How to see that $PSU(2)$ is same as $SO(3)$? Some background: 
We have an action of $SU(2)$ on the space of traceless Hermitian matrices, $\mathcal{H}$, via conjugation: $$SU(2)\times \mathcal{H}\to \mathcal{H}, \ (U,H)\mapsto UHU^{-1}.$$
The ineffectivity, $I$ , of this action is $I=\{+Id, -Id\}$.
We define $PSU(2)$ as the quotient of $SU(2)$ over this ineffectivity i.e. $$PSU(2):= SU(2)/\{+Id, -Id\}$$
We thus have an effective action $$PSU(2)\times \mathcal{H} \to \mathcal{H}, \ (UI, H)\mapsto UHU^{-1}.$$
Now, I want to show that $PSU(2)=SO(3).$ I am reading a file that says above effective action implies that we have $PSU(2)$ naturally as a subgroup of $SO(3)$. I don't understand this. 
Any help is appreciated.   
 A: Let me be more explicit: suppose you have an element
$$
\begin{bmatrix}
z & -\bar{w} \\
w & \bar{z}
\end{bmatrix}
$$
of $SU(2)$. Then $\|w\|^2 + \|z\|^2 = 1$. Writing $z = a + bi$ and $w = c + di$, you can form the vector $v = (b, c, d) / \sqrt{b^2 + c^2 + d^2}$ and the angle $\theta = \arccos(2a)$. 
Rodrigues' formula, applied to the axis-angle pair $(v, \theta)$ gives an element of $R(v, \theta)$ of $SO(3)$. If you look at the formula, it's clear that $R(-v, -\theta) = R(v, \theta)$, so this mapping passes to the quotient, providing a mapping from $PSU(2)$ to $SO(3)$. It takes only a little work to show that it's 1-1, and you're done. 
A: You constructed an action $PSU(2) \to GL(H)$. Using the fact that $H$ is $3$-dimensional, it's easy to show that your map actually defines an action $\rho:PSU(2) \to SO(3)$. The fact that it's effective means that for any $g\not = g'$ in $PSU(2)$, there exists some $x\in H$ such that $\rho(g)x \not = \rho(g')x$; that is, the maps $\rho(g), \rho(g')\in SO(3)$ differ. That just means that the map $\rho:PSU(2)\to SO(3)$ is injective and thus gives an embedding of $PSU(2)$ as a subgroup of $SO(3)$.
(To complete the proof and show that that subgroup is actually $SO(3)$, there are a couple of ways to proceed: show explicitly that suitable generators of $SO(3)$ lie in the image, use some Lie group machinery, etc.)
