# Find all polynomials in a vector space orthogonal to another polynomial with Gram Schmidt possibly

Here I am given the following.

Define an inner product in $$\mathcal{P}_2$$ by: $$\left=p(-1)q(-1)+p(0)q(0)+p(1)q(1).$$ a) Starting with the basis $$\left\{1,x,x^2\right\}$$ of $$\mathcal{P}_2$$, use the Gram-Schmidt process to find an orthonormal basis for $$\mathcal{P}_2$$ with respect to this inner product.

I already did this. The orthonormal basis is $$\left\{\frac{1}{\sqrt{3}},\frac{1}{\sqrt{2}}(x),\sqrt{\frac{3}{2}}\left(x^2-\frac{2}{3}\right)\right\}$$.

b) Find all the polynomials in $$\mathcal{P}_2$$ that are orthogonal to $$x^2-1$$.

I'm stuck here. I took the inner product of $$ax^2+bx+c$$ and $$x^2-1$$ which gives be that $$c=0$$ but how can I get what $$a$$ and $$b$$ are?

• You are done: the solutions are all $ax^2+bx$ with arbitrary numbers $a,b$. – Berci Dec 9 '20 at 6:30
• Thank you. I over thought this. I thought it was possible to find $a$ and $b$ somehow. – Future Math person Dec 9 '20 at 6:33
• It's a 3d vector space, so the orthogonals to one vector form a 2d subspace, meaning degree 2 of freedom among the coordinates $a,b,c$. – Berci Dec 9 '20 at 6:35
• That makes a lot of sense! Thank you! – Future Math person Dec 9 '20 at 6:40

You are done: simply verify that both $$x$$ and $$x^2$$ give zero inner product with $$x^2-1$$, and therefore so do any linear combination of them, that is, $$a$$ and $$b$$ can be arbitrary.
Also, the perpendicular of a nonzero vector in a 3d inner product space should be a 2d subspace, i.e. the degree of freedom of variables $$a,b,c$$ is $$2$$, which is clearly the case now.