TITLE: $A\in M_3(\mathbb R)$ orthogonal matrix with $\det (A)=1$.  Prove that $(\mathrm{tr} A)^2- \mathrm{tr}(A^2) = 2 \mathrm{tr} (A)$
QUESTION [11 upvotes]: $A\in M_3(\mathbb R)$ orthogonal matrix with $\det (A)=1$.
I need to prove that $(\mathrm{tr} A)^2-\mathrm{tr}(A^2) = 2 \mathrm{tr} (A)$ ; $\mathrm{tr}$=trace.
I know that if $A$ is orthogonal than $A^tA=I$ and that $A$ is diagonalizable and similar to $D=\begin{pmatrix}
\lambda_1 &  & \\ 
 & \lambda_2 & \\ 
 &  & \lambda_3
\end{pmatrix}$. We know as well that $\mathrm{tr} A=\mathrm{tr} D= \lambda_1+ \lambda_2+\lambda_3$  that  $\mathrm{tr} A^2=\mathrm{tr} D^2= \lambda_1^2+ \lambda_2^2+\lambda_3^2$ and that 
$\lambda_1 \lambda_2 \lambda_3=1$. It's not enough for solving the question.
What more should I know or use in order to solve it?
Thanks

REPLY [5 votes]: A rotation is diagonalisable over the complex numbers, but generally not over the reals. However, there always is an orthonormal basis $\mathcal{B}$ and a real number $\theta$ such that $$\mathrm{Mat}_{\mathcal{B}}(A)=\left(\begin{array}{ccc} \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0&0&1\end{array}\right)$$
and you'll have $$\mathrm{Mat}_{\mathcal{B}}(A^2)=\left(\begin{array}{ccc} \cos(2\theta) & -\sin(2\theta) & 0 \\ \sin(2\theta) & \cos(2\theta) & 0 \\ 0&0&1\end{array}\right)$$
Thus $\mathrm{Tr}(A)=1+2\cos(\theta)$ and $\mathrm{Tr}(A^2)=1+2\cos(2\theta)$. The rest follows from trigonometry.