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- TL;DR Summary
- 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

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.

##\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}##.\left(\begin{array}{rrr}

-2 & 1 & -2 \\

-2 & -2 & 1 \\

1 & -2 & -2

\end{array}\right)##

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.