Unitary $2 \times 2$ matrix - Question

Question

I am trying to prove that every unitary $2 \times 2$ matrix has the form \begin{bmatrix} a & -\overline{b} \\ b & \overline{a} \end{bmatrix} where $a, b \in \mathbb{C}$ such that $|a|^{2} + |b|^{2} = 1$

Assuming that $U$ is a unitary matrix, then it follows $$U^{*}U = I = UU^{*} $$ $$\pmatrix{x & y\\w & z} \pmatrix{\overline{x} & \overline{w}\\\overline{y} & \overline{z}} = \pmatrix{1 & 0\\0 & 1}$$ which means $$x\overline{x}+w\overline{w} = |x|^{2} + |w|^{2}= 1$$

From matrix multiplication above we also get the equations: $$w\overline{x} + z\overline{y} = 0$$ $$x\overline{w} + y\overline{z} = 0 $$ I kept trying different substitutions, but I am still stuck at proving that $z = \overline{x}$ and $y = -\overline{w}$

Help?

Answer 1

We have

$$\pmatrix{\overline{x} & \overline{w} \\ \overline{y} & \overline{z}} = U^*= U^{-1} = \frac1{\det U}\pmatrix{z & -y \\ -w & x}$$

so $z = (\det U) \overline{x}$ and $y = -(\det U) \overline{w}$, where $\left|\det U\right| = 1$.

Indeed, $\det U$ can be any $\lambda$ with $|\lambda| = 1$, just consider $U = \pmatrix{1 & 0 \\ 0 & \lambda}$.

Answer 2

You can't prove that because it is not true. All the matrices of the form$$\begin{bmatrix}a&-\overline b\\b&\overline a\end{bmatrix}$$such that $|a|^2+|b|^2=1$ have determinant $1$. The unitary $2\times2$ matrices are of the form$$\begin{bmatrix}a&-\omega\overline b\\b&\omega\overline a\end{bmatrix},$$with $|a|^2+|b|^2=1$ and $|\omega|=1$.