Which of the following sets of polynomials are vector spaces over R with the usual polynomial addition and scalar multiplication?
A: All polynomials of the form p ( x ) = a x 2 + b x + c with a , b , c ∈ R and a + b + c = 1 .
B: All polynomials of the form p ( x ) = a x 2 + b x + c with a , b , c ∈ R and a + b + c = 0 .
C: All polynomials of the form p ( x ) = a x 2 + b x + c with a , b , c ∈ R and p ( 1 ) = p ( 2 ) .
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The sets A,B,C are all subsets of the vector space of polynomials P 2 of degree ≤ 2 . To check if they form a vector space with the usual polynomial addition and scalar multiplication we need to check if they form a vector subspace of P 2 . This will be the case if the corresponding set is non-empty and if for any two polynomials p , q in the set the sum p + q and the scalar multiple γ p for each γ ∈ R is again in the same set. In particular, the zero polynomial 0 should be in the set.
For the set A, the polynomial 0 x 2 + 0 x + 0 = 0 ∈ / A since a = b = c = 0 and a + b + c = 0 = 1 so the set cannot be a vector space.
The set B is non-empty since 0 ∈ B . Suppose p , q ∈ B with p ( x ) = a p x 2 + b p x + c p and q ( x ) = a q x 2 + b q x + c q . Then a p + b p + c p = a q + b q + c q = 0 . We have p ( x ) + q ( x ) = ( a p + a q ) x 2 + ( b p + b q ) x + ( c p + c q ) ∈ B since ( a p + a q ) + ( b p + b q ) + ( c p + c q ) = a p + b p + c p + a q + b q + c q = 0 . Similarly, γ p ( x ) = γ a p x 2 + γ b p x + γ c p ∈ B since γ a p + γ b p + γ c p = γ ( a p + b p + c p ) = γ ⋅ 0 = 0 .
The set C is non-empty since for any constant polynomial the value at 1 equals the value at 2 .Suppose p , q ∈ C . Then p ( 1 ) = p ( 2 ) and q ( 1 ) = q ( 2 ) and ( p + q ) ( 1 ) = p ( 1 ) + q ( 1 ) = p ( 2 ) + q ( 2 ) = ( p + q ) ( 2 ) so p + q ∈ C . Similarly, ( γ p ) ( 1 ) = γ p ( 1 ) = γ p ( 2 ) = ( γ p ) ( 2 ) and γ p ∈ C .
Hence, only B and C are vector spaces.
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you will soon realise that these polynomials create a 3 dimensional linear vector space where a, b, and c are coordinates of position vectors. You must then realise that all of these equations describe 2 dimensional planes (C is a bit hard, you have to substitute the functions). If these planes include the origin, they can create a vector space (0+0+0 is unequal to 1 so A cannot work, but it works for the other ones).