In $\mathbb R[t]_2$ (vector space of all polynomials of up to a second degree), I need to compute the norm $||t^2 - 2t + 1||$, knowing only that:
$$\langle p(t),q(t)\rangle = p(-2)q(-2) + p(-1)q(-1) + p(0)q(0) + p(1)q(1)$$
I have absolutely no clue on how to compute it as I don't have the polynomials $p(t)$ and $q(t)$. Do I need to assume $p(t)$ and $q(t)$ generic polynomials, such as $p(t) = a + bt + ct^2$ and $q(t) = d + et + ft^2$, where $a,b,c,d,e,f$ would be scalars in $\mathbb R$?
On
As you have computed once for all the expression
$$\langle p,q \rangle=-2ea+4ad+6af+6eb-8ec-2bd+6cd-8bf+18cf,$$
said otherwise
$$(a \ \ b \ \ c)\underbrace{\left(\begin{array}{rrr}2&-1&3\\-1&3&-4\\3&-4&9\end{array}\right)}_A\begin{pmatrix}d\\e\\f\end{pmatrix},$$
it is not lost time if you are asked to calculate other norms.
Moreover, and this is the most important, due to the fact that the eigenvalues of symmetric matrix $A$ are positive ($0.5815,$$1.4351,$ $11.9834$) it proves that this inner product is associated with a "true" norm, which isn't a priori evident (triangular inequality is lengthy by direct means).
Just plug into your formula:
\begin{align} \|t^2-2t+1\|^2 &= \|(t-1)^2\|^2\\ &= \langle (t-1)^2, (t-1)^2\rangle\\ &= (-2-1)^2(-2-1)^2 + (-1-1)^2(-1-1)^2 + (0-1)^2(0-1)^2 + (1-1)^2(1-1)^2\\ &= 81+16+1\\ &= 98 \end{align} so $\|t^2-2t+1\|=\sqrt{98}$.