$A\in M_3(\mathbb R)$ orthogonal matrix with $\det (A)=1$.  Prove that $(\mathrm{tr} A)^2- \mathrm{tr}(A^2) = 2 \mathrm{tr} (A)$ $A\in M_3(\mathbb R)$ 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
 A: 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.
A: $\begin{align*}\mathrm{Tr}(A)^2-\mathrm{Tr}(A^2)&= (\lambda_1+\lambda_2+\lambda_3)^2-(\lambda_1^2+\lambda_2^2+\lambda_3^2)\\&=2(\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1)\\&=2\lambda_1\lambda_2\lambda_3\left(\frac{1}{\lambda_1}+\frac{1}{\lambda_2}+\frac{1}{\lambda_3}\right)\\&=2\det(A)\mathrm{Tr}(A^{-1})\\&=2\mathrm{Tr}(A).
\end{align*}$
(Edit: In the last step, we use the facts that $A^{-1}=A^T$ when $A$ is orthogonal and $\mathrm{Tr}(A^T)=\mathrm{Tr}(A)$ for any matrix $A$.)
A: Hint:
$$\det(I-A)=\det(A^tA-A)=\det(A^t-I)\det(A)=\det(A-I)\cdot1=(-1)^3\det(I-A),$$
so $\det(I-A)=0$. IOW one of the eigenvalues is equal to $1$. Does that help?
A: In $\mathbb Z[X,X^{-1}]$ we have 
$$(1+X+X^{-1})^2-(1+X^2+X^{-2})=2+2X+2X^{-1}.$$ 
In other words, your equality holds over any commutative ring, for any matrix similar to a diagonal matrix with diagonal entries $1,a,a^{-1}$, where $a$ is a unit. 
A: An old-fashioned argument
The matrix $A$ is the matrix of a rotation through some angle $\theta$.  It is a standard result that the trace of such a rotation is $1+2\cos\theta$.  The matrix $A^2$ is a rotation through $2\theta$ about the same axis.
So we want to show that 
$$(1+2\cos\theta)^2-(1+2\cos 2\theta)=2(1+2\cos\theta).$$
With a little manipulation this reduces to the familiar $\cos 2\theta=2\cos^2\theta -1$.
