solmidterm1official

solmidterm1official - Harvard University Solutions Midterm...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

solmidterm1official - Harvard University Solutions Midterm...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online