Orthonormal basis of a vector space with polynomials of the form $ax+b$

Question

Find the Orthonormal basis of vector space $V$ of the linear polynomials of the form $ax+b$ such that $\:$, $p:[0,1] \to \mathbb{R}$. with inner product

$$\langle p,q \rangle= \int_0^1 p(x)q(x) dx$$ for $q, p \in V$

The polynomial in $V$ are linear so the basis of $V$ are $\left \{1,x \right \}$ further i am not getting how to proceedes further

please help

Thankyou.

Answer 1

An orthonormal base can be

$\{1,\frac{(x-\frac{1}{2})}{\sqrt{\langle x-\frac{1}{2},x-\frac{1}{2}\rangle}}\}$

The idea is simple. The space has dimension $2$, so you must choose two polynomials of order at most $1$. The first polynomial can be $1$, that has norm $1$. Then the second polynomial $p$ must verifies the condition

$\langle 1,p\rangle =\int_0^1pdx=0$

Moreover $p$ must be unitary so you can simply consider

$q=\frac{p}{\sqrt{\langle p,p\rangle}}$

So

$\{1,q\}$ is an orthonormal base for your space.

How can you calcolate $p$? You can observe that

$\int_0^1 xdx=\frac{1}{2}$ so

$\int_0^1 xdx-\frac{1}{2}=\int_0^1( x-\frac{1}{2})dx=0$

Then you can choose

$p:=x-\frac{1}{2}$

You can observe that, fixed a polynomial $s$, you can always choose $p=s-\int_0^1sdx$

because

$\langle 1,p\rangle =\int_0^1 p dx=0$

This is not a mysterious result because it follows the same idea of the Gram-Schmidt orthogonalization

Answer 2

With Gram-Schmidt

$v_{1} = 1$, $v_{2} = \frac{x - <x,v_{1}>v_{1}}{ \| x - <x,v_{1}> v_{1}\|} $

So $<x,1> = \int_{[0,1]}xdx = \frac{1}{2}$

And $\| x - <x,1> \|^{2} = \int_{[0,1]} (x - \frac{1}{2})^{2} = \frac{1}{12}$

Then $v_{2} = \frac{x - \frac{1}{2}}{\sqrt{12}}$

$\{v_{1},v_{2} \}$ is a orthonormal basis

Answer 3

A standard approach for these types of problems is to use the Gram-Schmidt procedure:

---

You already found a basis of $V$, so let $\mathbf{v}_1=1$ and $\mathbf{v}_2=x$. Then an orthogonal basis $\{\mathbf{u}_1,\mathbf{u}_2\}$ is given by: $$\mathbf{u}_1=\mathbf{v}_1$$ $$\mathbf{u}_2=\mathbf{v}_2-\frac{\langle \mathbf{v}_2,\mathbf{u}_1\rangle}{\langle \mathbf{u}_1,\mathbf{u}_1\rangle}\mathbf{u}_1$$ So in your case, all you have to do is compute two simple integrals. An orthonormal basis $\{\mathbf{w}_1,\mathbf{w}_2\}$ can be found by normalizing the two vectors $\mathbf{u}_1$ and $\mathbf{u}_2$. Namely: $$\mathbf{w}_1=\frac{\mathbf{u}_1}{\|\mathbf{u}_1\|}=\frac{\mathbf{u}_1}{\sqrt{\langle \mathbf{u}_1,\mathbf{u}_1\rangle}}$$ $$\mathbf{w}_2=\frac{\mathbf{u}_2}{\|\mathbf{u}_2\|}=\frac{\mathbf{u}_2}{\sqrt{\langle \mathbf{u}_2,\mathbf{u}_2\rangle}}$$ This requires the computation of one additional (easy) integral.