# Thread: inner product find basis

1. ## inner product find basis

Let $C[0,\pi]$ be the complex inner product space of complex continuous functions on $[0,\pi]$, with inner product
$\langle f , g \rangle = \int_0^{\pi} f(x) \overline{g(x)} \, dx.$
Find an orthogonal basis for the subspace of $C[0,\pi]$ spanned by the functions ${1,2ix,x^3}$

I think I have to used Gram-Schmidt method but I am not sure where to start. Can you please show me the general method and I will then be able to solve it.
Many thanks

2. Originally Posted by charikaar
Let $C[0,\pi]$ be the complex inner product space of complex continuous functions on $[0,\pi]$, with inner product
$\langle f , g \rangle = \int_0^{\pi} f(x) \overline{g(x)} \, dx.$
Find an orthogonal basis for the subspace of $C[0,\pi]$ spanned by the functions ${1,2ix,x^3}$

I think I have to used Gram-Schmidt method but I am not sure where to start. Can you please show me the general method and I will then be able to solve it.
Many thanks

Apply EXACTLY Gram-Schmidt Method's proof...what's the problem?

For example, if $v_1=1\,,\,v_2=2ix\,,\,v_3=x^3$ , then

$u_i=\frac{v_1}{\|v_1\|}$ , where of course $\|v_1\|=\sqrt{}=\sqrt{\int\limits_0^\pi 1\cdot 1\,dx}=\sqrt{\pi}\Longrightarrow u_1=\frac{1}{\sqrt{\pi}}$

$w_2=v_1-u_1=2ix-\int\limits_0^\pi \frac{2ix}{\sqrt{\pi}}dx\cdot\frac{1}{\sqrt{\pi}}$ $=2ix-\frac{2i}{\pi}\cdot \frac{\pi^2}{2}=2ix-\pi i$ , and now $u_2=\frac{w_2}{\|w_2\|}$

Finally, evaluate $w_3=v_2-u_2-u_1$ and then $u_3=\frac{w_3}{\|w_3\|}$ , and then $u_1,u_2,u_3$ is your orthonormal basis.

Tonio