Find an orthogonal basis for $\mathbb P_2$

Question

The problem:

For polynomials $\mathbb{P_2}$ we define the inner product between p and q as:

$$ \langle p,q\rangle =p(t_0)q(t_0)+p(t_1)q(t_1)+p(t_2)q(t_2) $$

with $$t_0=0, t_1=1, \textrm{ and } t_2=2$$

Let three polynomials in $\mathbb{P_2}$ be defined as:

$$p(t)=1-t, q(t)=1+t, \text{ and } r(t)=t^2$$

1. Calculate $\,\langle p,q\rangle$ and the norm of p

$$\langle p,q\rangle =(1+0)*(1+0)+(1-1)*(1+1)+(1-2)*(1+2) = -2$$

The norm of $p$ is given by $\| p \|=\sqrt{\langle p,p\rangle}$ and is

$$\langle p,p\rangle = \sqrt{(1+0)*(1+0)+(1-1)*(1-1)+(1-2)*(1-2)}=\sqrt{2} $$

Part 2: Find an orthogonal basis for $\mathbb{P_2}$ by applying the Gram-Schmidt procedure to $p$,$q$ and $r$.

1. Can someone varify that the way of calculating the inner product and the norm of $p$ is correct?
2. Furthermore can you guys explain how to solve part 2 of the problem?

Thanks in advance.

Answer 1

Your computation of $\langle p,q\rangle$ and $\|p\|$ is correct.

To find an orthogonal basis, let the first element be $u_1=p$. Then compute \begin{align} u_2&=q-\operatorname{Proj}_{u_1}(q)\\ &= q - \frac{\langle u_1, q\rangle}{\langle u_1,u_1\rangle}u_1\\ &= 1+t - \left(\frac{-2}{2}\right)(1-t)\\ &= 2, \end{align} and further \begin{align} u_3&=r-\operatorname{Proj}_{u_1,u_2}(r)\\ &= t^2 - \left(\frac{\langle u_1,r\rangle}{\langle u_1,u_1\rangle}u_1 + \frac{\langle u_2,r\rangle}{\langle u_2,u_2\rangle}u_2 \right)\\ &= t^2 - \left(\left(\frac{-4}{2}\right)(1-t) + \left(\frac{10}{12}\right)2\right)\\ &= t^2 -\left(-\frac13+2t\right)\\ &= \frac13-2t+t^2. \end{align} Since $\langle u_i,u_j\rangle=0$ for $i\ne j$ and $\{u_1,u_2,u_3\}$ is linearly independent, this is an orthogonal basis for $\mathbb P_2$.