Let $V = \mathbb{R}^2$, show that $V$ is a vector space I am new to the concepts of vector spaces (as far as I remember), and I have some difficulties in understanding how can I show, in practice, if a set is a vector space or not.

I have an exercise in my last linear algebra assignment, which is the following:

Let $V = \mathbb{R}^2$, with addition $\oplus$ defined by:
$$\left(\begin{matrix} x \\ y \end{matrix}\right)
 \oplus\left(\begin{matrix} w \\ z \end{matrix}\right) =
 \left(\begin{matrix} x + w -3\\ y + z + 1 \end{matrix}\right)$$
and scalar multiplication $\odot$ defined by:
$$\alpha \odot \left(\begin{matrix} x \\ y \end{matrix}\right) =
 \left(\begin{matrix} \alpha x - 3\alpha +3\\ \alpha y + \alpha - 1
 \end{matrix}\right)$$
Show that $V$ is a vector space by showing that:
a) $V$ is closed with respect to $\oplus$ and $\odot$ and  i.e.
  
  
*
  
*$\forall$ v, w $\in V \rightarrow$ v $\oplus$ w $\in V$.
  
*$\forall \alpha \in \mathbb{R}$, $\forall$ v $\in V \rightarrow \alpha \odot$ v $\in V$.
b) All $8$ vector spaces axioms hold. The vector $\left(\begin{matrix}
 3 \\ -1 \end{matrix}\right)$ acts as a zero vector.


This can be all interesting if you have at least  an idea on how to do it, and maybe you can do it, but this is not exactly my case.
I know that to show a set is a vector space I have to show it's closed under addition and under scalar multiplication (nice!). I know, from my notes, what these axioms are.
But in this case, how can I show that:
$$\left(\begin{matrix} x \\ y \end{matrix}\right) \oplus\left(\begin{matrix} w \\ z \end{matrix}\right) = \left(\begin{matrix} x + w -3\\ y + z + 1 \end{matrix}\right)$$
is closed under addition, if we don't know the values of $x$, etc?
The same for all other options (show that is closed under scalar multiplication, and show that the axioms hold).
In general, how would you prove that a set like this is a vector space? Can I see an example? I am really not understanding what I have to prove, and the examples I have tried to see for me are like Chinese.
 A: Although you may not know what the values of your vector is, you do know what vectors of the set look like. In this case $\mathbb{R}^2 = \{ \begin{pmatrix} a \\ b \end{pmatrix} \colon b, a \in \mathbb{R} \}$. 
Given $\begin{pmatrix} x \\ y \end{pmatrix}$ and $\begin{pmatrix} z \\ w \end{pmatrix}$, is the vector $\begin{pmatrix} x + w - 3 \\ y + z + 1 \end{pmatrix}$ in $\mathbb{R}^2$ ?
A: To prove that $\mathbb{R}^2$ is closed under $\oplus$, you need to prove that ${x+w-3\choose y+z+1}\in \mathbb{R}^2$, regardless of the values of $x,w,y,z$.  Fortunately, this is easy, since $\mathbb{R}$ is itself closed under addition, so the sum of the three real numbers $x,w,-3$ is again a real number.  Similarly, the sum of the three real numbers $y,z,1$ is again a real number.  Thus ${x+w-3\choose y+z+1}$ is an ordered pair of real numbers, which is the definition of $\mathbb{R}^2$.
Proving closure under scalar multiplication is similar.
The hard parts are the vector space axioms; you should try them yourself rather than having people work through them all for you.
A: In the equation $$\left(\begin{matrix} x \\ y \end{matrix}\right) \oplus\left(\begin{matrix} w \\ z \end{matrix}\right) = \left(\begin{matrix} x + w -3\\ y + z + 1 \end{matrix}\right)$$
the right side is a vector in $\mathbb{R}^2$ and so it is closed under your specific type of addition. This is just regular addition with the specific modified zero vector added in as well.
The same goes for your modified version of scalar multiplication. $\alpha\odot\mathbf{v}=\alpha\mathbf{v}+(\alpha-1)\mathbf{v_0}$ where $\mathbf{v}_0$ is your modified zero vector. This still results in a vector in your space.
Checking the other 8 requirements should be fairly straightforward once you get past the notation.
A: For the first axiom, you have to prove that $$ \left(\begin{matrix} x + w -3\\ y + z + 1 \end{matrix}\right) \in V = \mathbb{R}^2$$ which is true for $x,y,z \in \mathbb{R}$, since you original vectors were in $\mathbb{R}^2$ you have $x,y,z \in \mathbb{R}$ and the result follows.
Similarly for the clousure with respect to scalar product, $$\alpha \odot \left(\begin{matrix} x \\ y \end{matrix}\right) = \left(\begin{matrix} \alpha x - 3\alpha +3\\ \alpha y + \alpha - 1 \end{matrix}\right) \in \mathbb{R}^2$$ if $\alpha, x, y \in \mathbb{R}$, which is true for the same argument above.
The rest are a little easier because you can use the propierties of $\mathbb{R}$ after you reach a certain place, like for example commutativity:
$$\left(\begin{matrix} x \\ y \end{matrix}\right) \oplus\left(\begin{matrix} w \\ z \end{matrix}\right) = \left(\begin{matrix} x + w -3\\ y + z + 1 \end{matrix}\right) = \left(\begin{matrix} w + x -3\\ z + y + 1 \end{matrix}\right) = \left(\begin{matrix} w \\ z \end{matrix}\right) \oplus\left(\begin{matrix} x \\ y \end{matrix}\right) $$
Note that in the middle equality we used the conmutativity of $\mathbb{R}$ to interchange $w,x$ and $y,z$.
