Let $\mathbb{F}$ be any field, and $P(\mathbb{F})$ the vector space of polynomials. A polynomial $p(x)\in P(\mathbb{F})$ is called a monic polynomial of degree $n$ if it is a polynomial of degree $n$, with leading term $x^n$. Let $S⊆P(F)$ be a subset of the form
$S=\{p_0(x),p_1(x),p_2(x),…\}$
where each $p_n(x)$ is a given monic polynomial of degree $n$.
a) Show that the set $S$ is linearly independent.
b) Show that the set $S$ spans $P(\mathbb{F})$.
(Thus, $S$ is a basis of $P(\mathbb{F})$.)
For this question, what does the monic polynomial of degree $n$ means, I want to prove $p_1(x)+p_2(x)+…+p_n(x)=0$, but I have no idea.
Best Answer
a) To check for linear independence, suppose that
$$\sum_{i=0}^∞ a_ip_i(x)=0$$
where all but finitely many ai are zero. 1 We claim that all ai are zero. If this were not the case, we could pick the largest m such that am = 0. Now compare the coefficient of $x^m$ on both sides. On the left hand side it is am (since pm(x) = $x^m$ + . . .), while on the right hnd side it is 0. Hence am = 0, a contradiction.
b) The check that the pi’s span P(F): We use induction to prove that all polynomials of degree ≤ n are linear combinations of $p_0$(x), . . . , $p_n$(x). This is immediate for n = 0 since polynomials of degree 0 are constant, p(x) = $a_0$, while $p_0$(x) = 1, so p(x) = $a_0$$p_0$(x). Suppose teh claim is known up to degree n, and let p(x) be a polynomial of degree n + 1. Then p(x) = $a_{n+1}$$x_n$ + . . . + $a_{1}x$ + $a_0$. But $p_{n+1}$(x) = $x_{n+1}$ + . . . + $b_1$x + $b_0$.
This shows that q(x) = p(x) − $a_{n+1}$ $p_{n+1}(x)$ is a polynomial of degree n, hence by induction is a linear combination of p0(x), . . . , pn(x). It follows that p(x) = $a_{n+1}$ $p_{n+1}$(x) + q(x) is a linear combination of $p_0$(x), . . . , $p_{n+1}$(x)