Are the set of all the polynomials of degree 2 a vector space?

[ironman1478]

Homework Statement

Let P denote the set of all polynomials whose degree is exactly 2. Is P a vector space? Justify your answer.

Homework Equations

(the numbers next to the a's are substripts
P is defined as ---->A(0)+A(1)x+A(2)x^2

The Attempt at a Solution

I really don't know how to do this problem. i want to say that it isn't a vector space because it violates the property of having an additive inverse. as in, there is no value of x such that

F(x) + (F(-x)) = F(x)+(-F(x)) = 0 if we keep all of the values for A the same

A(0) + A(1)x + A(2)x^2 + A(0) + A(1)(-x) + A(2)(-x)^2 == 2A(0) + 2A(2)x^2 != 0

therefore, there is no additive inverse.
i probably did it wrong, but i don't know. all i know is that the book says that it isn't a vector space, but it doesn't give the reason.

 

[jbunniii]

i want to say that it isn't a vector space because it violates the property of having an additive inverse.

If p(x) is a 2nd degree polynomial, isn't -p(x) also a 2nd degree polynomial? And what happens when you add these two together?

 

[jbunniii]

all i know is that the book says that it isn't a vector space, but it doesn't give the reason.

Can you think of two 2nd degree polynomials, p(x) and q(x), such that when you add them together, the resulting polynomial isn't 2nd degree?

 

[ironman1478]

so because P(x) + (-(P(x)) = 0 and therefore, the answer is not a 2nd degree polynomial, then it can't be a vector space because it isn't closed under addition? if so, then i guess i just forgot to check the first property for a set to be a vector space and assumed it to be true.

 

[jbunniii]

so because P(x) + (-(P(x)) = 0 and therefore, the answer is not a 2nd degree polynomial, then it can't be a vector space because it isn't closed under addition? if so, then i guess i just forgot to check the first property for a set to be a vector space and assumed it to be true.

Yes, any vector space has to contain 0, and 0 isn't a 2nd degree polynomial.

Another example would be p(x) = x^2 + x + 1, and q(x) = -x^2. Then p(x) + q(x) = x + 1, which is 1st order.