Finding the standard matrix after a reflection in the euclidean plane Let T be the reflection about the line $4x + 5y = 0$ in the euclidean plane. What is the standard matrix $A$ of $T$?
So by rearranging the equation I get $y = -4x/5$. So it will be a reflection on that line. But I am struggling to find the required ordered pairs of the matrix that can be reflected. And how exactly would those ordered pairs change after the reflection.
Any help?
 A: If you want to find the matrix based on the standard unit basis $\pmatrix{1\\0},\pmatrix{0\\1}$, then you will have to find the images of these vectors upon reflecting in the given line $y=-\frac{4}{5}x$. That's a little work, but doable.
Here is $\pmatrix{1\\0}$: First we need to set up the perpendicular line through $(1,0)$. Perpendicular slope is $\frac{5}{4}$ and line follows $y=\frac{5}{4}x-\frac{5}{4}$. Now intersecting this line with $y=-\frac{4}{5}x$, we find the following point of intersection: $S=(\frac{25}{41},-\frac{20}{41})$. It is now a matter of "simple counting" to see what point lies on the other side of $y=-\frac{4}{5}x$. We find the point $(\frac{9}{41},-\frac{40}{41})$. So this is essentially the image vector of $\pmatrix{1\\0}$ and thus forms the first column of the matrix, as AMD indicated in his comment. Now you need to go through exactly the same steps for $\pmatrix{0\\1}$. This image vector turns out to be $\pmatrix{-\frac{40}{41}\\-\frac{9}{41}}$. Can you try this? Ok, now we have the matrix:  
$M=\pmatrix{\frac{9}{41}&-\frac{40}{41}\\-\frac{40}{41}&-\frac{9}{41}}$
So if, for sake of argument you want to reflect a point, say $(3,2)$ in the line $y=-\frac{4}{5}x$, all you need to do is matrix multiplication $\pmatrix{\frac{9}{41}&-\frac{40}{41}\\-\frac{40}{41}&-\frac{9}{41}}$$\pmatrix{3\\2}$ and you will get an answer.
Note. While the above method works, it is quite cumbersome to do it that way. There is a better way, by choosing a different basis to begin with. Instead of choosing the standard unit basis, you can instead choose basis $\pmatrix{5\\-4}$,$\pmatrix{4\\5}$. Here is the reason: First vector lies on the given line and thus after reflection, the image vector is the same. The second vector is chosen perpendicular to the given line and to find that image vector is also simple: $\pmatrix{-4\\-5}$. Now setting up this matrix is a breeze: 
$M_2=\pmatrix{5&-4\\-4&-5}$. Now in case of wanting to reflect point $(3,2)$ you CANNOT do matrix multiplication by $M_2$. This is because $(3,2)$ is not expressed in our new basis. In order to do that, one would need to solve $p\pmatrix{5\\-4}$+$q\pmatrix{4\\5}$=$\pmatrix{3\\2}$ and solve for $p$ and $q$. This can be easily done through $\pmatrix{p\\q}$=$\pmatrix{5&-4\\-4&-5}^{-1}$$\pmatrix{3\\2}$ and use a graphing calculator. Now $\pmatrix{5&-4\\-4&-5}$$\pmatrix{p\\q}$ will give you the same answer. While the second method looks daunting at first, it is really the faster way once you get used to it.Hope this helps a bit. Good luck!
