Eigenvalues of 3D rotation matrix I'm having some trouble calculating the eigenvalues for this rotation matrix, I know that you subtract a $\lambda$ from each diagonal term and take the determinant and solve the equation for $\lambda$ but I think I'm having some trouble with trig identities since I can't seem to find the correct answer.
Since this is a rotation matrix i know that the $|\lambda|$ must = 1 I just can't seem to get it to work out.
Attempt:
$\begin{bmatrix}\cos(\theta)&0&-\sin(\theta)\\0&1&0\\\sin(\theta)&0&\cos(\theta)\end{bmatrix}$ 
$\begin{bmatrix}\cos(\theta)-\lambda&0&-\sin(\theta)\\0&1-\lambda&0\\\sin(\theta)&0&\cos(\theta)-\lambda\end{bmatrix}$ 
$((\cos(\theta)-\lambda)*det\begin{bmatrix}1-\lambda&0\\0&\cos(\theta)-\lambda\end{bmatrix})$ - $0$ +$((-\sin(\theta)*det\begin{bmatrix}0&1-\lambda\\\sin(\theta)&0\end{bmatrix})$ $=0$
$(\cos(\theta)-\lambda)*[(1-\lambda)(\cos(\theta)-\lambda)]$ - $0$ +$(-\sin(\theta)*[0-(1-\lambda)(\sin(\theta)]$ = $0$
$(\cos(\theta) - \lambda)(\cos(\theta)-\lambda-\lambda\cos(\theta)+\lambda^2)-[\sin(\theta)(-(\sin(\theta)-\lambda\sin(\theta)]$ =$0$
$[-\lambda^3+\lambda^2+\lambda^2(2\cos(\theta))-\lambda(\cos^2(\theta))-\lambda(2\cos(\theta))+\cos^2(\theta)]+[\sin(\theta)-\lambda\sin^2(\theta)]=0$
combining like terms i end up with
$[-\lambda^3+\lambda^2(1+2\cos(\theta))-\lambda(\cos^2(\theta)+2\cos(\theta)+\sin^2(\theta))+\cos^2(\theta)+\sin^2(\theta)$=$0$
I think this is correct so far, but its very possible i made a sign error or something like that. I can't seem to solve for $\lambda$ in any way that i can see.I think I'm very close, just not really good with my trig identities.
Help would be greatly appreciated, Thanks a lot!
 A: Sometimes its easier to work by inspection, noting any obvious properties a matrix may have, than to proceed with the full formalism of the characteristic polynomial, etc.; I think such is the case with the present matrix.  Setting
$A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}, \tag{1}$
it is very easy to see that, with
$\mathbf j = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \tag{2}$
we have
$A \mathbf j = \mathbf j =  1 \mathbf j, \tag{3}$
which shows us that $1$ is an eigenvalue of $A$ with eigenvector $\mathbf j$.  Next we observe that, setting
$\mathbf i = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \tag{4}$
and
$\mathbf k = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \tag{5}$
we have
$A \mathbf i= \cos \theta \mathbf i + \sin \theta \mathbf k \tag{6}$
and
$A\mathbf k = -\sin \theta \mathbf i + \cos \theta \mathbf k. \tag{7}$
(6) and (7) together show that the subspace $\mathcal S = \text{span}\{ \mathbf i, \mathbf k \}$ is invariant under the action of $A$; that is $A \mathcal S \subset \mathcal S$.  Therefore any eigenvalue of the restriction of $A$ to $\mathcal S$, $A_{\mathcal S}$, will necessarily be an eigenvalue of $A$.  It is easy to see that the matrix $A_{\mathcal S}$ is 
$A_{\mathcal S} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \tag{8}$
and the characteristic polynomial $p_{A_{\mathcal S}}(\lambda)$ of $A_{\mathcal S}$ is given by the simple quadratic
$p_{A_{\mathcal S}}(\lambda) = \det(A_{\mathcal S} - \lambda I) = (\cos \theta - \lambda)^2 + \sin^2 \theta$
$= \lambda^2 - 2(\cos \theta) \lambda + \cos^2 \theta + \sin^2 \theta = \lambda^2 - 2(\cos \theta)\lambda + 1; \tag{9}$
it is easy to see from the quadratic formula that the roots of $p_{A_{\mathcal S}}(\lambda)$ are 
$\lambda = \dfrac{1}{2}(2\cos \theta \pm \sqrt{4\cos^2 \theta - 4}) = \cos \theta \pm \sqrt{-\sin^2 \theta} = \cos \theta \pm i \sin \theta = e^{\pm i \theta}; \tag{10}$
we thus see these eigenvalues are in agreement with those given by Victor Liu in his answer.
In the above calculations, we have effectively replaced factoring the cubic characteristic equation of $A$ with the decomposition of $\Bbb R^3$ (or $\Bbb C^3$) into two separate invariant subspaces of $A$, $\text{span}\{\mathbf j \}$ and $\text{span} \{ \mathbf i, \mathbf k \}$.
Of course one can if one chooses exploit the cubic $p_A(\lambda)$ directly:
$p_A(\lambda) = \det (A - \lambda I) = \det(\begin{bmatrix} \cos \theta - \lambda & 0 & -\sin \theta \\ 0 & 1 - \lambda & 0 \\ \sin \theta & 0 & \cos \theta - \lambda \end{bmatrix})$
$ = (\cos \theta - \lambda)^2(1 - \lambda) + \sin^2 \theta (1 - \lambda) = ((\cos \theta - \lambda)^2 + \sin^2 \theta)(1 - \lambda)$
$= (\lambda^2 - 2(\cos \theta)\lambda + 1)(1 - \lambda), \tag{11}$
where evaluation of the determinant is made easy using Sarrus' rule.  It is easy to see from (11) that the roots of $p_A(\lambda)$ are the same as the eigenvalues discovered above, basically by "inspection".
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: Starting from your last line (assuming it's right),
$$-\lambda^3+\lambda^2(1+2\cos(\theta))-\lambda(\cos^2(\theta)+2\cos(\theta)+\sin^2(\theta))+\cos^2(\theta)+\sin^2(\theta)=0$$
Since $\sin^2\theta+\cos^2\theta=1$,
$$-\lambda^3+\lambda^2(1+2\cos\theta)-\lambda(1+2\cos\theta)+1=0$$
You can factor this as:
$$ (1-\lambda) (\lambda^2 -2\lambda \cos\theta +1) = 0 $$
So then you have eigenvalues $\lambda = 1,\cos\theta \pm\sqrt{\cos^2\theta-1}$. That last pair can be written as $\cos\theta \pm i\sin\theta = e^{\pm i\theta}$.
