Find angle of incomplete rotation matrix I'd like to find the angle of rotation of the following matrix 
$A= \begin{bmatrix} -\frac{1}{3} & \ast & \ast \\ \ast & -\frac{1}{3} & \ast \\ \ast & \ast & -\frac{1}{3} \end{bmatrix}$
I know that because it is an orthogonal matrix there exists some $T \in GL(3, \mathbb{C}$ so that $T^{-1}AT = \begin{bmatrix} \pm 1 & 0 &0 \\ 0 & cos(\theta) & -sin(\theta) \\ 0 & sin(\theta) & cos(\theta) \end{bmatrix} $ where $\theta$ is the angle of rotation. The problem is that I would've done this by finding the eigenvalues and now I'm unsure as to how else I can approach this.
If anyone could give me a hint as to what I need to look at that'd be greatly appreciated! 
 A: A rotation matrix $R$ in $SO_3(\Bbb{R})$ is conjugate to a matrix of the form
$$
A=\left(\begin{array}{crr}
1&0&0\\
0&\cos\theta&-\sin\theta\\
0&\sin\theta&\cos\theta
\end{array}\right).
$$
In other words there is a matrix $T$ such that $R=TAT^{-1}$.
Hint: Combine the following two facts (prove these first, if you haven't seen them before)


*

*$\operatorname{tr}(A)=1+2\cos\theta$,

*$\operatorname{tr}(TAT^{-1})=\operatorname{tr}(A).$


Recall that the trace of a square matrix $A$, $\operatorname{tr}(A)$, is the sum of its diagonal entries (and also the sum of its eigenvalues with correct multiplicities).

I guess that for the sake of completeness it behooves me to also show that rotation matrices with the prescribed diagonal actually exist. To that end consider
$$
R=\frac13\left(\begin{array}{ccc}
-1&2&2\\
2&-1&2\\
2&2&-1
\end{array}\right).
$$
It is easy to check that the rows (or columns) of $R$ form an orthonormal system. It is also easy to check that $\det R=1$, so the system is right-handed.
