# Finding an orthonormal basis for the space $P_2$ with respect to a given inner product

I am so confused on what to do for this question.

The questions asks to find an orthonormal basis of $P_2$, the space of quadratic polynomials, with respect to the inner product $$\langle p, q\rangle = 2\int_{0}^{1} p(x)q(x)\, dx.$$ I don't know how to this the question with integrals. All I know is that it has something do with the Gram-Schmidt process.

Any help is appreciated thanks!

• I've TeX-ified your question and added a clarifying comment based on context. Readers will be grateful if you format your future posts. :) – Andrew D. Hwang Mar 12 '14 at 22:03
• For starters, do you understand how the Gram-Schmidt algorithm works in, say, $\mathbf{R}^3$? – Andrew D. Hwang Mar 12 '14 at 22:04

The standard method of orthogonalisation works like this:

1) take a basis $\{a_n\}$

2) take first vector $a_1$ , normalise it, call it the first vector in the new basis: $$b_1:=\frac{a_1}{\sqrt{\langle a_1,a_1\rangle}}.$$

3) take second vector $a_2$, find its component orthogonal to the first vector in the new basis: $$a'_2=a_2- \langle a_2,b_1\rangle b_1 ,$$ normalise it, then call it the second vector in the new basis: $$b_2:=\frac{a'_2}{\sqrt{\langle a'_2,a'_2\rangle}}.$$

4) repeat for all consecutive vectors - orthogonalisation comes for all previous vectors in new basis: $$a_k'=a_k- \sum_{j=1}^{k-1} \langle a_k,b_j\rangle b_j,\quad b_k= \frac{a_k'}{\sqrt{\langle a_k',a_k'\rangle}}.$$

So, let's take a canonical basis in $P_2[x]$: $\{1,x,x^2\}$.

First vector is $1$. It's norm with respect to our inner product is $2\int_0^11\cdot 1dx =2$, hence the first vector in the new basis is $b_1=\frac{1}{\sqrt 2}$.

Second vector is $x$. The orthogonalisation yields $$x- \langle x,1/\sqrt 2\rangle 1/\sqrt 2= x-\frac{1}{\sqrt 2}\cdot2\int_0^1\frac{x}{\sqrt 2}dx =x-\frac 12.$$

Can you now normalise it (i.e. find the second vector in the new basis) and then find the third vector in the new basis?

• for the second vector do i do the formula thats something like x2 X v1/v1 x v1? – user123204 Mar 12 '14 at 22:18
• just wondering did you forget to multiply the 2 in front of the integral to solve for the second vector? – user123204 Mar 12 '14 at 23:00
• @user123204 I edited the post for clarity. If you still have questions, feel free to ask. – TZakrevskiy Mar 13 '14 at 7:24