Vector spaces of polynomials

Algebra Level 2

Which of the following sets of polynomials are vector spaces over R \mathbb{R} with the usual polynomial addition and scalar multiplication?

A: All polynomials of the form p ( x ) = a x 2 + b x + c p(x)=ax^2 + bx + c with a , b , c R a,b,c\in\mathbb{R} and a + b + c = 1 a+b+c=1 .

B: All polynomials of the form p ( x ) = a x 2 + b x + c p(x) = ax^2 + bx + c with a , b , c R a,b,c\in\mathbb{R} and a + b + c = 0 a+b+c=0 .

C: All polynomials of the form p ( x ) = a x 2 + b x + c p(x)=ax^2 + bx + c with a , b , c R a,b,c\in\mathbb{R} and p ( 1 ) = p ( 2 ) p(1) = p(2) .

A and B only B and C only C only All of them

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Affan Morshed
Apr 23, 2019

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).

Ognjen Vukadin
Jul 29, 2016

The sets A,B,C are all subsets of the vector space of polynomials P 2 P_2 of degree 2 \leq 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 P_2 . This will be the case if the corresponding set is non-empty and if for any two polynomials p , q p,q in the set the sum p + q p+q and the scalar multiple γ p \gamma p for each γ R \gamma \in \mathbb{R} is again in the same set. In particular, the zero polynomial 0 0 should be in the set.

For the set A, the polynomial 0 x 2 + 0 x + 0 = 0 A 0x^2 + 0x + 0 = 0\notin A since a = b = c = 0 a=b=c = 0 and a + b + c = 0 1 a+b+c = 0\neq 1 so the set cannot be a vector space.

The set B is non-empty since 0 B 0\in B . Suppose p , q B p,q\in B with p ( x ) = a p x 2 + b p x + c p p(x) = a_px^2 + b_p x + c_p and q ( x ) = a q x 2 + b q x + c q 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 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 p(x) + q(x) = (a_p+a_q)x^2 + (b_p+b_q)x + (c_p+c_q)\in 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 (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 \gamma p(x) = \gamma a_px^2 + \gamma b_p x + \gamma c_p \in B since γ a p + γ b p + γ c p = γ ( a p + b p + c p ) = γ 0 = 0 \gamma a_p + \gamma b_p + \gamma c_p = \gamma (a_p+b_p+c_p) = \gamma \cdot 0 = 0 .

The set C is non-empty since for any constant polynomial the value at 1 1 equals the value at 2 2 .Suppose p , q C p,q\in C . Then p ( 1 ) = p ( 2 ) p(1) = p(2) and q ( 1 ) = q ( 2 ) q(1) = q(2) and ( p + q ) ( 1 ) = p ( 1 ) + q ( 1 ) = p ( 2 ) + q ( 2 ) = ( p + q ) ( 2 ) (p+q)(1) = p(1) + q(1) = p(2) + q(2) = (p+q)(2) so p + q C p+q\in C . Similarly, ( γ p ) ( 1 ) = γ p ( 1 ) = γ p ( 2 ) = ( γ p ) ( 2 ) (\gamma p)(1) = \gamma p(1) = \gamma p(2) = (\gamma p)(2) and γ p C \gamma p\in C .

Hence, only B and C are vector spaces.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...