# how to prove if something forms a vector Space

I have difficulty understand how to proof if something is vector space. I miss lecture as I was sick and the book is not help a lot. Here is the question

Let $P$ be the set of all polynomials. Show that $P$, together with the usual addition and scalar multiplication of functions, forms a vector space.

I understand that we should use the 8 axioms to help us but how I would go about using those if the question is asking very generally. Any help is good help

P.S Pardon my English

EDIT: The 8 axioms are:

1. $x + y = y + x$ for any $x$ and $y$ in $V$.
2. $(x + y) + z = x + (y + z)$ for any $x$, $y$, and $z$ in $V$.
3. There exists an element $0$ in $V$ such that $x + 0 = x$ for each $x \in V$.
4. For each $x \in V$, there exists an element $−x$ in $V$ such that $x + (−x) = 0.$
5. $α(x + y) = αx + αy$ for each scalar $α$ and any $x$ and $y$ in $V$.
6. $(α + β)x = αx + βx$ for any scalars $α$ and $β$ and any $x \in V$.
7. $(αβ)x = α(βx)$ for any scalars $α$ and $β$ and any $x \in V$.
8. $1 · x = x$ for all $x \in V$.
• What eight axioms are you working with? – Brian Fitzpatrick Feb 23 '15 at 4:04
• Well, first off, you need a vector addition, and a scalar multiplication (presumably your scalar field is the real numbers, but it could be something else). Note for two polynomials $p,q$ we define their sum by: $(p+q)(x) = p(x) + q(x)$ and the scalar multiplication by: $(cp)(x) = c\cdot p(x)$. The first order of business is to show this addition is associative, commutative, and has an additive identity, and every polynomial has an additive inverse. – David Wheeler Feb 23 '15 at 4:05

Recall that a vector space is a set $V$ with an operation $+$ and scalar multiplication over a field (think $\mathbb R$) that satisfies some axioms. So, to prove that $P$ with the usual addition and scalar multiplication is a vector space, you just need to prove that each axiom holds. For instance, there is an axiom of closure under addition, i.e. if $p_1, p_2 \in P$, then $p_1 + p_2 \in P$. We can prove this:
Suppose $p_1 = \sum_{i=0}^n a_ix^i$ and $p_2 = \sum_{i=0}^m b_ix^i$, and assume without loss of generality that $n \geq m$. Then we have $$p_1 + p_2 = \sum_{i=0}^m (a_i + b_i)x^i + \sum_{i=m+1}^n a_ix^i,$$ which is a polynomial, so $p_1 + p_2 \in P$.
• Presumably (judging by the phrasing of the question itself) these are polynomials considered as a function space that is $p:x \mapsto p(x)$ is a function $F \to F$, not the abstract ring $F[x]$ (formal $F$-linear combinations of non-negative powers of $x$). While for most computational purposes there is no real difference, in this case the associative and commutative properties of the vector sum are "inherited" from that of the field, since $p(x)$ is just a scalar (the image of the input scalar $x$). – David Wheeler Feb 23 '15 at 4:24