For a vector $\bold{v}$, let $\bold{r}$ be the reflection of $\bold{v}$ over the line
\[\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}.\]
[asy]
unitsize(1 cm);
pair O, V, R;
O = (0,0);
V = (3,1);
R = reflect(O,(2,-1))*(V);
xaxis(-2,4);
yaxis(-3,1);
draw((-1)*(2,-1)--2*(2,-1));
draw(O--V,Arrow(8));
draw(O--R,Arrow(8));
dot(O);
label("$\mathbf{v}$", (O + V)/2, N);
label("$\mathbf{r}$", (O + R)/2, W);
[/asy]
There exists a $2 \times 2$ matrix $\bold{R}$ such that
\[\bold{r} = \bold{R} \bold{v}\]
for all 2-dimensional vectors $\bold{v}$. Find $\bold{R}$.
Let $\bold{v} = \begin{pmatrix} x \\ y \end{pmatrix}$. Then $\bold{v}$ lies on the line
\[\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix} + (1 - t) \bold{v},\]for a unique $t$. (In words, $\bold{x}$ belongs to the line between $\bold{v}$ and $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$, which uniquely determines $t$.) Rearranging, we get
\[t \bold{v} - (1 - t) \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \bold{x}.\]Then
\[\bold{v} = \frac{1}{2t - 1} \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \frac{1}{1 - 2t} \bold{x},\]which we can rewrite as
\[\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{t'} \begin{pmatrix} -1 \\ 1 \end{pmatrix} + \frac{1}{1 - 2t'} \bold{r}.\]The points $\bold{v}$ and $\bold{r}$ must have the same component in the direction orthogonal to $\begin{pmatrix} 1 \\ \frac{1}{2} \end{pmatrix}$, which is given by $2x + y = k$. Hence, for some $k$,
\begin{align*}
2x + y &= 2 \cdot \frac{1}{t'} + 1 \\
2(1 - 2t') &= (2t' - 1)(-2k - 1).
\end{align*}In other words,
\[4 - 8t' = (4t' - 2)(-2k + 1).\]Solving, we find $t' = \frac{1}{2}$, so $t = \frac{1}{4}$.
Then $\bold{x} = x \bold{v} - \frac{1}{3} (1 - x) \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ for all 2-dimensional vectors $\bold{v}$. We can write this as
\[\bold{x} = x \bold{v} - \begin{pmatrix} 2/3 \\ -1/3 \end{pmatrix} + \frac{1}{3} \begin{pmatrix} 2 \\ -1 \end{pmatrix} x.\]Let
\[\bold{R} = \begin{pmatrix} p & q \\ r & s \end{pmatrix}.\]Then
\[\bold{r} = \bold{R} \bold{v} = \begin{pmatrix} p & q \\ r & s \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} px + qy \\ rx + sy \end{pmatrix},\]so
\begin{align*}
px + qy &= \frac{4}{3} x + \frac{2}{3} y, \\
rx + sy &= -\frac{2}{3} x + \frac{1}{3} y.
\end{align*}Taking $x = 1$ and $y = 0$, we get
\[\begin{pmatrix} p \\ r \end{pmatrix} = \begin{pmatrix} 4/3 \\ -2/3 \end{pmatrix}.\]Taking $x = 0$ and $y = 1$, we get
\[\begin{pmatrix} q \\ s \end{pmatrix} = \begin{pmatrix} 2/3 \\ 1/3 \end{pmatrix}.\]Therefore,
\[\boxed{\bold{R} = \begin{pmatrix} 4/3 & 2/3 \\ -2/3 & 1/3 \end{pmatrix}}.\]
Well, since $\bold{r}$ is the reflection of $\bold{v}$ over the line $\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$, we can write $\bold{r} = \bold{v} - 2 \bold{p}$, where $\bold{p}$ is the projection of $\bold{v}$ onto the line.
To find $\bold{p}$, we can use the fact that the projection of a vector $\bold{v}$ onto a line with direction vector $\bold{d}$ is given by the formula
\[\bold{p} = \frac{\bold{v} \cdot \bold{d}}{\| \bold{d} \|^2} \bold{d}.\]
In this case, $\bold{d} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$, so
\[\bold{p} = \frac{\bold{v} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}}{\| \begin{pmatrix} 2 \\ -1 \end{pmatrix} \|^2} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \frac{\bold{v} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}}{5} \begin{pmatrix} 2 \\ -1 \end{pmatrix}.\]
Substituting this expression for $\bold{p}$ into the equation $\bold{r} = \bold{v} - 2 \bold{p}$, we have
\[\bold{r} = \bold{v} - 2 \cdot \frac{\bold{v} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}}{5} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \bold{v} - \frac{2}{5}(\bold{v} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}) \begin{pmatrix} 2 \\ -1 \end{pmatrix}.\]
So the matrix $\bold{R}$ is equal to
\[\bold{R} = \bold{I} - \frac{2}{5}(\bold{v} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}) \begin{pmatrix} 2 \\ -1 \end{pmatrix}^\top,\]
where $\bold{I}$ is the $2 \times 2$ identity matrix.
After expanding this expression, we find that
\[\bold{R} = \bold{I} - \frac{4}{5} \begin{pmatrix} 2 \\ -1 \end{pmatrix} \begin{pmatrix} 2 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \frac{4}{5} \begin{pmatrix} 4 & -2 \\ -2 & 1 \end{pmatrix} = \boxed{\begin{pmatrix} -\frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{pmatrix}}.\]
To find the matrix $\bold{R}$, we need to find the image of the standard basis vectors $\bold{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\bold{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ under the reflection $\bold{r}$.
Let's first find the image of $\bold{i}$. The vector $\bold{i}$ can be written as $\bold{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = 1 \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix} + 0 \cdot \begin{pmatrix} 0 \\ 1 \end{pmatrix}$. So, we have $\bold{r}(\bold{i}) = 1 \cdot \bold{r}\left(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\right) + 0 \cdot \bold{r}\left(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\right)$.
The reflection of $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ over the line $\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ can be found by considering the line segment connecting $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ to its reflection. The line segment connecting a point to its reflection over a line is perpendicular to that line. So, we need to find the line that is perpendicular to $\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$.
The direction vector of the perpendicular line is the negative reciprocal of the direction vector of $\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$. The direction vector of $\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ is $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$. So, the direction vector of the perpendicular line is $\begin{pmatrix} -1 \\ -2 \end{pmatrix}$.
Let's call the point where the perpendicular line intersects the line $\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ as $\bold{p}$. Since $\bold{p}$ lies on the line $\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$, we have $\bold{p} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$. The line connecting $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\bold{p}$ is perpendicular to the line connecting $\bold{p}$ and its reflection. So, the line connecting $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\bold{p}$ is parallel to the line $\bold{x} = \bold{r}\left(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\right)$.
The direction vector of the line connecting $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\bold{p}$ is $\bold{p} - \begin{pmatrix} 1 \\ 0 \end{pmatrix} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2t - 1 \\ -t \end{pmatrix}$. Since this line is parallel to $\bold{x} = \bold{r}\left(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\right)$, the direction vectors of both lines are scalar multiples of each other.
Comparing the coordinates, we get $\begin{pmatrix} 2t - 1 \\ -t \end{pmatrix} = c \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ for some scalar $c$. Equating the corresponding entries, we have $2t - 1 = c$ and $-t = 0$. The second equation gives us $t = 0$. Substituting $t = 0$ in the first equation, we obtain $c = -1$.
Hence, we have $\begin{pmatrix} 2t - 1 \\ -t \end{pmatrix} = -1 \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. This implies that $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is the reflection of $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$ over the line $\bold{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$.
Therefore, we have $\bold{r}\left(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\right) = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$.
Similarly, we can find that $\bold{r}\left(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\right) = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$.
Therefore, the matrix $\bold{R}$ is given by
\[\bold{R} = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}.\]
To find the matrix $\mathbf{R}$ that represents the reflection across the line $\mathbf{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$, we first need to determine where the standard basis vectors $\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ are mapped to under the reflection.
Let's start with the vector $\mathbf{i}$. We want to find $\mathbf{r}$ such that $\mathbf{r}$ is the reflection of $\mathbf{i}$ over the line $\mathbf{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$. Geometrically, this means that the line connecting $\mathbf{i}$ and $\mathbf{r}$ is perpendicular to the line $\mathbf{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$.
[asy]
unitsize(1 cm);
pair O, I, R;
O = (0,0);
I = (1,0);
R = reflect(O,(2,-1))*(I);
xaxis(-2,4);
yaxis(-3,1);
draw((-1)*(2,-1)--2*(2,-1),blue);
draw(I--R,Arrow(8),red);
dot((0,0));
dot(I);
dot(R);
label("$\mathbf{i}$", I, E);
label("$\mathbf{r}$", R, W);
label("$t\begin{pmatrix} 2 \\ -1 \end{pmatrix}$", (2,-1), N, blue);
[/asy]
So, to find $\mathbf{r}$, we can find the intersection point of the lines $\mathbf{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $\overrightarrow{\mathbf{i}\mathbf{r}} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix} = 0$.
Substituting $\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix} x \\ y \end{pmatrix}$ into the equation $\overrightarrow{\mathbf{i}\mathbf{r}} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix} = 0$, we get
\[\begin{pmatrix} x \\ y \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}.\]
This simplifies to $2x - y = 0$, or $y = 2x$.
Now let's do the same thing for vector $\mathbf{j}$. We want to find $\mathbf{r}$ such that $\mathbf{r}$ is the reflection of $\mathbf{j}$ over the line $\mathbf{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$. Geometrically, this means that the line connecting $\mathbf{j}$ and $\mathbf{r}$ is perpendicular to the line $\mathbf{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$.
[asy]
unitsize(1 cm);
pair O, J, R;
O = (0,0);
J = (0,1);
R = reflect(O,(2,-1))*(J);
xaxis(-2,4);
yaxis(-3,1);
draw((-1)*(2,-1)--2*(2,-1),blue);
draw(J--R,Arrow(8),red);
dot((0,0));
dot(J);
dot(R);
label("$\mathbf{j}$", J, N);
label("$\mathbf{r}$", R, E);
label("$t\begin{pmatrix} 2 \\ -1 \end{pmatrix}$", (2,-1), N, blue);
[/asy]
So, to find $\mathbf{r}$, we can find the intersection point of the lines $\mathbf{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $\overrightarrow{\mathbf{j}\mathbf{r}} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix} = 0$.
Substituting $\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix} x \\ y \end{pmatrix}$ into the equation $\overrightarrow{\mathbf{j}\mathbf{r}} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix} = 0$, we get
\[\begin{pmatrix} x \\ y \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}.\]
This simplifies to $2x - y = -1$, or $y = 2x + 1$.
Now we have the equations $y = 2x$ and $y = 2x + 1$ that represent the lines along which the reflections of $\mathbf{i}$ and $\mathbf{j}$ lie, respectively.
To find the matrix $\mathbf{R}$, we need to find the transformation that takes the standard basis vectors $\mathbf{i}$ and $\mathbf{j}$ to the corresponding reflection vectors $\mathbf{r}$. From the equations above, we have
\[\begin{pmatrix} 2 & -1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 2 & -1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}.\]
So, the matrix $\mathbf{R}$ that represents the reflection across the line $\mathbf{x} = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ is
\[\mathbf{R} = \begin{pmatrix} 2 & -1 \\ 0 & 1 \end{pmatrix}.\]
Therefore, we have $\mathbf{r} = \begin{pmatrix} 2 & -1 \\ 0 & 1 \end{pmatrix}\mathbf{v}$ for all 2-dimensional vectors $\mathbf{v}$.