I'm working on this problem:
Show that a matrix in $SU(2)$ always can be expressed in the form:
$$ A =
\begin{bmatrix}
\exp(i\psi)\cos(\theta) & \exp(i\psi)\sin(\theta) \\
-\exp(-i\psi)\sin(\theta) & \exp(-i\psi)\cos(\theta)
\end{bmatrix}
$$
where $\psi,\theta \in \Bbb R$, and $SU(2)$ is a set of $2 \times 2$ unitary matrix $A$ such that $\det(A) = 1$ .
Noticing that
$$A =
\begin{bmatrix}
\exp(i\psi) & \\
& \exp(-i\psi)
\end{bmatrix}
\begin{bmatrix}
\cos(\theta) & \sin(\theta) \\
-\sin(\theta) & \cos(\theta)
\end{bmatrix}
$$
After reading some useful conclusion (about eigenvalues and general form), I tried solving it by diagonalization, but got stuck on why there necessarily exist a real eigen basis for any matrix in $SU(2)$?
Moreover, is there a better way to prove the statement mentioned above?
Best Answer
This is false. Indeed, $\mathrm{SU}(2)$ is diffeomorphic to the sphere $S^3\subset\mathbb{C}^2=\mathbb{R}^4$ which is three-dimensional and hence there is no smooth surjection $\mathbb{R}^2\to\mathrm{SU}(2)$ by Sard's theorem.
In other words, Sard's theorem tells us that the set of elements of the form $$\begin{pmatrix}e^{i\psi}\cos\theta & e^{i\psi}\sin\theta \\ -e^{-i\psi}\sin\theta & e^{-i\psi}\cos\theta\end{pmatrix},\quad\psi,\theta\in\mathbb{R}$$ has measure zero in $\mathrm{SU}(2)$.
On the other hand, by the Hopf parametrization of the three-sphere, we can write every element $(\alpha,\beta)\in\mathbb{C}^2$ such that $|\alpha|^2+|\beta|^2=1$ as $$\alpha=e^{i\psi_1}\cos\theta,\quad\beta=e^{i\psi_2}\sin\theta$$ for $\psi_1,\psi_2,\theta\in\mathbb{R}$ and hence every element of $\mathrm{SU}(2)$ can be written $$A=\begin{pmatrix}\alpha & \beta \\ -\overline{\beta} & \overline{\alpha}\end{pmatrix}=\begin{pmatrix}e^{i\psi_1}\cos\theta & e^{i\psi_2}\sin\theta \\ -e^{-i\psi_2}\sin\theta & e^{-i\psi_1}\cos\theta \end{pmatrix}.$$