Can someone direct me to prove the following three claims about a rotation group $SO(3)$ and its Lie Algebra? In class my prof made three claims about a group and its Lie algebra. I cannot find direct reference to these claims because they are delivered in verbatim (im not even sure if I have them jogged down 100% correctly). Can someone please help me identify these three properties?
Claim 1: $R\in SO(3)$, so any rotation matrix $R$ is the matrix exponential skew symmetric matrix $M$
Claim 2: Any skew symmetric 3x3 matrix is determined by 3 distinct numbers i.e. a vector
Claim 3: $\omega$ be a vector in $R^3$. Let $v$ = normalized $\omega$ (?) and $\theta$ = length of $\omega$. Then, $\omega$ = $v$ $\theta$. Now, it can be shown that if you take the identity matrix and rotate it by the angle $\theta$ around the axis span{$v$} using the right-hand rule, the result is precisely the matrix $R$ 
Please direct me to appropriate reference or if these are in fact simple proofs, please show me how these are true. Thank you for your time!
 A: Let's address your questions one at a time. We will start with question 2 since that is the simplest. 
$2$. The fact that every skew-symmetric matrix is determined by $3$ numbers is simply the statement that the dimension of the subspace of $3\times 3$ skew-symmetric matrices $\mathrm{Skew}_3$ has dimension $3$. Explicitly, we have the following basis:
$$\left\{\begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix},\ \begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{pmatrix},\ \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{pmatrix}\right\},$$
This means that every $3\times 3$ skew-symmetric matrix $A$ can be written in the form
$$A = \begin{pmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{pmatrix}.$$
In fact, we can say a bit more than this. Every skew-symmetric matrix in fact represents a cross product. Let $\mathbf{v} = (a,\ b,\ c)^\mathrm{T}$ and let $\mathbf{u}\in\mathbb{R}^3$. Then we have
$$A\mathbf{u} = \mathbf{v}\times \mathbf{u},$$
which you can verify directly through matrix multiplication. There is in fact a very useful isomorphism between the Lie algebras $(\mathrm{Skew}_3=\mathfrak{so}(3),\ [\cdot,\ \cdot])$ and $(\mathbb{R}^3,\ \times)$ through the above correspondence (where $[\cdot,\ \cdot]$ is the commutator and $\times$ is the cross-product).
Explicitly, let $\Omega: \mathbb{R}^3 \rightarrow \mathfrak{so}(3)$ be given by
$$\Omega(\mathbf{v}) = \begin{pmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{pmatrix},$$
where $a,b,c$ are the components of $\mathbf{v}$. Clearly this is an bijection, so we must show that this map preserves the respective products, i.e.
$$[\Omega(\mathbf{u}),\ \Omega(\mathbf{u})] = \Omega(\mathbf{u}\times\mathbf{v}).$$
It is simplest to do this by remembering that the action of $\Omega(\mathbf{v})$ on a vector $\mathbf{u}$ is given by the cross product $\Omega(\mathbf{v})\mathbf{u}=\mathbf{v}\times\mathbf{u}$ as mentioned earlier. With that in mind, let $\mathbf{w}\in\mathbb{R}^3$ be an arbitrary vector. Then we have
$$\begin{align}\Omega(\mathbf{u}\times\mathbf{v})\mathbf{w} &= (\mathbf{u}\times\mathbf{v})\times \mathbf{w}\\
&=-(\mathbf{v}\times\mathbf{w})\times\mathbf{u} - (\mathbf{w}\times\mathbf{u})\times\mathbf{v}\\
&=\mathbf{u}\times(\mathbf{v}\times\mathbf{w}) - \mathbf{v}\times (\mathbf{u}\times\mathbf{w})\\
&=\Omega(\mathbf{u})\Omega(\mathbf{v})\mathbf{w}-\Omega(\mathbf{v})\Omega(\mathbf{u})\mathbf{w}\\
&= [\Omega(\mathbf{u}),\ \Omega(\mathbf{v})]\mathbf{w},\end{align}$$
where in the second step we use the Jacobi identity for the cross product. Since this holds for arbitrary $\mathbf{w}\in\mathbb{R}$, we must have 
$$[\Omega(\mathbf{u},\ \Omega(\mathbf{v})] = \Omega(\mathbf{u}\times\mathbf{v}).$$
This shows that $\Omega$ is a Lie algebra isomorphism between $(\mathbb{R}^3,\times)$ and $(\mathfrak{so}(3),[\cdot,\ \cdot])$. The existence of this correspondence is precisely the reason why we are able to represent rotations using a single vector, the direction of which represents the action of rotation (with the direction given by the right-hand rule) and the length of which represents the angle of rotation.
Next, let's address your first question. 
$1$. There is a well known theorem that the map $\exp : \mathfrak{g}\rightarrow G$ is a surjective map for any compact, connected Lie group $G$. Since $SO(3)$ is compact and connected, this immediately answers your first question (in fact this argument shows that the statement holds true for $SO(n)$ for any $n$). But this theorem is not particularly easy to prove, so let me give a proof specifically for the case of $SO(3)$. In fact, we will do more and give an explicit formula for $\exp: \mathfrak{so}(3)\rightarrow SO(3)$. Let us write elements of $\mathfrak{so}(3)$ as we did previously in terms of the parameters $a,b,c\in\mathbb{R}$.
Theorem: The matrix exponential $\exp:\mathfrak{so}(3) \rightarrow SO(3)$ is explicitly given by
$$\exp(A) = I + \frac{\sin\theta}{\theta}A + \frac{(1-\cos\theta)}{\theta^2}A^2,$$
where $\theta = \sqrt{\mathrm{tr}(A^\mathrm{T}A)} = \sqrt{a^2+b^2+c^2}$.
Proof: Note that we have
$$A^3 = -\theta^2 A$$
for any $3\times 3$ skew-symmetric matrix. This allows us to iteratively determine the remaining powers of $A$ in terms of $A$ and $A^3$, for example:
$$A^4 = -\theta^2 A^2,\ \ A^5 = -\theta^2A^3 = \theta^4 A,\ \ A^6 = -\theta^4 A^2,$$
and in general for $k\ge 0$ we have
$$A^{4k+1} = \theta^{4k}A,\ \ A^{4k+2} = \theta^{4k}A^2,\ \ A^{4k+2} = -\theta^{4k+2}A,\ \ A^{4k+4} = -\theta^{4k+2}A^2$$
Then we can calculate the matrix exponential through the power series:
$$\begin{align} \exp(A) &= \sum_{n=0}^\infty \frac{A^n}{n!}\\
&=I + \sum_{k=0}^\infty \left(\frac{A^{4k+1}}{(4k+1)!} + \frac{A^{4k+2}}{(4k+2)!} + \frac{A^{4k+3}}{(4k+3)!} + \frac{A^{4k+4}}{(4k+4)!}\right)\\
&=I + \sum_{k=0}^\infty \bigg(\frac{1}{\theta}\frac{\theta^{4k+1}}{(4k+1)!}A +\frac{1}{\theta^2}\frac{\theta^{4k+2}}{(4k+2)!}A^2\\ 
&\ \ \ \ \ -\frac{1}{\theta}\frac{\theta^{4k+3}}{(4k+3)!}A - \frac{1}{\theta^2}\frac{\theta^{4k+4}}{(4k+4)!}A^2\bigg)
\end{align}$$
Notice that the even terms in the above sum are precisely the power series for $\frac{1-\cos \theta}{\theta^2}A^2$ and that the odd terms are precisely the power series for $\frac{\sin\theta}{\theta}A$. Therefore we must have
$$\exp(A) = I + \frac{\sin\theta}{\theta}A + \frac{(1-\cos\theta)}{\theta^2}A^2,$$
as required. $\square$
The above theorem has a nice interpretation, because it is simply the statement of Rodrigues's formula, which shows that the above formula is surjective as a map from $\mathfrak{so}(3)$ to $SO(3)$ (see the link for details).
For your third question, I honestly have no idea what it means to rotate the identity matrix. If you can provide context and information, I'd be happy to provide an answer (assuming I know it of course).
