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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?

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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.

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    $\begingroup$ Thank you! The examples you've shown made it click. So we should simply look at this huge sum as an descriptive tool we use to express other polynomials just like conventional vectors. So, $P_n(\mathbb{F})$ can express anything in $P_{n-1}(\mathbb{F})$ by simply placing a 0 (or 0's if necessary) as a coefficient(s) in the term having nth degree (or all terms having less than a degree of n). $\endgroup$
    – Jaigus
    Sep 27 '18 at 8:14
  • $\begingroup$ How about the statement: "$p_0, p_1, ..., p_m$ in $P_{m}(\mathbb F)$ such that $p_j(5) = 0$ for each j"? This refers to all the polynomials in the set that evaluate to 0 at 5, yet does the subscript "j" in this statement mean each polynomial of degree j that evaluates to 0 at 5 or just each of the polynomials that evaluate to 0 at 5? $\endgroup$
    – Jaigus
    Sep 27 '18 at 8:31
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    $\begingroup$ No @Jaigus, the $j$ is just a tool to refer to all the $m$ polynomials $p_0,\dots,p_m$ at the same time. Since the polynomials are elements of $P_m(\mathbb F)$, they have degree $m$. Your interpretation of the evaluation is correct. $\endgroup$
    – Babelfish
    Sep 27 '18 at 8:47
  • $\begingroup$ Ah ok. So its in the same sense as saying "for all v that belongs to $ \mathbb R^{n}$." "j" used above is just some arbitrary element of the polynomials with 5 as a root. $\endgroup$
    – Jaigus
    Sep 27 '18 at 17:13
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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 $x \in \mathbb Z / 2\mathbb Z$).

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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.

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