Let $A$ and $B$ be complex $n \times n$ matrices.

(i) The commutator of $A$ and $B$ is defined to be

$$[A, B] \equiv A B-B A$$

Show that $[A, A]=0 ;[A, B]=-[B, A] ;$ and $[A, \lambda B]=\lambda[A, B]$ for $\lambda \in \mathbb{C}$. Show further that the trace of $[A, B]$ vanishes.

(ii) A skew-Hermitian matrix $S$ is one which satisfies $S^{\dagger}=-S$, where $\dagger$ denotes the Hermitian conjugate. Show that if $A$ and $B$ are skew-Hermitian then so is $[A, B]$.

(iii) Let $\mathcal{M}$ be the linear map from $\mathbb{R}^{3}$ to the set of $2 \times 2$ complex matrices given by

$$\mathcal{M}\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=x M_{1}+y M_{2}+z M_{3}$$

where

$$M_{1}=\frac{1}{2}\left(\begin{array}{cc} i & 0 \\ 0 & -i \end{array}\right), \quad M_{2}=\frac{1}{2}\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right), \quad M_{3}=\frac{1}{2}\left(\begin{array}{cc} 0 & i \\ i & 0 \end{array}\right) \text {. }$$

Prove that for any $\mathbf{a} \in \mathbb{R}^{3}, \mathcal{M}(\mathbf{a})$ is traceless and skew-Hermitian. By considering pairs such as $\left[M_{1}, M_{2}\right]$, or otherwise, show that for $\mathbf{a}, \mathbf{b} \in \mathbb{R}^{3}$,

$$\mathcal{M}(\mathbf{a} \times \mathbf{b})=[\mathcal{M}(\mathbf{a}), \mathcal{M}(\mathbf{b})]$$

(iv) Using the result of part (iii), or otherwise, prove that if $C$ is a traceless skewHermitian $2 \times 2$ matrix then there exist matrices $A, B$ such that $C=[A, B]$. [You may use geometrical properties of vectors in $\mathbb{R}^{3}$ without proof.]