How to prove that ${1,x,x^2}$ is a basis of a real polynomial functions space. Let $V$ be the real vector space of all polynomial functions from $\mathbb{R}$ to $\mathbb{R}$ at most second degree. That is, the space of all functions with form $f(x)=c_0+c_1x+c_2x^2$ with $c_i\in\mathbb{R}$
I need to prove that $\{1,x,x^2\}$ is a basis of $V$. But for that, first I need to prove the linear independence of those vectors, right? Well, here is where I get stuck. The only way I can think to do it, is defining an isomorphism $ T:V \to \mathbb{R^3} $ where each coeficient $ T(f)=(c_1,c_2,c_3) $ and proof the linear independece on $\mathbb{R^3}$.
Despite it work, probably it would be better if the proof is made without using transformations. Any suggestions?
 A: hint: Suppose that $a_1\cdot 1 + a_2\cdot x + a_3\cdot x^2 = 0, \forall x \in \mathbb{R}$, then substitute $x = 0,1,2$ into the equation and solve for $a_1,a_2,a_3$. What do you get?
A: Note that linear independence means you must prove that
$$ax^2 + bc + c = 0 \qquad\forall\ x\in\mathbb R$$
implies $a=b=c=0$. Chose some values of $x$ to get a very simple linear system with unique solution $a=b=c=0$ and you're done. 
To show that the basis spans $P_2$, just remark how the coefficients of the linear combination of a particular vector will look like (this is actually trivial).
A: This is both an extended comment and an answer, and I'm sorry if it is outside of the level of the questioner, but I think there is an interesting point to be made here. 
One needs to be somewhat careful when identifying polynomials to the natural functions coming from polynomials. For instance, in finite fields there are many examples of distinct polynomials which induce the same function. For example the polynomial $x^3+x^2+x+1$ thought of as a function $\Bbb Z/2 \Bbb Z \rightarrow \Bbb Z/2 \Bbb Z$ gives the same function as the $0$ polynomial. 
To see this cannot happen in infinite fields (like $\Bbb R$), note that if two polynomials $f$ and $g$ induce the same functions $\tilde f$ and $\tilde g$ then there difference $f-g$ is a polynomial with infinitely many roots (hence is zero as a polynomial).
Somewhat of an Aside: A field is algebraically closed if every non-constant (as a polynomial) polynomial has a root. Let a field be "functionally algebraically closed" if every non-constant (as a function) polynomial has a root. It is a nice (if somewhat easy) exercise to show the only functionally algebraically closed field which is not already algebraically closed is $\Bbb Z/2 \Bbb Z$.
A: First, $V$ is the space of polynomials of degree $\le 2,$ not the space of all polynomials. Suppose $c_0+c_1x+ c_2x^2$ is the zero function. Doesn't that imply all $c_j=0$?
