# Vector Spaces and polynomial functions

I'm really confused about a question I came across in my textbook.
It basically says this:

Consider the set of polynomial functions of degree 2. Prove that this set is not closed under addition or scalar multiplication (and therefore not a vectorspace).
I'm confused because I think it is closed under addition and scalar mult.
example:

f(x) = ax^2 + bx + c
g(x) = dx^2 + ex + f
(f+g)(x) = (a+d)x^2 + (b+e)x + (c+f)

(sf)(x) = (sa)x^2 + (sb)x + sc

both results should be in the set of polynomial functions of degree 2.
Why would the question say it is not closed under addition and scalar mult. ??
Am I missing something very basic here, or could it be a trick question or something?
Thanks!

## Answers and Replies

The set of polynomials of degree less than or equal to 2 is a vector space. If you only have polynomials of degree 2, this is no longer true.

I'm not sure I understand why this is.
So then the set of functions is only: ax^2
Is this what you mean? If it is, I'm still not sure why the set wouldn't be closed under addition and scalar mult. :

f(x) = ax^2
g(x) = bx^2
(f+g)(x) = (a+b)x^2
(sf)(x) = (sa)x^2

{where s is a scalar; a and b are coefficients)

I'm not sure what I'm missing in this problem. :(

And thanks for the quick reply master_coda

-Confused

x^2 + (-x^2) is not a degree 2 polynomial.

no but x^2 + (-x^2) = 0
0 is in the vector set.

then what is the additive zero element of your set?

I'm not sure what you mean by additive zero element.

And just so I'm clear, is 2x^2, for example, a degree 2 polynomial that would be in the set that the above question is asking?
i.e. degree 2 polynomials can still have coefficients, right?

every vector space must contain an element denoted O or 0 such that for all vectors u, u+O=O+u=u.

Oh, yeah, I get what you mean now.
Wouldn't the additive zero element be 0^2?

what's the degree of 0^2=0?

I'm sorry. But I don't see the point your trying to get across here.

f(x) = ax^2
g(x) = bx^2
(f+g)(x) = (a+b)x^2
(sf)(x) = (sa)x^2

{where s is a scalar; a and b are coefficients)

Why doesn't this show that its closed under addition and scalar mult?

when you said:

"x^2 + (-x^2) is not a degree 2 polynomial."

doesn't that support the following axiom:
"For every element u of V, there exists an element called the negative of u, denoted -u, such that u + (-u) = 0."
(This is one of the axioms from my text that needs to be satisfied in order for something to be a vector space)

Which further supports why I think this is a vector space.
Thanks for the hints so far, but don't seem to be grasping why this isn't a vector space. :(

The set of polynomials of degree less than or equal to 2 is a vector space. If you only have polynomials of degree 2, this is no longer true.

I'm not sure I understand why this is.

If I'm understanding this correctly...

First, you must realise that (for example) x + 1 is not a member of the set of second degree polynomials. It is however an element of the set of polynomials of degree <= 2. This is crucial.

Now, add two polynomials of degree 2. Can you say with certainty that this sum will also be a polynomial of degree 2 (unless you place certain restrictions on the coefficients)? No, consider (x^2 + 3x + 1) + (-x^2 + x + 3) = 4x + 4, which is a polyonomial of degree 1. Hence the set of second degree polynomials is not closed under addition.

However, you /can/ say that the sum of two second degree polynomials will be /a/ polynomial, and that its degree will be at most 2, hence the set of polyonomials of degree <= 2 is closed under addition (and is a vector space).

Last edited:
matt grime
Science Advisor
Homework Helper
And to REALLY get home something that you need ton understand: 0 is a constant polynomial, it has degree 0, as do all the polys f(x)=constant. How did you arrive at 0^2? Or if you like write down a polynomial in x with a non zero x^2 coeff that when added to every other polynomial leaves it unchanged.

Originally posted by endfx
"x^2 + (-x^2) is not a degree 2 polynomial."

doesn't that support the following axiom:
"For every element u of V, there exists an element called the negative of u, denoted -u, such that u + (-u) = 0."
(This is one of the axioms from my text that needs to be satisfied in order for something to be a vector space)

In order for you to use that axiom, you need to assume the set of degree 2 polynomials is a vector space. Since the set of degree 2 polynomials has no zero element, it is clearly not a vector space.

Thus you cannot draw any valid conclusions by assuming that it is a vector space, since that would be a false assumption.

Hurkyl
Staff Emeritus
Science Advisor
Gold Member
f(x) = ax^2
g(x) = bx^2
(f+g)(x) = (a+b)x^2
(sf)(x) = (sa)x^2

{where s is a scalar; a and b are coefficients)

Why doesn't this show that its closed under addition and scalar mult?

Because (a+b)x^2 isn't always a degree 2 polynomial.

Thanks very much to everyone who posted to my questions. Everybody's reply helped be understand and solve the problem.
Thanks again, I really appreciate it!