$SU(2)$ acting by conjugation, decomposition into irreducibles I am attempting past exam questions of the Cambridge Math Tripos. I know how to do the first few parts, which involves giving the irreducible representations of $U(1)$ and $SU(2)$. But I am not sure on how to do the last part. "Let $SU(2)$ act on the space $M_3(\mathbb{C})$ of $3 \times 3$ matrices by conjugation, where $A \in SU(2)$ acts by $$A: M \mapsto A_1MA_1^{-1},$$in which $A_1$ denotes the $3 \times 3$ block diagonal matrix $\left( \begin{array}{ccc} A & 0\\\ 0 & 1 \end{array} \right)$. Show that this gives a representation of $G$ and decompose it into irreducibles." For starters, I was told by my supervisor this representation is $9$ dimensional. Any help would be appreciated.

We give a more elementary solution than the one already provided. Let $V$ denote the representation on $M_3(\mathbb{C})$. Every $A \in SU(2)$ is conjugate to a diagonal matrix $D = \text{diag}(e^{i\theta}, e^{-i\theta})$ so $\chi_V(A) = \chi_V(D)$. For $X \in M_3(\mathbb{C})$, written as $X = (x_1, x_2, \dots, x_9)$ by concatenating the rows of the matrix, we may compute that$$D \cdot X = (x_1, z^2 \cdot x_2, z \cdot x_3, z^{-1} \cdot x_4, x_5, z^{-1} \cdot x_6, z^{-1} \cdot x_7, z \cdot x_8, x_9)$$where $z = e^{i\theta}$. Hence$$\chi_V(D) = 3 + z^2 + z^{-2} + 2(z + z^{-1}) = 3 + 2\cos 2\theta + 4\cos\theta = (2\cos\theta + 1)^2 = \chi_{(V_\text{triv} \oplus V_1)^{\otimes 2}}.$$Now, using the Weyl integration formula, we can do some annoying computations and decomopse$$V \cong 2V_{\text{triv}} \oplus 2V_1 \oplus V_2.$$