By reading some notes about the relationship between SU(2) and SO(3) (link to pdf), I came across the following statement:
Given any $A\in SU(2)$, the operator $R(A)$ defined as $$R(A)_{jk}=\frac{1}{2}\operatorname{Tr}(\sigma_j A\sigma_k A^\dagger),$$ where $\sigma_j$ are the Pauli matrices, belongs to $SO(3)$.
I can verify this to be true without too many problems by direct calculations, but this leaves me wondering as to where does this expression come from.
Is there an easy way to see why this particular expression should be a viable homomorphism of $SU(2)$ onto $SO(3)$? Or how one would think of building this particular expression?
Here is what I managed to understand so far.
Let $X$ be the traceless Hermitian matrix with decomposition $X=x_i\sigma_i$, where $\sigma_i$ are the Pauli matrices, and consider an arbitrary $A\in SU(2)$.
A natural way for $A$ to act on $X$ is by conjugation on the Pauli matrices. Another possibility is via its natural action on $\boldsymbol x$ as the $SO(3)$ matrix obtained through the $SU(2)\to SO(3)$ homomorphism.
Acting on $X$ by conjugation gives $$x_i A\sigma_i A^{-1}=\frac{1}{2}\operatorname{Tr}(A\sigma_i A^{-1}\sigma_j)x_i\sigma_j=[R(A)\boldsymbol x]_j \sigma_j,$$ where I defined in the last step the matrix $$R(A)_{ji}\equiv \frac{1}{2}\operatorname{Tr}(\sigma_jA\sigma_i A^{-1}).$$
Thus if we want the different actions of $A$ on $X$ via conjugation on $SU(2)$ and via natural action of $SO(3)$ to match, we need to map $A$ into an $SO(3)$ matrix as given above. I'm still not clear why this becomes the homomorphism in $SU(2)\to SO(3)$ that we are looking for though.