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Unformatted text preview: Math 136 Assignment 6 Solutions 1. Determine, with proof, which of the following are subspaces of the given vector space. Find a basis for each subspace. a) S 1 = x 1 x 2 x 3 ∈ R 3  x 1 + x 2 = 0 ,x 3 = x 2 of R 3 . Solution: We have 0 + 0 = 0 and 0 = 0, so ~ ∈ S 1 . Let ~x,~ y ∈ S 1 , so that we have x 1 + x 2 = 0, x 3 = x 2 , y 1 + y 2 = 0, and y 3 = y 2 . Then ~x + ~ y = x 1 + y 1 x 2 + y 2 x 3 + y 3 , satisfies ( x 1 + y 1 )+( x 2 + y 2 ) = ( x 1 + x 2 )+( y 1 + y 2 ) = 0+0 = 0, and x 3 + y 3 = ( x 2 )+( y 2 ) = ( x 2 + y 2 ). Hence, S 1 is closed under addition. Let ~x ∈ S 1 and t ∈ R , so that we have x 1 + x 2 = 0, x 3 = x 2 . Then t~x = tx 1 tx 2 tx 3 , satisfies ( tx 1 )+( tx 2 ) = t ( x 1 + x 2 ) = t (0) = 0, and ( tx 3 ) = t ( x 3 ) = t ( x 2 ) = ( tx 2 ). Hence, S 1 is also closed under scalar multiplication and hence is a subspace of R 3 . Let ~x ∈ S 1 . Then x 1 = x 2 and x 3 = x 2 , so any vector ~x ∈ S 1 has the form ~x = x 1 x 2 x 3 =  x 2 x 2 x 2 = x 2  1 1 1 . Hence, S 1 = span  1 1 1 and since this is clearly linearly independent, we have that  1 1 1 is a basis for S 1 . b) S 2 = { ax 2 + bx + c ∈ P 2  abc = 0 } of P 2 . Solution: Observe that ~ y = x 2 + 1 ∈ S 2 and ~ z = x ∈ S 2 , but ~ y + ~ z = x 2 + x + 1 6∈ S 2 . Hence, S 2 is not closed under addition and thus is not a subspace of P 2 . 1 2 c) S 3 = { p ( x ) ∈ P 2  p (1) = 0 } of P 2 . Solution: In P 2 the zero vector is the polynomial that satisfies z ( x ) = 0 for all x . Hence z (1) = 0, so ~ ∈ S 3 . Let p ( x ) ,q ( x ) ∈ S 3 . Then p (1) = 0 and q (1) = 0, so ( p + q )(1) = p (1) + q (1) = 0 + 0 = 0 ....
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This note was uploaded on 09/19/2011 for the course MATH 136 taught by Professor All during the Winter '08 term at Waterloo.
 Winter '08
 All
 Vector Space

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