Let $$A = \frac{1}{4} \left(\begin{matrix} \sqrt 3 + 2 & \sqrt 3 - 2 & -\sqrt 2 \\ \sqrt 3 - 2 & \sqrt 3 + 2 & - \sqrt 2 \\ \sqrt 2 & \sqrt 2 & 2\sqrt 3 \end{matrix} \right) \in \mathbb{R}^{3\times 3}.$$ Prove that $\phi: x \mapsto Ax$, $x\in \mathbb{R}^3$ is an isometry. Find the Euclidean normal form $B$ of $A$ and an orthogonal matrix $S\in O(3)$ such that $B=S^{-1}AS$.
Proving that $\phi$ is an isometry is simply verifying that $A^TA=AA^T= I$.
Now we know the Eucildean normal form of A must look something like this $$\left( \begin{matrix} \pm1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{matrix} \right)$$ Does this mean we need to find the eigenvalues of $A$ via the characteristic polynomial? There ought to be a better way of doing this.
How do we find the matrix $S$?
Hint: If the matrix is proper,i.e., has determinant $1$, there is a much easier approach using Rodrigues formula. For nonproper case, do the same for $-A$.