# Prove $\{1, x, x^2, …, x^n\}$ is a linearly independent set in $P_n(F)$

Where the vector space $P_n(F)$ is the set of all polynomials with coefficients from a field F (we'll say $R$ or any infinite field since the book's definition seems vague). A polynomial has coefficients from $F$, $n$ is the degree, a non-negative integer, and any polynomial function $f$ maps $F$ into $F$.

I chose to go by induction:

$a_0(x_0) = a_0(1)$
$a_0 = 0$ --> is linearly independent, trivially

$a_0(x_0) + a_1(x_1) = 0$
$a_0(1) + a_1(x) = 0$
$a_0 + a_1(x) = 0$
Pick $x$ not equal to $-a_0 / a_1$, then
$a_0 + a_1(x) = 0$ only if $a_0$ & $a_1 = 0$

Assume $\{1, x, ..., x^{n - 1}\}$ is linearly independent.
For case of set $\{1, x, ..., x^{n}\}$

Pick $x$ sufficiently large such that $|x * a_n| >> n$ & $|x * a_n| >> a_i$, the largest constant in the equation:
$a_0 + a_1(x) + ... + a_{n-1}(x^{n-1})$

Then:
$|an(x^n)| > |x^{n - 1} + x^{n - 1} + .... + x^{n - 1}| = |n * x^{n - 1}|$

From the above fact, $x_n = x^n$ cannot be constructed by a linear combination of the other vectors in the set, so the only solution to the equation is the one where all the coefficients of the vectors are 0. Hence, $\{1, x, ..., x^n \}$ is a linearly independent set for all n in Pn(F)

• Formatting tips here. – Em. Aug 5 '16 at 10:12
• Are u asking if you are correct? You are. But you are not using induction on $n$..... BTW click on Help, click on "How Do I Format Mathematics Here?" on the Help page, then click on the MathJax Quick Reference. – DanielWainfleet Aug 5 '16 at 10:16
• What's $F$?${}$ – user228113 Aug 5 '16 at 10:16
• 1. Sorry about the formatting, will edit later. 2. F is a field. 3 Yes, was wondering if the proof was sound – Rdesmond Aug 5 '16 at 10:19
• Is your $P_n(\mathbb F)$ a vector space of formal polynomials over $\mathbb F$ or a vector space of polynomial functions $\mathbb F\to\mathbb F$? Those are (effectively) the same when $\mathbb F$ is $\mathbb R$ or $\mathbb C$, but not over general fields! For example, over $\mathbb F^5$, the different formal polynomials $x^5$ and $x^1$ both correspond to the same polynomial function. – Henning Makholm Aug 5 '16 at 10:28