Finding orthonormal basis for inner product in P2(C) Let $p, q \in P_2(\Bbb C)$.  Define the inner product by
$$\langle p(x), q(x)\rangle = \int_0^1 p(x) \overline{q(x)} \, dx $$
How do I find orthonormal basis for inner product space?
 A: As mentioned in the comments, given a basis $\beta = \{v_1, \ldots, v_n\}$ of an inner product space $V$, applying the Gram-Schmidt orthogonalization process will produce an orthogonal basis $\gamma = \{u_1, \ldots, u_n\}$ for $V$.  We can compute the $u_i$ using the formula given in the link:
$$
u_k = v_k - \sum_{j=1}^{k-1} \frac{\langle u_j, v_k \rangle}{\langle u_j, u_j \rangle} u_j \, .
$$
So we begin with the standard basis $\beta = \{1, x, x^2\}$.  We can take $u_1 = 1$, so to find $u_2$ we need to compute
$$
u_2 = x - \frac{\langle 1, x \rangle}{\langle 1, 1 \rangle} 1 \, .
$$
Now
\begin{align*}
\langle 1, 1 \rangle &= \int_0^1 1 \, \overline{1} \ dx = 1&
&&\langle 1, x \rangle &= \int_0^1 1 \, \overline{x} \ dx = \left. \frac{x^2}{2} \right|_0^1 = \frac{1}{2}
\end{align*}
so
$$
u_2 = x - \frac{\langle 1, x \rangle}{\langle 1, 1 \rangle} 1 = x - \frac{1/2}{1} 1 = x - \frac{1}{2} \, .
$$
Can you use the formula to compute $u_3$?  Once you've done that, don't forget to normalize your vectors so they all have length $1$.
