Show isometry and find geometric meaning

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    Geometric Isometry
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SUMMARY

The matrix A, defined as \(\frac{1}{3}\begin{pmatrix}-2 & 1 & -2 \\ -2 & -2 & 1 \\ 1 & -2 & -2\end{pmatrix}\), is confirmed to be an isometry since \(AA^t=I\). The geometric interpretation of the transformation G, which is a rotation about the line \(t(1,1,1)\), is derived from the kernel of \(A+I\), yielding the vector \(t(1,1,1)\). The rotation is clockwise with an angle of \(\frac{2\pi}{3}\) when viewed from the point \((1,1,1)\) towards the origin. The transformation F represents an anti-clockwise rotation of \(\frac{\pi}{3}\) followed by reflection in the origin, establishing a relationship between G and F as G being the negative of F.

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TL;DR
With respect to an orthonormal basis in 3-space, show that the matrix of ##F## given below is an isometry and find its geometric meaning.
The matrix ##A## in question is

##\dfrac{1}{3}
\left(\begin{array}{rrr}
-2 & 1 & -2 \\
-2 & -2 & 1 \\
1 & -2 & -2
\end{array}\right)##
One can easily verify that ##AA^t=I##, hence an isometry. To find its geometric meaning, one can proceed to find ##U=\text{ker} \ (F-I)=\text{ker} \ (A-I)##, i.e. ## (A-I)\textbf{x}=\textbf{0}##. This yields ##\textbf{x}=\textbf{0}##. Put ##G=-F## and find ##U=\text{ker} \ (G-I)=\text{ker} \ (A+I)## instead, which yields ##\textbf{x}=t(1,1,1)##. Thus ##G## is a rotation about the line ##t(1,1,1)##. To determine the orientation of the rotation, one can define ##\textbf{w}=(1,1,1)##, ##\textbf{u}=(1,-1,0)## (any vector perpendicular to ##\textbf{w}##) and ##\textbf{v}=A\textbf{u}=(-1,0,1)##. Evaluating the determinant with the columns given by ##\textbf{u},\textbf{v},\textbf{w}## shows that they are negatively oriented and thus the rotation is clockwise seen from the point ##(1,1,1)## down on the origin. The angle of rotation is deduced from the dot product of ##\textbf{u}## and ##\textbf{v}##, which is ##\frac{2\pi}{3}##.

This is the geometric meaning of ##G##. What is that of ##F## and its matrix? The answer given is a rotation ##\frac{\pi}{3}## anti-clockwise from the point ##(1,1,1)## followed by reflection in the origin. Since ##G=-F##, isn't ##F## the rotation of ##G## followed by reflection in the origin? That is, a clockwise rotation ##\frac{2\pi}{3}## from the point ##(1,1,1)## looking down on the origin followed by reflection. The answer seems to reverse this.
 
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If ##G = -F##, let ##v \in \mathbb{R}^3##. Geometrically, ##Fv = -Gv## and thus ##v## transforms first under ##G## and the result ##Gv## gets an opposite sign, meaning that we reflect the vector ##Gv## around the origin.

But the order shouldn't matter actually: since the maps involved are linear, we also have ##Fv = G(-v)## and then ##F## first reflects the vector ##v## around the origin and then performs a rotation.
 
Okay, so the answers are equivalent?
 
Could it be that the answer tries to describe the transformation ##G## in terms of ##F##? Then it would make sense. However, that wouldn’t be the geometric meaning of ##F## and its matrix.
 

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