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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?

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  • $\begingroup$ Do you know Gram-Schmidt orthogonalization? $\endgroup$ Dec 6, 2014 at 3:43
  • $\begingroup$ Yes, I do but I think I only know how to do it with specified vectors like {[1,2,3],[3,2,1],[2,2,1]} or something. $\endgroup$
    – User
    Dec 6, 2014 at 3:48
  • $\begingroup$ If you know a basis of an inner product space $V$, Gram-Schmidt gives you an orthogonal basis of $V$. So do you know of a basis for $P_2(\Bbb C)$? $\endgroup$ Dec 6, 2014 at 3:49
  • $\begingroup$ I know of the standard basis [1,x,$x^2$], so is it {[1,0,0],[0,x,0],[0,0,$x^2$]}? $\endgroup$
    – User
    Dec 6, 2014 at 3:53
  • $\begingroup$ No, you need to use the standard basis in the orthogonalization process. See my answer below. $\endgroup$ Dec 6, 2014 at 4:32

1 Answer 1

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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 $1$.

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