Inner Product of two polynomials Show that the function $<p(x), q(x)>$ = $p(−1)q(−1) + p(0)q(0) + p(1)q(1)$ defines an inner product on $P_2(R)$
I know that an inner product has to hold three conditions, but I am not exactly sure if I use $<1 + x + x^2, 1 +x + x^2>$.  Any help would be greatly appreciated.
 A: An inner product  satisfies three properties: conjugate symmetry, linearity, and positive-definiteness. You need to show that these properties are satisfied for every pair of elements from the vector space of polynomials of degree 2.
That is, you need to show that 
$$\langle p(x),q(x) \rangle = p(−1)q(−1) + p(0)q(0) + p(1)q(1)$$
 satisfies the above properties for any two polynomials $p(x) = a_2x^2 + a_1x + a_0$ and $q(x) = b_2x^2 + b_1x + b_0$.
Here is a proof for the symmetry property:
$$\langle p(x),q(x) \rangle = p(−1)q(−1) + p(0)q(0) + p(1)q(1) = q(−1)p(−1) + q(0)p(0) + q(1)p(1) = \langle q(x),p(x) \rangle$$
Can you try proving the other two properties? You will need to find explicit values for $p(-1)$, $p(0)$, $p(1)$, etc. 
For example: $$p(-1) = a_2(-1)^2 + a_1(-1) + a_0 = a_2 - a_1 + a_0$$
A: @eigenchris
Show that $$<p,q>=\sum_{i=-1}^{1}p(i)q(i)$$ is an inner product on P2. Calculate an orthogonal basis for ${\{P_{0},P_{1},P_{2}\}}$  with respect to this inner
product. Start with the natural basis ${\{1, x, x_{2}}\}$ for P2.
