Rotation matrix

Give 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.

Solution

To examine if the matrix is a rotation matrix it is sufficient to check whether $\det A=1$ and $AA^t =\mathbb{I}_3$. We begin by evaluating the determinant of the matrix. Hence:

$$\det A= \frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}+ \frac{1}{2}\cdot \frac{1}{2}= \frac{3}{4}+ \frac{1}{4}=1$$

which satisfies the first condition. Now on to check the second one. Successively we have:

$$A A^t = \begin{pmatrix}
\sqrt{3}/2 &0  &1/2 \\
 0& 1 &0 \\
 -1/2& 0 &\sqrt{3}/2
\end{pmatrix}\cdot \begin{pmatrix}
\sqrt{3}/2 &0  &-1/2 \\
 0& 1 & 0\\
 1/2&0  &\sqrt{3}/2
\end{pmatrix}\overset{{\rm calculations}}{=\! =\! =\! =\! =\! =\! =\!} \begin{pmatrix}
1 &0  &0 \\
 0& 1 &0 \\
 0&0  &1
\end{pmatrix}=\mathbb{I}_3$$

and the second condition is actually satisfied.Hence the matrix is a rotation  matrix. Furthermore the matrix is orthogonal, meaning that it has the number $1$ as an eigevalue.

We calculate the eigenvalues as well as the eigenvectors of the matrix $A$. For the eigenvalues we have:

$$\begin{vmatrix}
\sqrt{3}/2 -\ell & 0 &1/2 \\
 0& 1-\ell & 0\\
 -1/2& 0 &\sqrt{3}/2-\ell 
\end{vmatrix}=0 \Leftrightarrow  \left ( \frac{\sqrt{3}}{2}-\ell \right )^2 \left ( 1-\ell \right ) + \frac{1}{4}\left ( 1-\ell \right )=0 \Leftrightarrow \\\Leftrightarrow \left ( 1-\ell \right )\left [ \frac{1}{4}+ \left (\frac{\sqrt{3}}{2}-\ell\right )^2 \right ]=0$$

Hence, this confirms that $1$ is an eigenvalue of $A$. The other two eigenvalues are conjugate complex numbers.To find the axis we seek (or much better the vector parallel to the axis) we just evaluate the eigenvectors of the matrix $A$ and we are done. (Calculations are easy)

Now, to complete the exercise we need to compute the angle of rotation. Using the formula:

$${\rm Tr}(A)=1+2\cos  \theta$$

it suffices only to solve the equation with respect to $\theta.$ Successively we have:

$$\sqrt{3}+1 = 1+2\cos \theta \Leftrightarrow \cos \theta= \sqrt{3}/2 \Leftrightarrow  \theta =\frac{\pi}{6}$$