# How to find the matrix of reflection operator

I got a 3D space. In got a canonical equation of a plane: $ax + by + cz+ dt = 0$

How I can find the matrix of symmetric transformation transforming a point to its reflection?

-
Find out how the basis vectors are transformed. Or choose a basis so the plane is, for example, the $x$-$y$ plane, and then the form of the matrix is particularly simple. –  copper.hat Feb 4 '13 at 23:15
Is $t$ time? Is the plane moving in time and so you are looking to reflect across this moving plane? –  Maesumi Feb 4 '13 at 23:23

To find the matrix of the transformation, you need to find the images of the three basis vector $e_1,e_2,e_3$.

Recall that a vector normal to the plane $ax+by+cz+d=0$ is given by $v_1=(a,b,c)$. Of course, $a,b,c$ are not all zero, so we suppose without loss of generality that $a\not=0$.

Note that $v_2=(-b,a,0)$ is a non-zero vector normal to $v_1$ (compute the dot product), so it is "parallel to the plane". Moreover, $v_3=v_1\wedge v_2$ is a third non-zero vector normal to both $v_1$ and $v_2$.

Therefore, you get a basis $(v_1,v_2,v_3)$ of $\mathbb{R}^3$ constituted of

1. One vector normal to the plane
2. Two vectors parallel to the plane

Therefore, in this basis, the matrix of the transformation is very simple: $v_1$ is sent to $-v_1$ and $v_2,v_3$ are fixed!

Now, two possbilities:

• You know about change of basis matrices: then you compute the change of basis matrix from $(e_1,e_2,e_3)$ to $(v_1,v_2,v_3)$ and you will get the matrix in the canonical basis after a short calculation.
• You don't know about that: you give $(e_1,e_2,e_3)$ as linear combinations of $(v_1,v_2,v_3)$, apply the transformation (now it is very simple) and then go back as linear combinations of $(e_1,e_2,e_3)$. Actually that's what we do in the previous item.
-