Is the set of all quadratic, real-valued functions a vector space?

Question

I know there are already some similar questions but I still do not understand how to prove that the set of all quadratic functions defined on the real line is a vector space.

The question I got was: "Consider the quadratic, real-valued functions defined on the real line. Prove that it is a vector space. Of what dimension? Produce a basis."

If I read the exercise I would assume that it is a vector space.

But how is it possible since a quadratic function like ax^2+bx+c can be never closed under scalar multiplication I think because if the scalar would be zero I do not get a quadratic function as result.

Cheers for any help or hint.

Answer 1

Let's be careful by what we mean by a quadratic function. Typically, one defines a quadratic function $p:\mathbb{R} \to \mathbb{R}$ as a function given by the equation $p(x) = ax^2 + bx + c$ for some $a,b,c \in \mathbb{R}$ with $a \neq 0$. If we enforce the condition $a \neq 0$, then the set of quadratics are not a subspace of ${C}(\mathbb{R})$, since it would not contain the zero function.

If we don't enforce this condition and allow $a=0$, then it is indeed a subspace of ${C}(\mathbb{R})$, since clearly we have closure under scalar multiplication and addition. The dimension is $3$ with the canonical basis $\{1, x, x^2\}$. However, in this case, it is not ideally described as the set of quadratic functions, but rather the set of polynomials of degree $\leq 2$.

Answer 2

Your doubt is a legitimate one, but I'm sure that whoever created that exercise had in mind that a “quadratic function” is any function from $\mathbb R$ into itself of the type $x\mapsto ax^2+bx+c$. In articular, if $a=b=c=0$, then you get the null function.