Htykuut

Please, can you explain me how do we get this formula
$$ A = I - 2nn^{T} $$ in $$ R^{3} $$? This should be matrix of reflection, but I don't know how to prove that.

Let $a,b,n$ be unit vectors orthogonal to each other - $a,b$ basis for the plane, $n$ orthogonal to the plane.

You can easily check that any vector $v$ is represented in the basis {$a,b,n$} as

$v=a(a^Tv)+b(b^Tv)+n(n^Tv)=(aa^T+bb^T+nn^T)v$

($a^Tv, b^Tv, n^Tv$ are here scalars)

hence $aa^T+bb^T+nn^T=I$.

During the reflection components in the plane are unchanged, component orthogonal to the plane is negated.

Hence we have $w= (aa^T+bb^T-nn^T)v=(I-2nn^T)v$.

This is maybe more of a trick than of a proof, but I find this approach much easier to memorize.

Let $x_1, x_2, x_3$ be the Cartesian coordinates of the space. It is clear that the reflection around the plane $x_1x_2$ is the map
$$
(x_1, x_2, x_3)mapsto (x_1, x_2, -x_3), $$
and this is a linear map that can be written in matrix form as
$$
A_{(0,0,1)}begin{bmatrix} x_1 \ x_2 \ x_3end{bmatrix} = left( I-begin{bmatrix} 0 & 0 &0\ 0 & 0 & 0 \ 0 & 0 & 2end{bmatrix}right)begin{bmatrix}x_1\ x_2 \ x_3end{bmatrix},$$
that is,
$$
A_{(0,0,1)}=I-2begin{bmatrix} 0 \ 0 \ 1end{bmatrix}begin{bmatrix} 0 &0 &1end{bmatrix}.$$

We write
$$tag{1}
A_{(0,0,1)}vec x = vec y.$$

Now, to obtain the formula for the reflection around the plane having $vec{n}$ as normal vector, we change variables;
$$
vec x = Rvec x ', quad vec y = Rvec y', $$
where $R$ is a rotation matrix such that
$$
Rvec n = begin{bmatrix} 0 \0 \ 1end{bmatrix}.$$
Applying this change of variable in (1) yields
$$tag{2}
A_{vec n}vec x ' =vec y ',$$
where $A_{vec n}$ is the reflection matrix we want to compute. And now it is a matter of unwrapping $vec x', vec y'$ in (2) and computing, using the fact that $R^TR=I$, because $R$ is a rotation.