Special Orthogonal Group and Cayley-Hamilton theorem One of the questions on my University's algebra qual had us prove that given an arbitrary $\mathbf{A}\in SO_{3}(\mathbb{R})$, there is some constant, $-1\leq\alpha\leq3$, such that $$\mathbf{A}^{3}-\alpha\mathbf{A}^{2}+\alpha\mathbf{A}-\mathbf{I}_{3}=0.$$
Appealing to the Cayley-Hamilton theorem, and the fact that $\det\mathbf{A}=1$, I was able to (through good old fashioned number crunching) show that $$\mathbf{A}^3-(\text{tr }\mathbf{A})\mathbf{A}^2+\beta\mathbf{A}-\mathbf{I}_{3}=0,$$ where $\beta=-a_{12}a_{21} + a_{11}a_{22} - a_{13}a_{31} - a_{23}a_{32} + a_{11}a_{33} + a_{22}a_{33}$. So naturally I assumed that the $\alpha$ they were looking for was $\alpha=\text{tr }\mathbf{A}$, and that (by the restrictions placed on $\mathbf{A}$ by its orthogonality) $\beta=\alpha$.
I also suspected that since $\text{tr }\mathbf{I}_{3}=3$ and $\text{tr }\begin{bmatrix}
  1 & 0 & 0\\
  0 & -1 & 0\\
  0 & 0 & -1
\end{bmatrix}=-1$ (both of which are in $SO_{3}(\mathbb{R})$) then these might contribute to the proposed bounds on $\alpha$, however I was not able to formally show anything.
Any help in proving the bounds on $\alpha$ or on showing that $\alpha=\beta$ (of which I'm pretty certain is indeed true, after trying a number of specific examples) is much appreciated.

Mariano's hint is the key to solving the problem.  Here is a solution which explains why the hint is true.
Proposition: If $A\in U_n$, the unitary group, then every eigenvalue of $A$ is of norm $1$.  
proof. If $Av=\lambda v$, then because $A$ is unitary $|v|=|Av|=|\lambda v|=|\lambda||v|$.
Note that we are viewing orthogonal matrices as unitary because it is a more natural to think about complex eigenvectors in this case.
Proposition:  If $A\in SO_3(\mathbb R)$, then $1$ is an eigenvalue of $A$.
proof.  Since the characteristic polynomial of $A$ is real and of degree $3$, it must have some real root, and since every eigenvalue of $A$ is of norm $1$, this root must be $\pm 1$.  Suppose that the eigenvalues are $-1, \alpha$, and $\beta$.  Then $\alpha\beta=-1$.  However, since $|\alpha|=|\beta|=1$, we must have $\beta=-\overline{\alpha}$.  Since $\operatorname{tr} A=-1+(\alpha-\overline{\alpha})$ is real, $\alpha$ must be real too, and hence $\alpha=\pm 1$.
By the proposition, the eigenvalues of $A\in SO_3$ are therefore $1, \alpha, \overline{\alpha}$ for some $\alpha\in C$ with $|\alpha|=1$.  The characteristic polynomial of $A$ is then
$(x-1)(x-\alpha)(x-\overline{\alpha})=(x-1)(x^2-2rx+1)=x^3-(2r+1)x^2+(2r+1)x-1$ 
where $r=\operatorname{Re}(\alpha)$. The result now follows by Cayley-Hamilton.
