Let $P_3(\mathbb{R})$ be a real vector space of real polynomials of degree $\le 2$
1) Show that $$ \langle p,q\rangle = \int_{-1}^{1} p(t)g(t)\,dt,$$
Defines an inner product on $P_3(\mathbb{R})$.
Now let $ V = (1,X,X^2)$ be a basis for $P_3(\mathbb{R})$ and let $$ L_V:\mathbb{R^3}\to\mathbb{R^3} \\ \begin{pmatrix} a \\ b \\ c \end{pmatrix}\mapsto a+bX+cX^2$$ be the corresponding linear isomorphism.
2) Find an inner product on $\mathbb{R^3}$ such that $L_V$ is a linear isometry.
I believe I have done 1 correctly: Have to show that the properties of the inner product are true, so: a) $\langle p,p\rangle \ge 0$, so: $\int_{-1}^1 p(t)p(t) \, dt=\int_{-1}^1 p(t)^2 \, dt \ge 0$
b) $\langle p,p\rangle = 0 \iff p=0$, so: $\int_{-1}^1 p(t)^2=0 \, dt$
c) $\langle p,q\rangle = \int_{-1}^1 p(t)g(t) \, dt = \int_{-1}^1 q(t)p(t) \, dt = \langle q,p\rangle$
d) $\langle\alpha p + \beta q, r\rangle = \int_{-1}^1(\alpha p+\beta q(t) r(t) \, dt = \alpha \int_{-1}^1 p(t)+\int_{-1}^1 dt \beta q(t) r(t) \, dt = \alpha \langle p,r\rangle + \beta\langle q,r\rangle$
2) I know that a linear transformation is a linear isometry if $$\langle L(v_1),L(v_2)\rangle = \langle v_1,v_2\rangle$$ We have that: $L(v_1) = a+bx+cx^2$ and $L(v_2) = d+ex+fx^2$, what am I suppose do to now?
Calculate \begin{eqnarray} \langle L({\bf v_1}) | L({\bf v_2}) \rangle = \big\langle \sum_{k=0}^2 v_{1,k} x^k \big| \sum_{l=0}^2 v_{2,l} x^k\big\rangle = \sum_{k,l} v_{1,l} v_{k,l} \color{blue}{\langle x^k|x^l\rangle} = \sum_{k,l} v_{1,k} \color{blue}{\alpha_{k,l} }v_{1,l} = {\bf v_1}^T A {\bf v_2} \end{eqnarray}
Where I have defined the matrix $A$ with entries
$$ \alpha_{k,l} = \langle x^k | x^l \rangle $$
So $L$ is a linear isometry if ${\bf v_1}^T A {\bf v_2}$ is an inner product in $\mathbb{R}^3$, can you take it from here? (Note that $A = A^T$)