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
Best Answer
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.