Orthogonal and orthonormal basis in the vector space of polynomials

Find an orthogonal basis with integer coefficients in the vector space of polynomials $$f(t)$$ of degree at most $$2$$ over $$\mathbb{R}$$ with inner product $$\langle f, g\rangle=\int_0^1f(t)g(t)\, dt$$.

In addition, find an orthonormal basis for the above space.



I have done the following:

Let $$S=\{1,x,x^2\}$$.

We normalize the first vector of the basis.

$$v_1=1$$

$$\langle v_1, v_1\rangle=\langle 1, 1\rangle=1$$

$$\tilde{v}_1=\frac{v_1}{\|v_1\|}=1$$

$$w_2:=v_2-\langle v_2,v_1\rangle v_1=x-\langle x,1\rangle 1=x-\frac{1}{2}$$

$$v_2=\frac{w_2}{\|w_2\|}=2\sqrt{3}\left (x-\frac{1}{2}\right )$$

$$w_3:=v_3-\langle v_3,v_1\rangle v_1-\langle v_3,v_2\rangle v_2=x^2-\langle x^2,1\rangle 1-\langle x^2,2\sqrt{3}x-\sqrt{3}\rangle \left (2\sqrt{3}x-\sqrt{3}\right )=\ldots =x^2-x+\frac{1}{6}$$

$$v_3=\frac{w_3}{\|w_3\|}=6\sqrt{5}\left (x^2-x+\frac{1}{6}\right )$$

Therefore an orthonormal basis is $$\left \{1 \ , \ 2\sqrt{3}x-\sqrt{3} \ , \ 6\sqrt{5}\left (x^2-x+\frac{1}{6}\right )\right \}$$ that is also orthogonal.

Is everything correct?

• $\{1,2x-1,6x^{2}-6x+1\}$ is an orthogonal set with integer coefficients. – Kavi Rama Murthy Jun 14 at 23:34