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. 

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 
