Let $\mathfrak{su}(2)$ be the Lie algebra of the Lie group $SU(2)$.
If $g\in SU(2)$ I was wondering what is the form of $Ad(g):\mathfrak{su}(2)\to\mathfrak{su}(2)$ defined as the tangent map of $i_g:SU(2)\to SU(2)$, given by $i_g(x)=gxg^{-1}$, at the identity element.
I'd like to know this because I want to prove that the image of $Ad$ is in the orthogonal group $O(3)$, viewed as the group of locally invertible maps $\theta:\mathfrak{su}(2)\to\mathfrak{su}(2)$ that preserve the Killing form, and specifically that it is the connected component of the identity of $O(3)$.
I have already seen that $\mathfrak{su}(2)$ is a real vector space with base the Pauli matrices, and that $O(3)$ can be viewed as the group of locally invertible maps $\theta:\mathfrak{su}(2)\to\mathfrak{su}(2)$ that preserve the Killing form but I don't know how to calculate explicitly the form of $Ad(g)$ in order to prove the inclusion of $Ad(SU(2))$ in $O(3)$.
Thanks
Let $G$ be a subgroup of $\mathrm{GL}(n,\mathbb{C})$. Then, for all $X\in\mathrm{Lie}(G)\subseteq\mathrm{Mat}(n,\mathbb{C})$ and $g\in G$ we have $$\mathrm{Ad}(g)X=(di_g)_1(X)=\frac{d}{dt}\Big|_{t=0}i_g(e^{tX})=\frac{d}{dt}\Big|_{t=0}ge^{tX}g^{-1}=g\left(\frac{d}{dt}\Big|_{t=0}e^{tX}\right)g^{-1}=gXg^{-1}.$$