if $\{1, x, x^2, x^3, \dots, x^{n-1}\}$ is the base of a vector space, can I say that is $\bar{P} = [ P_1, P_2, \dots, P_m]$ a tensor? Where each $P_i$ is a combination of the monomial base. If it's, What is its dimension?
I'd like to thank in advance.
Ok, if you bound the degree of polynomials in an array by, say $d$, and since $$P_i=P_i(x)=a_{i0}+a_{i1}x+...+a_{id}x^d,$$ where the $a_{ij}$ are taken from a field, say $\mathbb R$, then the vector space of your $\tilde P$ objects, collected into a set, let's called $V$, will have $\dim V=(d+1)^m$.
For, if we denote by $\mathbb R_d[x]$ the abelian group of polynomials of degree $d$ at most, then $$V=\mathbb R_d[x]\times\cdots\times \mathbb R_d[x],$$ is the cartesian product of $m$ factors, with each factor having $\{1,x,x^2,...,x^d\}$ as basis.
Elements of a vector space are also dubbed rank one contravariant tensors.