Okay so I'm reading a book that says we can express a matrix in $A \in SU(2)$ uniquely as $A=\begin{bmatrix} u & v\\ -\overline{v} & \overline{u}\\ \end{bmatrix}$
where $u,v \in \mathbb{C}$ and $|u|^{2}+|v|^{2}=1$.
He then goes on to say we can express the matrix $A$ in the form $A=\cos \theta I+S$ where $I$ is the identity matrix and $S$ is skew hermitian and Re $u = \cos \theta$ for $\theta \in [0,\pi]$.
Can someone explain why we can do this? In particular, I don't comprehend why, for an arbitrary complex number $u$, we can always get $\theta \in [0,\pi]$ and why is it $\cos \theta I$ rather than $r \cos \theta I$ where $r$ is the modulus of $u$?
In fact, I'm not sure I believe the author that you can actually do this...
The author is correct. Let $u={r(\cos \psi + i\sin \psi )}.$ Then we can write $A=r\cos \psi I+S$ where
$$S = \left( {\begin{array}{*{20}c} {ir\sin \psi } & v \\ { - \bar v} & { - ir\sin \psi} \\ \end{array}} \right)$$
Note that $r\cos\psi\in[-1,1]$ since $0 \le r = \left| u \right|=\sqrt {1 - \left| v \right|^2 } \le 1$, so we can write $\cos\theta=r\cos\psi$ for some $\theta\in [0,\pi]$ (since $\cos\theta$ is surjective over its range on $[0,\pi]$).