There exists a $2 \times 2$ matrix $R$ such that $r = R v$ for all 2-dimensional vectors $v$. Find $R$. For a vector $v$, let $r$ be the reflection of $v$ over the line
$$x = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}.$$
There exists a $2 \times 2$ matrix $R$ such that
$$r = R v$$
for all 2-dimensional vectors $v$. Find $R$.

I know that $$\text{proj}_{w} v = \begin{pmatrix} \frac{4}{5} & -\frac{2}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} v$$ for all 2 dimensional vectors $v$ and where $w=\begin{pmatrix} 2 \\ -1 \end{pmatrix}$.  But how is that going to help me?
 A: If $u$ is the projection of $v$ onto $w$, the reflection of $v$ over $w$ is given by $2u-v$.  See the diagram below.
Hence $$v'=2u-v=2\begin{pmatrix}\frac{4}{5}&-\frac{2}{5}\\-\frac{2}{5}&\frac{1}{5}\end{pmatrix}v-\begin{pmatrix}1&0\\0&1\end{pmatrix}v = \begin{pmatrix}\frac{3}{5}&-\frac{4}{5}\\-\frac45&-\frac35\end{pmatrix}v.\ \blacksquare$$
Another way of doing the problem, if you are given that such a matrix $R$ exists: if the matrix is $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, then $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}a\\c\end{pmatrix},$$ and $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}b\\d\end{pmatrix}.$$  So all you need to do is figure out where $(1,0)$ and $(0,1)$ go after the reflection, and the transformation matrix can be formed using the two results!  This method works in general for transformation matrices.
A: Finding the matrix that performs a reflection over a line through the origin.
Consider: a) A vector $\vec{v}$ , 
b) a straight line passing through the origin with a normalized direction vector $\vec{w}$ = $(1/√5)\begin{pmatrix} 2 \\ -1\end{pmatrix}$, and 
c) the reflected vector $\vec{v'}$.
We have: 
1) $\vec{v} + \vec{v'} = \lambda \vec{w}$ or 
$\vec{v'} = \lambda \vec{w} - \vec{v}$ .
Squaring both sides:
$|\vec{v'}|^2 = (\lambda)^2 -2 \lambda \vec{w} \cdot \vec{v} + |\vec{v}|^2$.
Note: $ |\vec{v}| = |\vec{v'}|$, I.e. reflected vector and original vector have the same length.
Hence:
2) $\lambda = 2 \vec{w} \cdot \vec{v}$.
Combining eqs. 1) and )2:
$\vec{v'} = [2 \vec{w} \cdot \vec{v}]\vec{w} - \vec{v}$.
Let $\vec{v} = \begin{pmatrix} h \\ k\end{pmatrix}$.
The expression in the brackets is: 
$2 (1/√5)(2h - k)$.
$\vec{v'}$ =
$(2/5)(2h - k)\begin{pmatrix} 2 \\ -1\end{pmatrix} - \begin{pmatrix} h \\ k\end{pmatrix}$.
Input in the above formula: $\vec{v}$, output: the reflected vector $\vec{v'}$.
If we prefer to describe the above operation by a 2×2 matrix 
we consider its action on the basis vectors :
$\begin{pmatrix} 1 \\ 0\end{pmatrix}$ and 
$\begin{pmatrix} 0 \\ 1\end{pmatrix}$.
Choose:
A) $\begin{pmatrix} h \\ k\end{pmatrix}$ =$\begin{pmatrix} 1 \\ 0\end{pmatrix}$ :
$\vec{v'} = (4/5) \begin{pmatrix}2 \\ -1\end{pmatrix} -\begin{pmatrix}1 \\ 0\end{pmatrix}$ =
$\begin{pmatrix} 3/5 \\ -4/5\end{pmatrix}$.
B) $\begin{pmatrix} h \\ k\end{pmatrix} = \begin{pmatrix} 0 \\ 1\end{pmatrix}$ :
$\vec{v'} = (-2/5)\begin{pmatrix} 2 \\ -1\end{pmatrix} - \begin{pmatrix} 0 \\ 1\end{pmatrix}$ =
$\begin{pmatrix} -4/5 \\ -3/5\end{pmatrix}$.
We find for the 2×2 matrix as noted by Fractal in his answer:
$\\ \begin{pmatrix} a&b \\ c&d\end{pmatrix} = \begin{pmatrix} 3/5&-4/5 \\ -4/5&-3/5\end{pmatrix}$
