What can we do with the isomorphism between $\mathbb{F}^{n+1}$ and $\mathbb{P}_n^\mathbb{F}$?

Question

A classic linear algebra example of isomorphism is that $\mathbb{P}_n^\mathbb{F}$, the set of polynomials of degree at most $n$ over a field $\mathbb{F}$, is isomorphism to $\mathbb{F}^{n+1}$. The idea of an isomorphism is that most (all?) relations and properties carry over from one to the other, and vice versa. But consider the aspect of solving an equality of two polynomials in $\mathbb{P}_n^\mathbb{F}$ (with set, predefined coefficients). I can't seem to think of any equivalent action in $\mathbb{F}^{n+1}$.

Is there an equivalent action there, or is there no analogy? If there is a relation, is it common to use geometric intuitions on $\mathbb{F}^{n+1}$ to make claims about solving relations with polynomials in $\mathbb{P}_n^\mathbb{F}$?

Answer 1

Polynomials have more structure than just that of a vector space. Namely, the space $\mathbb{P}$ of polynomials is not only a vector space, but rather a ring, or even more an algebra. This algebra has a normed ring structure, given by degree.

Solving a polynomial equation is more related to the multiplicative structure of $\mathbb{P}$, that is, finding monomials which divide a given polynomial.

So to answer your question: no, there is no good analogue of "solving a polynomial equation" in $\mathbb{F}^n$.

On the other hand, the isomorphism you mentioned is usually given by $a_0+a_1x+\cdots+a_nx^n\mapsto(a_0,\ldots,a_n)$, and the coefficients of a polynomial may give some information about how to find a root, e.g. with the rational root theorem. But this result is not usually proved based on the vector space structure of $\mathbb{F}^{n+1}$, so this is not a good analogy.

Answer 2

In that example it's all the vector space properties that "carry over". The isomorphism says, essentially, that two polynomials are the same just when their coefficients are the same.

In elementary linear algebra that example lets you discuss differentiation as a linear transformation. Evaluation at a point is a linear functional.

Other algebraic properties of polynomials - for example, multiplication, are not preserved by the vector space isomorphism.

The vector space of all polynomials (not just those with a fixed bounded degree) is also interesting. It's infinite dimensional, so often not part of the elementary linear algebra curriculum.

Answer 3

Here is an example of using linear algebra arguments to prove something about degree $\le n$ polynomials.

Let us consider $n+1$ distinct elements $x_1, x_2, \ldots, x_{n+1} \in \mathbb{F}$. Based on this choice, we can define a function $\mathbb{P}_n^{\mathbb{F}} \to \mathbb{F}^{n+1}$ by: $$T(p) := (p(x_1), p(x_2), \ldots, p(x_{n+1})).$$ Then we can check that $T$ is a linear transformation. Furthermore, $T$ has trivial kernel (or null space): namely, if $T(p) = 0$, then $p(x_1) = p(x_2) = \ldots = p(x_{n+1}) = 0$. By the assumption that $x_1, \ldots, x_{n+1}$ are distinct, then $p$ has $n+1$ distinct roots; but $p$ having degree $\le n$ also then implies that $p$ is the zero polynomial.

So, we now have a linear transformation $T : \mathbb{P}_n^{\mathbb{F}} \to \mathbb{F}^{n+1}$ which is one-to-one; and the domain $\mathbb{P}_n^{\mathbb{F}}$ and the codomain $\mathbb{F}^{n+1}$ have equal (finite) dimension $n+1$. Therefore, in fact $T$ is also onto. In other words, for any $y_1, y_2, \ldots, y_{n+1} \in \mathbb{F}$, there exists a polynomial $p$ of degree $\le n$ such that $p(x_1) = y_1, p(x_2) = y_2, \ldots, p(x_{n+1}) = y_{n+1}$.

(This fact is related to the general notion of polynomial interpolation. There are other ways to prove the same fact with an explicit construction, for example via the Leibniz interpolation formula.)