Reflections

Quote:

Originally Posted by vcf323

Find the matrix http://courses.webwork.maa.org:8080/...0fcdd79cd1.png of the reflection in the line http://courses.webwork.maa.org:8080/...6819fae171.png in http://courses.webwork.maa.org:8080/...7237a801b1.png that consists of all scalar multiples of the vector http://courses.webwork.maa.org:8080/...82ad4d9741.png5http://courses.webwork.maa.org:8080/...706f6033a1.png 6 .

Since your image is messed up lets do it for an arbitrary vector
$\vec{v}=\begin{pmatrix} a \\ b\end{pmatrix}=ae_1+be_2$

So we need to know that transform of the basis vectors $e_1,e_2$

First we need to project $e_1$ on the vector $v$

$\displaystyle \text{proj}_{v}e_1=\frac{\vec{v} \cdot e_1}{||v||^2}\vec{v}=\frac{a}{a^2+b^2}\vec{v}$

Now to find the the direction vector we subtract

$\frac{a}{a^2+b^2}\vec{v}-e_1$

The above vector takes us from the tip of $e_1$ onto the line but we want to reflect across it so we need to multiply it by 2. This gives

$\displaystyle T(e_1)=e_1+2\left( \frac{a}{a^2+b^2}\vec{v}-e_1\right)= \frac{2a}{a^2+b^2}(ae_1+be_2)-e_1=\left( \frac{a^2+2b^2}{a^2+b^2}\right)e_1+\left(\frac{2ab }{a^2+b^2}\right)e_2$

Now just do the exact same thing with $e_2$