Orthogonal matrices Ok i'll reformulate my question. The real thing I have to prove is that for any M in SO(3), there is a basis $e_1 , e_2, e_3$ and a real $\alpha$ such as:
M$e_1$ = cos$\alpha$$e_1$ + sin$\alpha$$e_2$
M$e_2$ = -sin$\alpha$$e_1$ + cos$\alpha$$e_2$
M$e_3$ = $e_3$
And I wanted to know if that was equivalent to say that there is a matrix P in SO(3) and a real \alpha such as 
PM$P^-1$ = $$
        \begin{matrix}
        cos\alpha  & sin\alpha & 0 \\
        -sin\alpha & cos\alpha & 0 \\
             0       &     0       & 1 \\
        \end{matrix}
$$
And if not, what is the link with such a matrix?
 A: $\newcommand{\Basis}{\mathbf{e}}$It's true that if $M \in SO(3)$, there exists a real number $\alpha$ and an (ordered) orthonormal basis $(\Basis_{i})_{i=1}^{3}$ such that
\begin{align*}
M\Basis_{1} &= \phantom{-}(\cos\alpha)\Basis_{1} + (\sin\alpha)\Basis_{2}, \\
M\Basis_{2} &= -(\sin\alpha)\Basis_{1} + (\cos\alpha)\Basis_{2}, \\
M\Basis_{3} &= \Basis_{3}.
\end{align*}
(Prahlad Vaidyanathan's comment is perfectly correct for the wording of the question. But again, more is true than is stated in the question: orthonormality of the basis $(\Basis_{i})_{i=1}^{3}$ guarantees $P \in O(3)$. Further, if necessary you can replace $\Basis_{2}$ by $-\Basis_{2}$ and replace $\alpha$ by $-\alpha$ to ensure $P \in SO(3)$.)
If $P = \left[\begin{array}{@{}ccc@{}}\Basis_{1} & \Basis_{2} & \Basis_{3}\end{array}\right]$ is the $3 \times 3$ matrix whose $i$th column is $\Basis_{i}$, then $P$ is an orthogonal matrix, i.e., $P^{-1} = P^{t}$, and
$$
P^{-1}MP = P^{t}MP = \left[\begin{array}{@{}ccc@{}}
    \cos\alpha & -\sin\alpha & 0 \\
    \sin\alpha & \phantom{-}\cos\alpha & 0 \\
    0 & 0 & 1 \\
  \end{array}\right].
\tag{1}
$$
(These assertions do require proof, but the wording of your question only requests clarification of what is true. Everything above should be covered in a first course in linear algebra.)
As for your original question about rotations in $n$-dimensional space, if $M \in SO(n)$, it turns out there exist orthogonal unit vectors $\Basis_{1}$ and $\Basis_{2}$ such that $M$ acts on the plane $\operatorname{Span}(\Basis_{1}, \Basis_{2})$ as rotation by some real number $\alpha$, and $M$ stabilizes the orthogonal complement (see G. C.'s comment). One can therefore do induction on dimension, expressing $P^{-1}MP$ is a form analogous to (1). The story differs depending whether $n$ is even or odd; Wikipedia has a good discussion. But arguably, our intuition of the nature of "rotation" as a geometric operation, shaped by experience with dimensions $2$ and $3$, differs substantially from mathematical actuality. That's likely why you had so much trouble trying to generalize to arbitrary dimension. Particularly:


*

*"Rotating four-dimensional space in a particular direction generally does not return to the identity matrix." Precisely, if $n \geq 4$, then $SO(n)$ contains non-compact $1$-parameter subgroups. For example, if $R(\alpha)$ denotes the upper left-hand $2 \times 2$ block in (1), then the $4 \times 4$ block diagonal matrices
$$
R(t) \oplus R(\sqrt{2}t)
  = \left[\begin{array}{@{}cc@{}}
    R(t) & 0 \\
    0 & R(\sqrt{2}t) \\
  \end{array}\right]
$$
form a non-compact subgroup of $SO(4)$; the image is a curve of irrational slope in a torus.

*An $n$-dimensional rotation is guaranteed to have an axis if and only if $n$ is odd. (At Math.SE, one occasionally comes across questions of the type, "What (hyper-)volume is swept out when some solid is revolved about an axis in $n$-dimensional space?" Sounds reasonable, but on closer inspection doesn't make sense.)
