# Basis of second degree polynomials

• Oct 2nd 2012, 04:00 PM
christianwos
Basis of second degree polynomials
Find a basis for the set of quadratic polynomials p(x) that satisfy p(1)=0.
• Oct 3rd 2012, 12:46 AM
chiro
Re: Basis of second degree polynomials
Hey christianwos.

Is this an orthogonal basis in terms of a vector space of three two degree polynomials?

Also if this the above type of basis you need to specify the range of the polynomial if you wish to use an integral as the definition of the inner product.
• Oct 3rd 2012, 01:36 AM
johnsomeone
Re: Basis of second degree polynomials
Quote:

Originally Posted by christianwos
Find a basis for the set of quadratic polynomials p(x) that satisfy p(1)=0.

Basis implies it's a vector space (or module). But it's obviously not, since if f and -f are such quadratics, then, if it were a vector space, f+(-f) = 0 would again be such a quadratic - and it clearly is not. I assume what's intended is "Find a basis for the set of polynomials p of degree <= 2 such that p(1) = 0". Also, the base field/ring is unspecified. I'll assume it's just some arbitrary field F.

Have: "Let V = {p in F[x] | deg(p) <=2 and p(1) = 0 }. Then V is a vector space over F. Find a basis for V."

It's easy enough to show that V actually is a vector space over F.

Main observation is: V = {(ax+b)(x-1) in F[x] | a, b in F}, which includes all the degenerate cases (a=0, or a=b=0).
So have p in V iff p(x) = (ax+b)(x-1) = a(x)(x-1) + b(x-1) = a(x^2-x) + b(x-1) for some a, b in F.
Thus p in V iff p(x) = ar(x) + bs(x) for some a, b in F, where r(x) = x^2 - x and s(x) = x-1.
Therefore V = LS{ r, s }.

If a, b in F and ar(x) + bs(x) = 0 in F[x], then a(x^2-x) + b(x-1) = ax^2 + (b-a)x -b is the 0 polynomial in F[x],
which implies that a = b = 0.
Therefore r and s are linearly independent over F.

Therefore { r, s } is a basis for V.