Linear independence for polynomials. I see this in my textbook for linear algebra and the topic is linear independence:

Why is S a basis? I can't understand why it's linearly independent? Are all the coefficients 0 when we set it equal to the 0 vector? All the x's can be 0 too and the coefficients may not be right?
Is all the keyword here?
 A: With the unfortunate formulation in the quoted text, the given polynomials may indeed fail to be independent: Instead of polynomials (formal(!) sums of scalar multiples of powers of an unknown), they seem to consider polynomial functions (as suggested by their formulation "the polynomial $\mathbf 0(x)=0$ for all $x$"). If we consider only the field with two elements $0$, $1$ instead of the field of real (or complex, or rational, ...) numbers, it turns out that $x^2$ is the same polynomial function as $x$ (because $0^2=0$ and $1^2=1$ is all there is to be checked), hence over that field, $x$ and $x²$ are not linearly independent as polynomial functions.
The linear independence as polynomials, on the other hand, is immediate from the very definition as formal sums of scalar multiples of powers of an unknown: Two such formal sums are equal - by definition - if and only if all corresponding coefficients are equal, in particular, a polynomial is the zero polynomial if and only if all coefficients are zero.
Nevertheless, the above fineprint is moot (and may be inspiring for some, but confusing for others). Either way, the zero vector n the vector space of polynomials is the zero polynomial, which is the polynomial with all coefficients zero. It is easy to see that a (real or complex) polynomial with not all coefficients zero is not the zero function:
If $f(x)=a_0+a_1x+\ldots +a_nx^n$ is any non-zero polynomial, i.e., at least one coefficient (without loss of generality, $a_n$) is non-zero, then for all real $x>n\max\{|\frac {a_0}{a_n}|, \ldots,|\frac {a_n}{a_n}|\} $, we have for $0\le i<n$, $$|a_ix^i|= |\frac{a_0}{a_n}||a_nx^{i}|<\frac1n|a_nx^{i+1}|\le\frac1n |a_nx^n|$$ and hence
$$|f(x)|=|a_0+\ldots +a_nx^n|\ge |a_nx^n|-|a_0|-|a_1x|-\ldots -|a_{n-1}x^{n-1}|>0 $$
A: When dealing with polynomials as abstract entities, we don't evaluate $x$ but handle it too as a separate entity. 
A linear combination uses scalar coefficients, so that every linear combinations of $1,x,x^2, \dots, x^n$ is a polynomial of degree $\le n$. 
In particular, $x^{n+1}$ is linearly independent from them. 
