Matrix of reflection in $R^3$

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.

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• This form is called "Householder transform" in the domain of Numerical Analysis. – Jean Marie Nov 26 '18 at 10:32

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

• Very nice! Thank you. – Giuseppe Negro Dec 3 '18 at 11:06

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 $$A_{(0,0,1)}\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} = \left( I-\begin{bmatrix} 0 & 0 &0\\ 0 & 0 & 0 \\ 0 & 0 & 2\end{bmatrix}\right)\begin{bmatrix}x_1\\ x_2 \\ x_3\end{bmatrix},$$ that is, $$A_{(0,0,1)}=I-2\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}\begin{bmatrix} 0 &0 &1\end{bmatrix}.$$

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; $$\vec x = R\vec x ', \quad \vec y = R\vec y',$$ where $$R$$ is a rotation matrix such that $$R\vec n = \begin{bmatrix} 0 \\0 \\ 1\end{bmatrix}.$$ 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.