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?

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.
