# 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.

• should I write the zero polynomial as a linear combination of the polynomials in S.i have no idea Commented Oct 20, 2018 at 1:09
• Do you know the $p_x$? There aren't in general linearly independent. For example, ${x, x^2, and x^2 +2x}$ are not linearly independent. Commented Oct 20, 2018 at 1:28

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) = $$a_0p_0$$(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)

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.

• what should i plugging in to wronskian , P(x)=x2+αx+β ?? Commented Oct 20, 2018 at 1:48
• You'd put in the elements of S. Have they been specified? Possible options for spanning all polynomials are the Hermite polynomials, the Legendre Polynomials, the list goes on. You can't proceed unless you know S. Commented Oct 20, 2018 at 1:51
• I think I figured it out. S is given. Each p_n is some monic polynomial of degree n. Commented Oct 20, 2018 at 1:54
• To show linear independence, you want to ask if a monic polynomial, say $p_t(x)$, can be written as a linear combination of monic polynomials of distinct degrees not equal to $t$. Can you use only polynomials of degree $<$ t? If not, how do you cancel out the terms with exponent $>$ t? Commented Oct 20, 2018 at 2:32