# Inner Product Proof - Verify on L2[-1,1]

1. Feb 5, 2015

### ElijahRockers

1. The problem statement, all variables and given/known data
This question has two parts, and I did the first part already I think.

If B = {u1, u2, ..., un} is a basis for V, and
$v = \sum_{i=1}^n a_i u_i$
and $w = \sum_{i=1}^n b_i u_i$

Show $<v,w> = \sum_{i=1}^n a_i b_i^* = b^{*T}a$

Here's how I did it:

$<v,w> = <\sum_{i=1}^n a_i u_i , w> = \sum_{i=1}^n a_i<u_i , w>$
$= \sum_{i=1}^n a_i b_i^* <u_i , u_i> = \sum_{i=1}^n a_i b_i^*$
Thus proved... however in class he mentioned $a_i = <v,u_i >$ for doing this but I'm not sure how... I've tried to examine it but I can't seem to justify it. And I think I did the proof without that, since <v,aw> = a*<v,w>

Second part of the question, where I'm confused, is, verbatim:

Verify $V = L^2 [-1,1]$, where B is the set of orthonormal Legendre polynomials,

$p_0 (x) = \frac{1}{\sqrt{2}}$
$p_1 (x) = \sqrt{\frac{3}{2}}x$
$p_2 (x) = \sqrt{\frac{5}{8}}(3x^2 -1)$

and v,w are replaced by $x-x^2$ and $12+x-3x^2$

2. Relevant equations

3. The attempt at a solution
Not really sure where to start... he mentioned a_i = <v, u_i > in class but I don't really feel comfortable with using that here because I don't understand how that's true. ( I feel like it's really simple, and that's why it's bothering me so much) If somebody could point me in the right direction as to why that expression is true, I could probably finish the question..

2. Feb 5, 2015

### Ray Vickson

Write $u(x) =x-x^2$ and $v(x) = 12 + x - 3x^3$ as constant-coefficient linear combinations of $p_0(x), p_1(x), p_2(x)$. You can do it using your instructor's hint, or you can do it for $u(x)$ the hard way, by getting three equations for $a_0, a_1,a_2$ from the identities $u(x) = a_0 p_0(x) + a_1 p_1(x) + a_2 p_2(x) \; \forall x$. Do the same type of thing for $v(x)$. Then, if you want to, you can verify explicitly that $a_i = \langle u,p_i \rangle$, etc.