What exactly is in the polynomial vector space?

Question

I am having some trouble understanding the polynomial vector space notation: $P_{m}(\mathbb F)$ which means the set of all polynomials that take inputs from the space $\mathbb F$ and output to the space $\mathbb F$ with coefficients in $\mathbb F$ if there exits $a_{0}, ...,a_{m} \in \mathbb F$ such that $p(z)=a_{0}+a_{1}z+a_{2}z^{2}+...+a_{m}z^{m}$

What does it really mean "if there exist $a_{0}, ...,a_{m} \in \mathbb F $? Can't we always include some arbitrary coefficient values to a polynomial?

And, since $p(z)=a_{0}+a_{1}z+a_{2}z^{2}+...+a_{m}z^{m}$ is just one huge sum, is it just one huge polynomial? I'm also confused as to whether or not it includes just the polynomials (i.e. x^2) or the polynomials evaluated at all possible values as well? (i.e. $(0)^2, (1)^2, (2)^2, etc.$)

And, following polynomials of this form, it doesn't seem to inlclude something like "$x+5$", as it seems to refer to functions in the form of $x^0, x^1,...x^n$".

I'm new to linear algebra so please bear with me. Can someone please explain this notation and the contents of this set?

Answer 1

There are a lot of questions and potential confusions here, I will try to clarify some of them with the use of examples.

$P_1(\mathbb{R})$ is the set of all linear functions with real coefficients, i.e. $ax^1+b x^0$ where $a$ and $b$ are real. The polynomial $x+5$ is an element of this space because $x+5= 1x^1+ 5 x^0$.

On the other hand, $P_2(\mathbb{C})$ is the set of all quadratic polynomials with complex coefficients. An example element in this space is the polynomial $3x^2+(2+i)x+3$. Another example is $3x^2+0x+1$, which is usually written as $3x^2+1$. This example is actually also an element of $P_2(\mathbb{R})$ too, because all the coefficients are real.

Answer 2

A polynomial with coefficients in the field $F$ in the variable $X$ is an expression of the sort $$ P(X)=a_o+a_1X+a_2X^2+\cdots+a_dX^d $$ where the coefficients $a_i\in F$. Polynomials can be added and multiplied in the usual way to give other polynomials as result. A polynomial can also be multiplied by a scalar (i.e. an element of $F$) and a result is a polynomial.

(In fact there's a more formal definition of polynomials, but let's ignore it for now)

There's a notion of degree of a polynomial. The polynomial $P(X)$ written above has degree $\deg(P(X))=d$ provided $a_d\neq0$. The degree of a polynomial is well-behaved under the operations described above, namely $$ \deg(P+Q)\leq\max(\deg(P),\deg(Q)),\ \deg(PQ)=\deg(P)+\deg(Q),\ \deg(aP)=\deg(P) $$ as long as $a\neq0$. The first and last such relations for the degree imply that if we let $$ P_m(X)=\{\text{polynomials $P(X)$ such that $\deg(P)\leq m$}\} $$ then $P_m(X)$ is closed under addition and multiplication by scalars. A straightforward exercise shows then that $P_m(X)$ is a linear space.

Answer 3

You should not interprete polynomials over a field $F$ as a map/ function from $F$ to $F$.

A polynomial is a formal linear combination over $F$ of the monomials $x^i$. So, first of all, all monomials are also polynomials. Examples are $x^2, x^3, x$ or $0$. Linear combinations of those monomials are finite sums of such monomials, where you are allowed to multiply each monomial with a coefficient $a_i \in F$. For $F=\mathbb R$, examples are $x^2, 3x^2-5, x^3-7x$ or $x^6+x^2-4$. One important point is that you do not allow infinite sums of monomials.

You can add polynomials (as it is neccesary for a vector space). If you have polynomials $p(x)=a_0 + a_1 x + a_2 x^2 + \dots a_n x^n$ and $q(x) = b_0 + b_1x + \dots +b_mx^m$ (so all $a_i$ and $b_i$ are elements of $F$), you can add those two to obtain $$(p+q)(x) = \sum_{i=0}^{\max(m,n)} (a_i+b_i)x^i$$ where we set all $a_i$ and $b_i$ not already defined to $0$.

You can also multiply $p$ with a scalar $\alpha \in F$: $$(\alpha p)(x) = \sum_{i=1}^n (\alpha a_i)x^i$$

So why is a polynomial not a map?

Given a polynomial $p(x)\in F[x]$, we can define a map $\sigma_p\colon F \to F, a \mapsto p(a)$. If you think of the polynomial $x^2$ as a parabola, you consider this map. But this map is not the same as $p(x)$. For example, take $F=\mathbb Z / 2\mathbb Z$, the field with two elements. Then the polynomials $p(x) = x^2$ and $q(x) = x$ are different (since they have different coefficients), but the evaluation maps $\sigma_p\colon \mathbb Z / 2\mathbb Z \to \mathbb Z / 2\mathbb Z, x\mapsto x^2$ and $\sigma_q \colon \mathbb Z / 2\mathbb Z \to \mathbb Z / 2\mathbb Z, x\mapsto x$ are the same (since $x^2=x$ for all $x \in \mathbb Z / 2\mathbb Z$).