Linear Transformation with eigenvectors and eigenvalues Let $\;T : R_2 \to\ R_2\;$ be the linear transformation that reflects 
$\left(  
\begin{array}{c}
    x \\
    y \\
  \end{array}
\right)$
in the
line $\;y = mx\;$, that is, in the line $\left(  
\begin{array}{c}
    x \\
    y \\
  \end{array}
\right)$
= $\;t\;$
$\left(  
\begin{array}{c}
    1 \\
    m \\
  \end{array}
\right)$
for $\;t ∈ R\;$.

i) Explain why
$\left(  
\begin{array}{c}
    1 \\
    m \\
  \end{array}
\right)$
and 
$\left(  
\begin{array}{c}
    -m \\
    1 \\
  \end{array}
\right)$
are eigenvectors of $T$ and find the
corresponding eigenvalues.
ii) Let $A$ be the matrix representation for $T$. By finding matrices $P$, $D$,
with $D$ diagonal such that $P^{−1}AP = D$, (or otherwise), find an
explicit formula for the matrix $A$.
Thank you
 A: An eigenvector is a vector such that $T \vec v=\lambda \vec v$. This means that such a vector, when transformed by $T$, change its length but non chang its direction, i.e. $T \vec v$ stay on the same straight line as $\vec v$.
Helping with a drawing you can easily see that the eigenvectors of your transformation can only stay on the given line $r$ of equation $y=mx$ or on its orthogonal line passing by the origin, i.e. the line $y=-\frac{1}{m}x$. All other vectors in the plane, when reflected on $r$, ''jump''  on an other straight line. And since 
$$
\vec v_1= 
\left[
\begin {array}{ccccc}
 1\\
 m\\
\end {array} 
\right]
\qquad
\vec v_2= 
\left[
\begin {array}{ccccc}
 -m\\
 1\\
\end {array} 
\right]
$$
are  vectors on those two lines, they are eigenvectors.
Now note that $\vec v_1$ is a fixed point of $T$ so that its eigenvalue must be $\lambda_1=1$ and $\vec v_2$ is changed by $t$ in a vector that has opposite coordinates, so that its eigenvalue is $ \lambda_2=-1$.
The matrices of your second question is:
$T=P^{-1}AP$ where $A$ is the diagonal matrix having $a_{11}=\lambda_1$ and $a_{22}=\lambda_2$, and $P$ is the matrix that have as columns the eigenvectors, so you have:
$$
P=
\left[
\begin {array}{ccccc}
 1&-m \\
 m& 1 
 \end {array} 
\right]
\qquad
P^{-1}=
\dfrac{1}{1+m^2}
\left[
\begin {array}{ccccc}
 1&m \\
 -m& 1 
 \end {array} 
\right]
$$
$$
A=
\left[
\begin {array}{ccccc}
 1&0 \\
 0& -1 
 \end {array} 
\right]
$$
and if you perform the product $T=P^{-1}AP$ you find the searched matrix.
To complete the exercise use  $m=\tan \theta$ to show that the matrix $T$ has the form:
$$
T=
\left[
\begin {array}{ccccc}
\cos 2\theta&\sin 2\theta \\
\sin 2\theta& -\cos 2\theta 
 \end {array} 
 \right]
$$
that is the general form of a reflection in the plane.
