Behavior of Matrices in SO(3) 
Show that each matrix in SO$(3)$ equals $e^X$ for some skew-symmetric $X$.

Here, SO(3) refers to the special orthogonal group that is the rotation group of $\mathbb{R}^3$. I am also supposed to use the following facts that I have derived in proving the above result.

*

*Given $B = \begin{pmatrix} 0 & -\theta & 0 \\ \theta & 0 & 0 \\ 0 &0 &0 \end{pmatrix}$, $e^B =\begin{pmatrix}
\cos \theta & -\sin \theta & 0\\
\sin \theta & \cos \theta & 0\\
0 & 0 & 1
\end{pmatrix}$, where $e^B$ denotes the exponential of $B$. (Note that this is what I calculated the exponential to be - is this correct?)


*For any orthogonal matrix $A$, we have $Ae^BA^T = e^{ABA^T}$. (I have already proven this result.)
Again, I need to use results 1 and 2 in the proof of my question, but I'm not seeing how these two results combine to show my desired result. Since $e^B$ is one of the 3-dimensional rotation matrices (assuming that I calculated my exponential correctly), do I need to make some sort of argument based on rotation? Or is there a simpler way?

Let $g\in SO(3,\mathbb{R})$ be any element.
(1) We first show that $1$ is an eigenvalue of $g$. Consider the characteristic polynomial $f_g$ of $g$. It is clear that $f_g\in \mathbb{R}[x]$ and $\deg(f_g)=3$. Thus $f_g$ has a real root, say $\lambda_1$. Let $\alpha_1$ be the unit eigenvalue of $\lambda_1$. Since $g$ preserve the length, we get $\lambda_1^2=\pm 1$. If $\lambda_1=-1$, let $\lambda_2, \lambda_3$ be the other complex eigenvalues. Then $\lambda_2\lambda_3=-1$. Since $\lambda_2,\lambda_3$ satisfies a monic real quadratic equation $x^2+ax-1,$ which has only real roots, we have $\{\lambda_2,\lambda_3\}=\{\pm 1 \}$. Thus $1$ is an eigenvalue of $g$.
(2) Let $\alpha_1$ be an unit eigenvector of $g$ and let $W=Span\{\alpha_1\}$. Consider $W^{\perp}$, which has dimension 2 and is invariant under $g$. Thus $g|_{W^{\perp}}\in SO(2,\mathbb{R})$. Thus there exists orthonormal basis $\alpha_2,\alpha_3$ and an angle $\theta$ such that
$$g\begin{pmatrix}\alpha_2 \\ \alpha_3 \end{pmatrix}=\begin{pmatrix}\alpha_2 & \alpha_3 \end{pmatrix}\begin{pmatrix} \cos(\theta)&\sin(\theta)\\ -\sin(\theta)& \cos(\theta)\end{pmatrix}.$$
The rest is easy.
