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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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    $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – José Carlos Santos Nov 26 '18 at 9:43
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    $\begingroup$ This form is called "Householder transform" in the domain of Numerical Analysis. $\endgroup$ – Jean Marie Nov 26 '18 at 10:32
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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 $w= (aa^T+bb^T-nn^T)v=(I-2nn^T)v$.

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

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