4 x - 10 x 2 ) + c ( - 2 - 5 x - 6 x 2 ) = 0 is a = b = c = 0. Hence these three polynomials are linearly independent. And since we know that dim( P 2 ( R )) = 3. Then we know these three polynomials
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Solution02_121 - Problem Set 2 Solutions J Scholtz Question...

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