Prove polynomial set $S=\{p_0(x), p_1(x), p_2(x), \cdots\}$ is linearly independent 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.
 A: 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)
A: Monic here means the coefficient on the highest order term of the polynomial is +1.
A) The first few members of S are $\{1, x+c, x^2+ex+f, ...\}$
Prove S restricted to polynomials of degree equal to or less than $n$ span all polynomials of degree less than or equal $n$. Prove that this implies linear independence for polynomials of order n+1 if one additional polynomial is added to S. Theorem follows by induction. 
Test linear independence by creating a matrix with each row composed of the coefficients of the polynomials. Non vanishing determinant implies independence.
B) it can be proven any $n$ linearly independent vectors span a vector space. Here you'll need infinite basis vectors unless the degree of polynomials is limited.
