Do rational functions form a vector space?

Question

Consider rational functions of the form $f(x) = \frac{p(x)}{q(x)}$ where $p$ is a polynomial of degree not exceeding $n$ and $q$ is a particular polynomial $q(x) = (x−1) \cdot (x−2) \cdot \ldots \cdot (x−m)$. Do these functions form a vector space? I think these function forms a vector space because they are closed under addition and scaling by a number. In case they do, what is the dimension of this vector space? Could you suggest a basis and write down an explicit formula for the coordinate functions?

Answer 1

Let $V:=\mathbb{R}_{\deg \leq n}[x]$ be the vector space of polynomials with degree $\leq n$.

Let $W$ the set of polynomials you describe. It is clearly a vector space.

In short: "Whether you divide your polynomials by $q(x)$ or not, is doesn't change anything".

More precisely, $$\tag{1}p(x) \in V \ \ \longleftrightarrow \ \ p(x)/q(x) \in W $$

is an isomorphism (immediate proof). In particular, $V$ and $W$ share the same dimension $(n+1)$.

Isomorphism (1) provides as well a bases correspondence between the canonical basis $\{1,x,x^2,\cdots x^n\}$ of $V$ and:

$$\left\{\frac{1}{q(x)}, \frac{x}{q(x)},\frac{x^2}{q(x)},\cdots \frac{x^n}{q(x)}\right\}$$

which is thus a basis for $W.$