Prove all 8 axioms of a vector space? $\newcommand{\u}{{\bf u}} \newcommand{\v}{{\bf v}} \newcommand{\w}{{\bf w}} \newcommand{\V}{{\bf V}} \newcommand{\L}{{\bf L}}  \newcommand{\W}{{\bf W}} $ 
I suppose that this question has been asked before but:
Suppose that $\u,~ \v, ~ \w \in V$. where $\u,~ \v, ~ \w$ are vectors and $\V$ is a vector space
$$\u + \v \in \V \tag{Closure under addition}$$
$$\u + \v = \v + \u \tag{Commutative property}$$
$$\u + (\v+\w)=(\u+\v)+\w \tag{Associative property}$$
$\V$ has a zero vector $0$ such that for every $\u \in \V$, $\u+0=\u$. $\tag{Additive identity}$
For every $\u \in \V$, there is a vector in $\V$ denoted by $−\u$ such that $\u+(−\u)=0$. $\tag{Additive inverse}$
Now let's also assume that $c,d \in \mathbb R$
$$c\u \in V \tag{Closure under scalar multiplication}$$
$$c(\u+\v)=c\u+c\v \tag{Distributive property}$$
$$(c+d)\u=c\u+d\u \tag{Distributive property}$$
$$c(d\u)=(cd)\u \tag{Associative property}$$
$$1(\u)=\u \tag{Scalar identity}$$
I guess my questions are: 


*

*Is there like any property that says if one axiom fails, is it still a vector space?

*Also is a combination of any number of vectors in space $\V$ also a vector space?
a. Suppose $\W = \{a\v + b\w ~:~ a,b \in \mathbb R$} for some $\v, \w \in \V$. Is $\W$ a vector space?
If so explain, if not explain why not.
b. Now suppose $\L = \{a_1 \v_1 + a2 \v_2 + \dots + a_n \v_n ~:~ a_1, a_2, \dots a_n \in \mathbb R\}$. Is $\L$ a vector space?

*Is there a way to prove all 8 axioms without going through the process of tedious proofs? 
Thank you for the help. (Thank you everybody) ありがとう 皆さん
 A: *

*The eight axioms define what a vector space is. If $(V,+,.)$ fails in at least one of these axioms, it's not a vector space. If $(V,+,.)$ satisfy all the axioms, it's a vector space. You can see these axioms as what defines a vector space.

*a. Guess $W=\{ av+bw:a,b\in\mathbb{R}\}$ so that it's the set of combinations of $v,w\in V$ where $V$ is a vector space as I understood. $W$ is a vector space and you can prove it easly using what I wrote bellow in 3.
b. Same remark.

*You can prove that $(S,+,.)$ is a vector space (i.e., satisfies all the 8 axioms) in a much easier way if you notice that $S$ is a subset of a set $V$ such as $(V,+,.)$ is a vector space. For example, we prove using the 8 axioms that $(E,+,.)$ is a vector space (there are a lot of examples like $E=\mathbb{R}^n$). Now if you notice that $V\subset E$ then $(V,+,.)$ is a vector space if and only if:


*

*$V\neq\emptyset\,\,\,\,\,\,\,\,\,\,\,\,(1)$

*$\forall u,v\in V,\forall\alpha ,\beta\in\mathbb{R},\,\alpha.u+\beta.v\in V\,\,\,\,\,\,\,\,\,\,\,\,(2)$
A good way to prove that $V\neq\emptyset$ is to prove that $0_E\in V$ where $0_E$ is the neutral element in the abelian group $(E,+)$, because if $0_E\notin V$, $V$ isn't a vector space since $(2)$ isn't correct.
A: I didn't understand your first question, but I'll answer the other two.
2.
It's enough to prove that L is a vector space since W is a special case of L with n = 2
The first thing you should do is to check that for every $v,w \in L$ and every $\alpha \in \mathbb{R}$, $v+w \in L$ and $\alpha v \in L$. This is necessary because in a vector space you should be able to add two vectors or multiply by a scalar and get another vector in this vector space.
You also need to check that $L \ne \emptyset$.
Both of this things are very easy to prove and I'll leave it to you.
Notice that $L$ is a subset of $V$. That's why it is obvious that all the axioms will remain valid in $L$.
For example, say we want to show that for every $u,v,w \in L:  (u+v)+w = u+(v+w)$. we can think of $u,v,w$ as vectors in $V$ and than since $V$ is a vector space $(u+v)+w = u+(v+w)$, and you're done.
In a similar way you can show that all the other axioms remain true in $L$.
By the way,
L is usually called The span of the vectors $v_1, ... v_n$ and is denoted $sp\{v_1, ..., v_n\}$.
3.
In 2, we did have sort of a shortcut because $L$ was a subspace of $V$ and that made everything "easier to see", but most of the time there are no shortcuts and you have to check each and every axiom. the good news is that it's usually very easy, and not needed very often. You need to prove it only once for a certain space and than you have all the power of linear algebra in your hands.
