Be $M \in M_2(\mathbb{R})$ with $det(M) = 1$ Be $M \in M_2(\mathbb{R})$ with $\det(M) = 1$
If $|\text{tr}(M)| < 2$, Prove that exist $P \in M_2(\mathbb{R})$ with $\det(P)=1$ and exist $\alpha \in \mathbb{R}$ such that:
$ M = P\cdot \begin{bmatrix} \cos(\alpha) & -\sin(\alpha)  \\ \sin(\alpha) & \cos(\alpha) \end{bmatrix} \cdot P^{-1}$
$\\$
My attempt to solution:
I could not solve the problem. And I do not know how to continue.
$1)$ I started analising the caracteristic polynomial of M:
$P(x)_M = x^2-\text{tr}(M)\cdot x + \det(M)$
And we know that $\det(M) = 1$ and $|\text{tr}(M)| < 2$, implies that $\Delta = (\text{tr}(M))^2-4 \det(M)<0$
So, both eigenvalues of $M$ are complexes and distinct, and one is conjugate of another.
$2)$ This matrix remember me the rotation matrix, but i don't know how to use that fact.
I do not know what to do. I tried for a long time, but I really could not resolve.
 A: Denote by $\lambda_1$ and $\lambda_2 = \overline{\lambda_1}$ the two complex eigenvalues. Since
$$\det(M) = \lambda_1 \lambda_2 = |\lambda_1|^2 = 1$$ this means that we can write $\lambda_1 = e^{i\alpha}$ and $\lambda_2 = e^{-i\alpha}$ for some $\alpha \in \mathbb{R}$. Since the eigenvalues are distinct, this means that $M$ is diagonalizable over $\mathbb{C}$. Let $v_1 \in \mathbb{C}^2$ be a complex eigenvector corresponding to $\lambda_1$. Then since $M$ is real we have
$$ M \overline{v_1} = \overline{Mv_1} = \overline{\lambda_1 v_1} = \overline{\lambda_1} \overline{v_1} = \lambda_2 \overline{v_1} $$
so $\overline{v_1}$ is an eigenvector of $M$ corresponding to $\lambda_2$. Set
$$ u_1 := \frac{v_1 - \overline{v_1}}{2}, u_2 := \frac{v_1 + \overline{v_1}}{2i}. $$
That is, $u_1$ and $u_2$ are the real and imaginary parts of the eigenvector $v_1 = u_1 + iu_2$. Note that $u_1,u_2$ must be linearly independent over $\mathbb{R}$. Why? If $au_1 + bu_2 = 0$ for $a,b \in \mathbb{R}$ then
$$ au_1 + bu_2 = a \cdot \frac{v_1 - \overline{v_1}}{2} + b \cdot \frac{v_1 + \overline{v_1}}{2i} = \frac{1}{2} \left( (a - ib) v_1 -(a + ib) \overline{v_1} \right) = 0. $$
Since $v_1$ and $\overline{v_1}$ are linearly independent, we must have
$$ a - ib = a + ib = 0 \implies a = b = 0. $$
Let's see how $M$ acts on $u_1$:
$$ Mu_1 = \frac{\lambda_1 v_1 - \overline{\lambda_1} \overline{v_1}}{2} = \frac{\lambda_1 (u_1 + iu_2) - \overline{\lambda_1}(u_1 - iu_2)}{2} \\
= \frac{\lambda_1 - \overline{\lambda_1}}{2}u_1 + \frac{\lambda_1 + \overline{\lambda_1}}{2i} u_2 
= \cos \alpha \cdot u_1 + \sin \alpha \cdot u_2. $$
A similar calculation shows that
$$ Mu_2 = -\sin \alpha \cdot u_1 + \cos \alpha \cdot u_2. $$
This means that if you take $P \in M_2(\mathbb{R})$ whose columns are $u_1,u_2$ then $P$ will be invertible and
$$P^{-1} M P = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}. $$
