# Thread: Inner product

1. ## Inner product

V is the space of polynomials with real coefficients with degree ≤ n.
The inner product is defined like this: <p,q> = the integral from -1 to 1 of pq(x)
I need to prove that if: p ≠ 0, <p,x^i> = 0, i < k then: deg p ≥ k

Thanks for any kind of help

2. By contradiction, assume $\deg p < k$ then $p$ belongs to the subspace of the polynomials of degree less or equal to $k-1$, which is generated by: $1,...,x^{k-1}$.
But $p$ is orthogonal to each of the elements of a set that generates it, hence it must be the $\bold{0}$ polynomial $(*)$! contradiction!

$
(*)$
$p(x)=a_0\cdot 1 + a_1\cdot x^1+...+a_{k-1}\cdot x^{k-1}$ now : $=a_0\cdot <1,p> + ...+a_{k-1}\cdot =0$ hence $p=\bold{0}$