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,
\[\boxed{\bold{R} = \begin{pmatrix} 4/3 & 2/3 \\ -2/3 & 1/3 \end{pmatrix}}.\]
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 $\bold{R}$.
1 answer
1 answer