# 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

• Hi and welcome! Can you please show some of what you tried so far? Commented Feb 7, 2015 at 12:21
• in i) , I can't even explain why the 2 vectors are eigenvectors of T Commented Feb 7, 2015 at 12:23
• Have you tried drawing a picture (for $m=2$, say)? Commented Feb 7, 2015 at 13:00
• 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. Commented Feb 7, 2015 at 14:01

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: $$T=PAP^{-1}$$ 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 $$T=PAP^{-1}$$ 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.