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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 = { xp ( x )  p ( x ) ∈ P 2 } of P 4 . Solution: Observe that xp ( x ) is a polynomial of degree at most 3, so S 1 is a subset of P 4 . Since the zero vector in P n is the polynomial 0, we have that 0 ∈ P 2 , so 0 = x (0) ∈ S 1 . So the zero vector of P 4 is in S 1 . Let q 1 ( x ) , q 2 ( x ) ∈ S 1 . Then there exists p 1 ( x ) , p 2 ( x ) ∈ P 2 such that q 1 ( x ) = xp 1 ( x ) and q 2 ( x ) = xp 2 ( x ). We have q 1 ( x ) + q 2 ( x ) = xp 1 ( x ) + xp 2 ( x ) = x ( p 1 ( x ) + p 2 ( x )) , and p 1 ( x ) + p 2 ( x ) ∈ P 2 since P 2 is closed under addition. Hence q 1 ( x ) + q 2 ( x ) ∈ S 1 . Similarly, for any t ∈ R we get tq 1 ( x ) = t ( xp 1 ( x )) = x ( tp 1 ( x )) ∈ S 1 , since tp 1 ( x ) ∈ P 2 as P 2 is closed under scalar multiplication. Thus, by the subspace test S 1 is a subspace of P 4 . Observe that every vector in S 1 has the form xp ( x ) = x ( c + bx + ax 2 ) = cx + bx 2 + ax 3 . Hence, S 1 = span { x, x 2 , x 3 } . Clearly { x, x 2 , x 3 } is also linearly independent, and hence it is a basis for S 1 . b) S 2 = a 1 a 2 a 3  a 1 a 2 a 3 = 0 of M (2 , 2). Solution: Observe that A = 1 1 0 0 and B = 0 0 0 1 are both in S 2 , but A + B = 1 1 0 1 6∈ S 2 , so S 2 is not closed under addition and hence is not a subspace. c) S 3 = { p ( x ) ∈ P 2  p (2) = 0 } of P 2 . Solution: In P 2 the zero vector is the polynomial that satisfies z ( x ) = 0 for all x . Hence z (2) = 0, so z ( x ) ∈ S 3 ....
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This note was uploaded on 11/25/2010 for the course MATH 136 taught by Professor All during the Spring '08 term at Waterloo.
 Spring '08
 All
 Linear Algebra, Algebra, Vector Space

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