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.

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    $\begingroup$ A matrix in $SO_3$ is diagonalizable over $\mathbb C$, with one eigenvalue equal to $1$ and the other two complex conjugate numbers of modulus $1$. Can you go on from there? $\endgroup$ – Mariano Suárez-Álvarez Sep 14 '11 at 4:30
  • $\begingroup$ Thanks, that helps me prove the bounds on $\alpha$. Since similar matrices have the same trace, then the trace of any matrix in $SO_{3}$ must be equal to $1+2\Re(z)$, with $|z|=1, z\in\mathbb{C}$, hence $-1\leq\Re(z)\leq1$. I don't see how this helps me get that $\alpha=\beta$ however. $\endgroup$ – Patch Sep 14 '11 at 4:40
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    $\begingroup$ Well, you can look more :) $\endgroup$ – Mariano Suárez-Álvarez Sep 14 '11 at 4:41
  • $\begingroup$ Similar matrices also have the same characteristic polynomials (not just the same trace and determinant). $\endgroup$ – Aaron Sep 14 '11 at 4:48
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    $\begingroup$ Then you should put down the bag for a while and think more about this... $\endgroup$ – Mariano Suárez-Álvarez Sep 14 '11 at 5:11

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

where $r=\operatorname{Re}(\alpha)$. The result now follows by Cayley-Hamilton.


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