How to determine the Standard Matrix

Solution 1:

Have you heard of the Rodrigues' rotation formula? Let $k=\dfrac{1}{\sqrt{3}}(1,1,1)$ in the formula and $$ K=\left(\begin{array}{ccc} 0 & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}} & 0 & -\frac{1}{\sqrt{3}}\\ -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & 0 \end{array}\right) $$

Then, $$ \begin{align} R&=I+(\sin\frac{2\pi}{3})K+(1-\cos\frac{2\pi}{3})K^2\\ &=\left(\begin{array}{ccc} 0&0&1\\ 1&0&0\\ 0&1&0 \end{array}\right). \end{align} $$

You should check what $R$ does to $(1,1,1)^T$ and vectors orthogonal to this vector. Also, when taking $R^6$, remember that this rotates by $\frac{2\pi}{3}$ six times. Without doing the matrix calculation, what should you get? What about $R^3$?