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.