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.
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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?
Best Answer
Yes, everything is well and good. You followed the Gram-Schmidt process and got a right result. Well done!
Edit: for the orthogonal part. It remains to get integer coefficients (multiply by the pertinent square roots to do away with them) and to re-normalise to get norm-one vectors.