Vector space when the operations are non-standard I am taking a linear algebra course and one of the things they have shown us is how to determine that V is a vector space according to eight axioms. When the operations for o-plus and o-dot are standard the axioms make sense however I am having trouble understanding how to show V is a vector space if the operations are non-standard. The issue is a lot of examples I have looked at state 'it is easy to show that axiom [X] is true' but clearly I missing something because it isn't feeling that easy.
Here is the example I have.
Let V be the set of vectors in $\mathbb{R}^2$ with the following operations for addition and scalar.
$$\begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix}\bigoplus \begin{bmatrix}
y_{1}\\ 
y_{2}
\end{bmatrix}=\begin{bmatrix}
0\\ 
x_{2}+y_{2}
\end{bmatrix}$$
$$\alpha \bigodot \begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} = \begin{bmatrix}
\alpha x_{1}\\ 
\alpha x_{2}
\end{bmatrix}$$
Axiom 1 is to show that x + y = y + x so
$$\begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} + \begin{bmatrix}
y_{1}\\ 
y_{2}
\end{bmatrix} = \begin{bmatrix}
0\\ 
x_{2} + y_{2}
\end{bmatrix} = 
\begin{bmatrix}
y_{1}\\ 
y_{2}
\end{bmatrix} + \begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix}$$
Axiom 1 seems to hold true and I cannot think of two vectors in the space that wouldn't work.
Axiom 2 is to show that (x + y) + z = x + (y + z) so
$$\left (  \begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} +
\begin{bmatrix}
y_{1}\\ 
y_{2}
\end{bmatrix}
\right ) + \begin{bmatrix}
z_{1}\\ 
z_{2}
\end{bmatrix} = 
\begin{bmatrix}
0\\
x_{1} + y_{1}
\end{bmatrix} +
\begin{bmatrix}
z_{1}\\
z_{2}
\end{bmatrix} =
\begin{bmatrix}
0\\ 
x_{2} + y_{2} + z_{2}
\end{bmatrix}$$
$$ \begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} + \left ( \begin{bmatrix}
y_{1}\\ 
y_{2}
\end{bmatrix} + \begin{bmatrix}
z_{1}\\ 
z_{2}
\end{bmatrix} \right ) =
\begin{bmatrix}
x_{1} \\
x_{2}
\end{bmatrix} + 
\begin{bmatrix}
0\\
y_{2} + z_{2}
\end{bmatrix} =
\begin{bmatrix}
0\\
x_{2} + y_{2} + z_{2}
\end{bmatrix}$$
Axiom 2 is similar to Axiom 1 for the o-plus operation so it seems to hold true.
Axiom 3 is to show there exists a $0_{v}$ such that
$$\begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} + 0_{v} = \begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix}$$
I don't think Axiom 3 could be true because
$$\begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} + \begin{bmatrix}
0\\ 
0
\end{bmatrix} = \begin{bmatrix}
0\\ 
x_{2}
\end{bmatrix}
\neq \begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix}$$
as long as $x_{1} \neq 0$.
Axiom 4 is to show that the additive inverse results in the $0_{v}$ therefore
$$\begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} + \begin{bmatrix}
-x_{1}\\ 
-x_{2}
\end{bmatrix} = \begin{bmatrix}
0\\ 
0
\end{bmatrix}$$
this will always be true because $x_{1}$ plus any value will always be $0$ which means Axiom 4 holds.
Axiom 5 I think holds because
$$\alpha \left ( \begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} + \begin{bmatrix}
y_{1}\\ 
y_{2}
\end{bmatrix} \right ) = \begin{bmatrix}
0\\ 
\alpha(x_{2} + y_{2})
\end{bmatrix}=\begin{bmatrix}
0\\ 
\alpha x_{2} + \alpha y_{2})
\end{bmatrix} = \alpha \begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} + \alpha \begin{bmatrix}
y_{1}\\ 
y_{2}
\end{bmatrix}$$
Axiom 6 I don't think holds because
$$(1 + 2) \bigodot \begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} = \begin{bmatrix}
3 x_{1}\\ 
3x_{2}
\end{bmatrix} \neq \begin{bmatrix}
0\\ 
3x_{2}
\end{bmatrix} = 1\bigodot\begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix} + 2\bigodot\begin{bmatrix}
x_{1}\\ 
x_{2}
\end{bmatrix}$$
I think Axioms 7 and 8 hold because the definition of o-dot didn't change and the definition of o-plus is not used.
So I conclude that Axioms 3 and 6 do not hold and that would mean that V equipped with these operations is not a vector space of $\mathbb{R}^2$.
I need some help because this is marked incorrect however I don't know what I am missing. Any insights would be greatly appreciated.
 A: You don't say just what it is that's marked incorrect. I think your answer shows that you understand the underlying ideas, but your exposition has many flaws. I have to infer that you understand - the words don't tell me directly.
Here are some ways you could improve your answer.
Your argument for Axiom 1 begins by stating that axiom - but then you say "therefore". That word says you are going to use the axiom to prove something, not that you are going to check the axiom in this context. Instead of "therefore" you want something like "Consider two vectors $x$ and $y$ ..." - followed by a general argument using algebra, not just one case with numbers.
Your argument for the failure of Axiom 3 is on the right track. You show that $[0 0]$ can't work as the zero vector. But what you should show is that nothing can possibly work as the zero vector.
Once you have shown there is no zero vector you know you don't have a vector space. You have answered the question. But if you insist on going on to check the other axioms, you should skip 4. If there's no zero vector then it makes no sense to ask about negatives.
You get 5 right, with the kind of argument you should have used for 1 and 2.
