Find ${T^{*}}(p(x))$ for an arbitrary $p(x) = a+bx+cx^2$ Consider the vector space $P^2(\mathbb{R})$ of real quadratic polynomials with inner product
$$\left\langle p(x),q(x) \right\rangle  = p(-1)q(-1)+p(0)q(0)+p(1)q(1).$$
Let $T:P^2(\mathbb{R}) \to P^2(\mathbb{R})$ be defined by $T(p(x)) = p^\prime(x)$.
Find $T^\ast(p(x))$ for an arbitrary $p(x) = a+bx+cx^2$.
 A: You can probably brute force this using the definition of the adjoint, but if you have Lagrange polynomials at your disposal, you might find it useful to use this to give a neq basis for $P^2(\mathbb{R})$, with respect to which your inner product takes a much simpler form.
Recall that a real quadratic $p$ is completely determined by its values at three distinct points, e.g., $-1$, $0$, and $1$. By the method of Lagrange polynomials, one therefore has that $p \in P^2(\mathbb{R})$ is given by
$$
 p(x) = p(-1)\frac{(x-0)(x-1)}{(-1-0)(-1-1)} + p(0)\frac{(x+1)(x-1)}{(0+1)(0-1)} p(1)\frac{(x+1)(x-0)}{(1+1)(1-0)}\\ = p(-1)f_{-1}(x) + p(0)f_0(x) + p(1)f_1(x),
$$
where
$$
 f_{-1}(x) = -\tfrac{1}{2}x + \tfrac{1}{2}x^2, \quad f_0(x)= 1 - x^2, \quad f_1(x) = \tfrac{1}{2}x + \tfrac{1}{2}x^2.
$$
One can therefore check that $\gamma = \{f_{-1},f_0,f_1\}$ is an orthonormal basis for $P^2(\mathbb{R})$ with, for any $p \in P^2(\mathbb{R})$,
$$
 p(x) = p(-1)f_{-1}(x) + p(0)f_0(x) + p(1)f_1(x).
$$
By finding the matrix $[T]_\gamma$ of $T$ with respect to the orthonormal basis $\gamma$, one can compute $[T^\ast]_\gamma = [T]_\gamma^\ast$, and from there obtain compute the linear transformation $T^\ast$ itself.
