Determine if a function is linear, $F: \mathbb{R^{2}} \rightarrow \mathbb{R^{2}}$ I am reading up on linearity in my linear algebra textbook and I can't seem to find a good example on how to solve this problem:
Determine if the following function $F: \mathbb{R^{2}} \rightarrow \mathbb{R^{2}}$ is linear:
$$
F\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x - y \\ x + y \end{pmatrix}
$$
The answer in the back of the book says that the above function is linear. I know that a function $F: V \rightarrow W$ is linear if:
Given $V$ and $W$ are real vector spaces and $c$ is a scalar
$$
F[\vec{v} + \vec{w}] = F[\vec{v}] + F[\vec{w}] \quad\text{and}\quad F[c\vec{v}] = cF[\vec{v}]
$$
However I don't really know how to apply these rules to the above problem, so if anyone could either show me how to solve this problem or direct me to an already worked out example similar to mine I would greatly appreciate it.
EDIT - To further elaborate on where I am struggling: What is $\vec{v}$ and $\vec{w}$ in the original problem?  To me it seems like $\vec{v} = x - y$ and $\vec{w} = x + y$, but I'm not sure. 
 A: All you have to do is check if the function itself satisfies the properties you have given! You may be confused by the notation used. Consider $v,w \in \mathbb{R}^2$
We have
$$F(v + w) = F \begin{pmatrix}v_1 + w_1 \\ v_2 + w_2 \end{pmatrix} = \begin{pmatrix} v_1 + w_1 - (v_2 + w_2) \\ v_1 + w_1 + (v_2 + w_2)\end{pmatrix}$$
Can you now show that the above is equivalent to
$$ F(v) + F(w) = F \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} + F \begin{pmatrix} w_1 \\ v_w \end{pmatrix} $$
A: Yes it is. You can see that because the map is given by a matrix, i.e.
$$F(\vec{x})=M\vec{x},\qquad M=\begin{pmatrix} 1 & -1 \\ 1 & 1\end{pmatrix}$$
here $\vec{x}=\begin{pmatrix} x \\ y \end{pmatrix}$ and $M\vec{x}$ is the matrix $M$ multiplied by the matrix $\vec{x}$.
But then you know that if you choose two vectors $\vec{u},\vec{v}$ and any constant $c$ you have $M(c\vec{u}+\vec{v})=cM\vec{u}+M\vec{v}$, by rules of matrix multiplication, so
$$F(c\vec{u}+\vec{v})=M(c\vec{u}+\vec{v})=cM\vec{u}+M\vec{v}=cF(\vec{u})+F(\vec{v})$$
i.e. the function is linear.
A: Yes, you found your real question:

What would be $\vec v$ and $\vec w$ be in the original problem?

Well, $\vec v$ and $\vec w$ denote arbitrary vectors in the given vector space. Now our vector space is $\Bbb R^2$ and we can give names to the coordinates of the general vectors $\vec v$, $\vec w$, e.g. 
$$\vec v=\pmatrix{x\\y}\quad\quad \vec w=\pmatrix{x'\\y'}$$
or e.g.
$$\vec v=\pmatrix{v_1\\v_2}\quad\quad \vec w=\pmatrix{w_1\\w_2}$$
(or anything).
A: The definition of linearity you give is:
$$F[\vec v + \vec w] = F[\vec v] + F[\vec w] \quad \text{  and  } \quad F[c \vec v] = c F[\vec v]$$
Consider that $\vec v$ and $\vec w$ are objects that $F$ operates on. So you must ask yourself: what objects does $F$ operate on in your example? Does it follow this rule?
You've defined $F$ to operate on elements of $\mathbb R^2$, so they are the $\vec v$ and $\vec w$ in this example.
Checking the first property of linearity:
$$\begin{align}F \left[ \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} \alpha \\ \beta \end{pmatrix}\right] & = F \left[ \begin{pmatrix} x + \alpha \\ y + \beta \end{pmatrix}\right] \\ &= \begin{pmatrix} x+ \alpha - y - \beta \\ x + \alpha + y + \beta \end{pmatrix}\end{align} $$
And then:
$$F \left[ \begin{pmatrix} x \\ y \end{pmatrix} \right] + F \left[\begin{pmatrix} \alpha \\ \beta \end{pmatrix}\right] = \begin{pmatrix} x - y \\ x + y \end{pmatrix} + \begin{pmatrix} \alpha - \beta \\ \alpha + \beta \end{pmatrix} = \begin{pmatrix} x + \alpha - y - \beta \\ x + \alpha + y + \beta \end{pmatrix}$$
We see that these two things are indeed equal, as linearity requires. Checking the other defining property of linearity is even simpler.
