Polynomial Basis without determinant Suppose $s=<1,x,x^2>$ in the vector space P of polynomials in the variable $x$.
I am proving that $B=\left\{x^2,x,1\right\}$ is a basis of S. I have been told I am not allowed to use the determinant when solving.
I have started by supposing thre are scalars $\alpha, \beta, \gamma \in \mathbb{R}$ so that $\alpha x^2+\beta x+\gamma1=\vec 0$ Now $\alpha=\beta=\gamma=0$ for us to be linearly independent. I'm not sure how to show this though. 
 A: Your zero vector on the right side is just the zero polynomial. By definition of the zero polynomial all coefficients must be zero, e.g. Definition of Zero Polynomial 
A: If $x$  is a real number then you can derivate the equation twice and easily conclude that all the coefficients must be $0$
A: So, I guess this depends on the definition you are using. In some contexts, we just define algebraic symbols $(1=X^0,X^1,\ldots, X^n,\ldots)$ and declare a polynomial with coefficients in $R$ (usually some ring) to be something of the form $P(X)=\sum_{i=0}^n a_iX^i$, where the $a_i\in R$. In this case, it is true by definition that the $X^i, X^j$ are independent if and only if $i\ne j$. 
If we are in the circumstance where we are regarding $p(x)=\sum_{i=1}^n a_ix^i$ to be some function $p: \mathbf{R}\to \mathbf{R}$, with $a_i\in \mathbf{R}$ for all $i$, then showing linear independence of $\{1,x,x^2,\ldots,x^n\}$ can be done using different methods. For example, in the $\mathbf{R}$ case, suppose that for some $n$,
$$ x^n=\sum_{i=0}^{n-1}a_i x^i=p(x).$$
Then, it ought to be the case that for all $x\in \mathbf{R}$,
$$ \frac{x^n}{p(x)}=1. $$
However, passing to the limit, we observe that 
$$ \lim_{n\to \infty} \frac{x^n}{p(x)}=\infty$$
which implies that eventually, $x^n>p(x)$. This is a contradiction. So, $x^n$ can not be represented as such a linear combination.
