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

  • $\begingroup$ Hi and welcome! Can you please show some of what you tried so far? $\endgroup$ Feb 7, 2015 at 12:21
  • 1
    $\begingroup$ in i) , I can't even explain why the 2 vectors are eigenvectors of T $\endgroup$
    – samheihey
    Feb 7, 2015 at 12:23
  • $\begingroup$ Have you tried drawing a picture (for $m=2$, say)? $\endgroup$ Feb 7, 2015 at 13:00
  • $\begingroup$ Doing basic geometry (line perpendicular to other line, middle point of a line segment, intersection point of two lines, etc.), can you prove that, provided $\;m\neq 0\;$ , a point $\;(a,b)\in\Bbb R^2\;$ is mapped by $\;T\;$ to $\;\left(a-\frac{2b}m\,,\,\,2ma+b\right)\;$ ? This is the first, basic step. Then apply this to a a basis of $\;\Bbb R^2\;$ and etc. $\endgroup$
    – Timbuc
    Feb 7, 2015 at 14:01

1 Answer 1


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.


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