Rotation matrix

[Tolaso J Kos]

Given the matrix

$$A= \begin{pmatrix}
\sqrt{3}/2 &0 &1/2 \\
0& 1 &0 \\
-1/2& 0 & \sqrt{3}/2
\end{pmatrix}$$

examine if it is a rotation of a plane around an axis that is perpendicular to it. If so, determine the angle of rotation and the axis.

[Papapetros Vaggelis]

The matrix \(\displaystyle{A}\) is orthogonal, that is \(\displaystyle{A\,A^t=I_{3}=A^t\,A}\) .

Also, \(\displaystyle{\rm{det}(A)=1}\) . We define \(\displaystyle{f=f_{A}:\mathbb{R}^3\longrightarrow \mathbb{R}^3}\) by

\(\displaystyle{f_{A}(x,y,z)=A\,(x,y,z)^t=\left(\dfrac{\sqrt{3}}{2}\,x+\dfrac{z}{2},y,-\dfrac{1}{2},x+\dfrac{\sqrt{3}}{2}\,z\right)}\) .

We observe that

\(\displaystyle{f_{A}(x,0,z)=\left(\dfrac{\sqrt{3}}{2}\,x+\dfrac{z}{2},0,-\dfrac{1}{2},x+\dfrac{\sqrt{3}}{2}\,z\right)\,,\forall\,x\,,z\in\mathbb{R}}\) .

so: \(\displaystyle{f_{A}(x,0,z)=\begin{pmatrix}
\sqrt{3}/2&1/2 \\
-1/2& \sqrt{3}/2
\end{pmatrix}\,(x,z)^t}\) .

Therefore, the given matrix is a rotation of \(\displaystyle{x\,z}\) - plane around the \(\displaystyle{y-axis}\)

which is perpendicular to it.

The angle of rotation is \(\displaystyle{\theta=\dfrac{\pi}{6}\,\,,\sin\,\dfrac{\pi}{6}=1/2\,\,,\cos\,\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}}\) .