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
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
By contradiction, assume $\displaystyle \deg p < k$ then $\displaystyle p$ belongs to the subspace of the polynomials of degree less or equal to $\displaystyle k-1$, which is generated by: $\displaystyle 1,...,x^{k-1}$.
But $\displaystyle p$ is orthogonal to each of the elements of a set that generates it, hence it must be the $\displaystyle \bold{0}$ polynomial $\displaystyle (*)$! contradiction!
$\displaystyle
(*)$$\displaystyle p(x)=a_0\cdot 1 + a_1\cdot x^1+...+a_{k-1}\cdot x^{k-1}$ now : $\displaystyle <p,p>=a_0\cdot <1,p> + ...+a_{k-1}\cdot <x^{k-1},p>=0$ hence $\displaystyle p=\bold{0}$