solmidterm1official

# solmidterm1official - Harvard University Solutions Midterm...

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Harvard University Solutions Midterm 1 for Math 121, Fall 2007 Monday, October 22, 2007 1. (20 points) a) True b) False c) False d) True e) True 2. (40 points) a) Clearly, the zero vector lies in each of the subsets in (i), (ii). Thus, check if for any two vectors v , w from the subset and for any λ R the vector v + λw is in the subset as well. (i) { (0 ,x, 2 x, 3 x ) | x R } is a subspace of R 4 . Proof: Let v = (0 ,x, 2 x, 3 x ) and w = (0 ,y, 2 y, 3 y ) be two arbitrary vectors from this subset. Then it is v + λw = (0 , x + λy, 2 x + 2 λy, 3 x + 3 λy ) = (0 ,z, 2 z, 3 z ) with z = x + λy . Thus, v + λw is in the subset. This proves that this subset is indeed a subspace of R 4 . (ii) { ( x 4 ,x 3 ,x 2 ,x ) | x R } is not a subspace of R 4 ? Proof: Consider the vector v = (1 , 1 , 1 , 1), which belongs to this subset. The vector 2 v = (2 , 2 , 2 , 2) is not in this subset, as there does not exist any x R with (2 , 2 , 2 , 2) = ( x 4 ,x 3 ,x 2 ,x ). This proves that this subset is not a subspace of

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## This note was uploaded on 10/23/2010 for the course MATH Math 121 taught by Professor Bieri during the Fall '07 term at Harvard.

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solmidterm1official - Harvard University Solutions Midterm...

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