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.


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


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

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

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

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

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?


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.

  • $\begingroup$ By "missed one part of the question" do you mean that I didn't find an orthogonal basis separatetly from finding an orthonormal one? @KaviRamaMurthy $\endgroup$ – Mary Star Jun 14 at 23:29
  • $\begingroup$ Asked about orthogonality, not orthonormality. In all honesty, I missed the 'orthonormal' bit since I was doing the corresponding integrals by hand to check that it was, in effect, orthogonal. $\endgroup$ – Sam Skywalker Jun 14 at 23:29
  • $\begingroup$ @MaryStar There are two parts. One is to find an orthogonal set with integer coefficients and the other is to find an orthonorml basis. Did you do the first part? $\endgroup$ – Kavi Rama Murthy Jun 14 at 23:33
  • $\begingroup$ $\{1,2x-1,6x^{2}-6x+1\}$ is an orthogonal set with integer coefficients. $\endgroup$ – Kavi Rama Murthy Jun 14 at 23:34
  • $\begingroup$ The second part, about the orthonormal basis, is correct, isn't it? So, we cannot just say that the orthonormal basis is also orthogonal, we have to find an other one? @KaviRamaMurthy $\endgroup$ – Mary Star Jun 14 at 23:35

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