User:IssaRice/Linear algebra/Dual space of vector space of polynomials

Given a vector space $V$ , its dual space $V^{'}$ is the set of all linear functionals $V \to F$ , i.e. all linear functions that "evaluate" each vector into a constant.

For the space $P (R)$ of single-variable polynomials on $R$ , an element $T$ of $(P (R))^{'}$ takes a polynomial $p$ and returns a real number $T (p)$ . What does this space look like?

If $p$ is defined by $p (x) = a_{n} x^{n} + \dots + a_{1} x + a_{0}$ , we want to say something like that T "acts linearly" on p. except... i don't think axler defines what a basis is for an infinite-dimensional space. want to say that $(1, i d, {i d}^{2}, \dots)$ (using functional notation here for type-pedantry, where $i d = (x \mapsto x)$ ) is a basis for $P (R)$ .

so given $p = a_{n} {i d}^{n} + \dots + a_{1} i d + a_{0}$ , we have $T (p) = a_{n} T ({i d}^{n}) + \dots + a_{1} T (i d) + a_{0} T (1)$ .

now the interesting thing is that even though each polynomial must have finite length (because not all infinite sums converge), each machine eating up a polynomial can have infinite length -- it can get away with this because each input is finite. mentally, when i think about this situation, it's sort of like the inputs have chickened out of being infinite, so the linear functionals can "afford to" stay infinite.