Show that the given transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ is linear by showing it is a matrix transformation $R$ rotates a vector $45^{\circ}$ counterclockwise about the origin. 
So, I have a vector and its tail part begins at the origin. It is $45^{\circ}$ and will rotate to the left. I don't have any particular points like $(3,2)$, so I assume that my vector is going to $(x,y)$. While rotating, the $x$ value does not change and stays fixed at $0$. I notice that the $y$ will change though which is why I get $(0,y)$. 
$R\left(\begin{matrix} x \\ y \end{matrix}\right) = \left(\begin{matrix} 0 \\ y \end{matrix}\right)$
$e_1 = \left(\begin{matrix} 1 \\ 0 \end{matrix}\right)$ and $e_2=\left(\begin{matrix} 0 \\ 1 \end{matrix}\right)$
$\left(\begin{matrix} 0 \\ y \end{matrix}\right) = 0\left(\begin{matrix} 1 \\ 0 \end{matrix}\right) + y\left(\begin{matrix} 0 \\ 1 \end{matrix}\right)$
$=\left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right)\left(\begin{matrix} 0 \\ y \end{matrix}\right)$
So
$R\left(\begin{matrix} x \\ y \end{matrix}\right) = A\left(\begin{matrix} 0 \\ y \end{matrix}\right)$
where
$A = \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right)$
I don't think I did this correctly, so I appreciate any feedback. 
 A: We have to start with an arbitrary point in the plane $(x,y)$.  I think it might be easiest to do a polar transformation $$x \mapsto r\cos(t) \\ y \mapsto r\sin(t)$$ first though.  That way rotating counterclockwise by $45^\circ = \frac {\pi}{4} \text{ rad}$ is simply a matter of adding the angle to $t$. 
So we'll "parametrize" your vector $(x,y)$ by $({x(t)}, {y(t)}) = ({r\cos(t)}, {r\sin(t)})$.  Then what the transformation $R$ does is $$\begin{align}R(x,y) &= (r\cos\left(t+\frac {\pi}{4}\right), r\sin\left(t+\frac {\pi}{4}\right)) \\ &= (r\cos(t)\cos\left(\frac {\pi}{4}\right)-r\sin(t)\sin\left(\frac {\pi}{4}\right), r\sin(t)\cos\left(\frac {\pi}{4}\right)+r\cos(t)\sin\left(\frac {\pi}4\right)) \\ &= \frac {\sqrt{2}}{2}(r\cos(t)-r\sin(t),r\sin(t)+r\cos(t)) \\ &= \frac{\sqrt{2}}{2}[(x-y,x+y)]\end{align}$$
Now we can see which matrix corresponds to this transformation.  Forgetting about the $\frac{\sqrt{2}}{2}$ for a second, the matrix which transformation $(x,y)$ into $(x-y,x+y)$ is just $\begin{bmatrix} 1 & -1 \\ 1 & 1\end{bmatrix}$ isn't it?
Thus, reinserting the factor of $\frac {\sqrt{2}}{2}$, we get $$\begin{align}R\begin{bmatrix} x \\ y \end{bmatrix} &= \frac{\sqrt{2}}2\begin{bmatrix} 1 & -1 \\ 1 & 1\end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} \\ &= \frac{\sqrt{2}}2\begin{bmatrix} x-y \\ x+y \end{bmatrix}\end{align}$$
Q.E.D.
