# Inner product - orthogonal basis

In my textbook there is the following example of how to find an orthonomal basis with Gram-Schmidt process.

Let P2, the vector space of all polynomials of degree two or less, with the inner product defined as: $$<p, q> = p(−1)q(−1) + p(0)q(0) + p(1)q(1)\, for\ p, q\in P_2 )$$

Find an orthonormal basis for P2.

The basis is given by {$1,X,X^2$}. Let $V$ be an inner product space with the basis $V = (v_1,v_2,...v_n)$ and let $p_k$ for $k = 1,2,3,...n-1$ is the orthogonal projection of $v_{k+1}$ on the subspace $Span(v_1,v_2,...v_k)$. Then we have that $W = (v_1,v_2-p_1,v_3-p_2,...,v_n-p_{n-1})$ is the orthogonal basis for V.

$u_1 = \frac {1}{{\left\|v_1\right\|}}v_1 = \frac {1}{\sqrt3}\cdot 1$

It is now possible to find $p_1 = <x,\frac {1}{\sqrt3}>\cdot \frac{1}{\sqrt3}=0$...

And then it continues to calculate the remaning $u_k$'s and $p_k$'s. However, my question is just about the computation itself.

So for $u_1$ the norm of $v_1$ is calculated by $\sqrt{<1,1>} = \sqrt{1\cdot1 + 1\cdot1 + 1\cdot1} = \sqrt3$ and if I had to the it for $X$ then it would be $\sqrt{<X,X>} = \sqrt{1+0+1} = \sqrt2$ right? What I do not understand is, how $v_1$ is simply 1 and for $p_1$ how does $<x,\frac{1}{\sqrt3}> \cdot \frac{1}{\sqrt3}$ = 0 ?

Hope someone can clarify

I don't understand your first question. I'd say that $v_1$ is simply $1$ because whoever created that exercise decide to start with the basis $(v_1,v_2,v_3)=(1,X,X^2)$.
And $\left\langle X,\frac1{\sqrt3}\right\rangle=(-1)\times\frac1{\sqrt3}+0\times\frac1{\sqrt3}+1\times\frac1{\sqrt3}=0$.
• That makes sense, thank you! For the first question I was just confused why it was $1$ as it is done with the basis you described. – Mads Jeppesen May 19 '18 at 14:24