Sorry guys, I have a problem about finding an orthogonal matrix $C \in Mat_{nxn}(\mathbf R)$ such that $$ C^{tr}MC= \begin{bmatrix} \lambda &0&0\\ 0 &cos(\alpha)&sin(\alpha)\\ 0 &-sin(\alpha)&cos(\alpha) \end{bmatrix} $$ where $\lambda$ is an eigenvalue of an orthogonal matrix $M \in Mat_{nxn}(\mathbf R)$, and $$ M= \begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix} $$
I tried firstly by looking for a matrix which is formed by normalized eigenvector of $M$, but I only found one eigenvector $$ \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$ Then I have no idea what should I continue.
Thanks in advance, for your attention and your help.
In fact, there are many ways to go about this question. Here is a quick (but unconventional way) to get an answer.
In fact, it suffices to get any orthogonal matrix $C$ whose first column is the normalized eigenvector $\frac 1{\sqrt 3}(1,1,1)$. One convenient such matrix is the Householder matrix that reflects across the plane perpendicular to $v = \frac 1{\sqrt 3}(1,1,1) - (1,0,0) = \frac 1{\sqrt{3}}(1-\sqrt{3},1,1)$. That is, it suffices to take $$ C = I - 2 \frac{vv^T}{v^Tv}, \quad v = \frac 1{\sqrt{3}}(1-\sqrt{3},1,1). $$ With computation, that comes out to $$ C = \pmatrix{1/\sqrt{3} & 1/\sqrt{3} & 1/\sqrt{3}\\ 1/\sqrt{3} & 1/2 - \sqrt{3}/6 & 1/(\sqrt{3} - 3)\\ 1/\sqrt{3} & 1/(\sqrt{3} - 3) & 1/2 - \sqrt{3}/6}. $$