The task is to find a kernel of representation
$P: SU_2 → \text{Aut}_\mathbb{R}(E), P(A) → X → AXA^{−1}$, where $E$ is traceless Hermitian matrix $$\begin{bmatrix}x_1&x_2+ix_3\\x_2+ix_3&-x_1\end{bmatrix}$$ where $x_1, x_2, x_3$ are in $\mathbb{R}$.
As far as I understand, I am supposed to find $A$ such that $AXA^{−1} = X$ for each $X$. How can I proceed with that? Thank you.
Question: "Are you suggesting that if the representation is irreducible, kernel is trivial, right? Since there is a correspondence between irreducible repre. and simple modules."
Answer:
Lemma: If $\mathfrak{g}$ is a Lie algebra and if $\rho$ is any $\mathfrak{g}$-module it follows $ker(\rho) \subseteq \mathfrak{g}$ is an ideal. Hence if $\mathfrak{g}$ is simple (it has no non-trivial ideals) it follows $ker(\rho)=(0)$.
Proof: Let $x\in ker(\rho)$ and let $y \in\mathfrak{g}$. It follows $\rho([y,x])=[\rho(y), \rho(x)]=[\rho(y),0]=0$ hence $[\mathfrak{g},ker(\rho)] \subseteq ker(\rho)$. Hence if $\mathfrak{g}$ is simple it follows $ker(\rho)=(0)$.
The Lie group $SU(n)$ is simple and so is its Lie algebra $\mathfrak{su}(n)$.