# General form of $SU(2)$ matrix

I am pretty sure it's duplicate but on the other hand it's hard to find by a title.

I am having hard time showing that every matrix in $$SU(2)$$ is of the form $$\begin{pmatrix}z & -\overline w \\ w & \overline z\end{pmatrix}$$ for some $$z, w \in \mathbb{C}$$ such that $$|z|^2 + |w|^2 =1$$. In most of the sources (including answers here) it is stated as something well-known. It is indeed well-known but I have never seen a proof and couldn't come up with one.

• Seems like you also need $|z|^2+|w|^2=1$, right? For instance $\begin{bmatrix}2&0\\0&2\end{bmatrix}$ isn't in $\mathrm{SU}(2)$. – carmichael561 May 19 at 20:04
• Hint: Do you know the explicit form of the inverse of $\begin{pmatrix}a&b\\c&d\end{pmatrix}$? – Toffomat May 19 at 20:06

Let $$\begin{bmatrix}x_1 & x_2\\x_3 & x_4\end{bmatrix}\in SU(2),$$ then $$x_1\cdot x_4-x_3\cdot x_2=1$$ because determinant is $$1$$(this you did not mention) and $$\begin{bmatrix}\overline{x_1} & \overline{x_3}\\\overline{x_2} & \overline{x_4}\end{bmatrix}=\begin{bmatrix}x_4 & -x_2\\-x_3 & x_1\end{bmatrix}$$ so $$x_1=\overline{x_4}$$ and $$\overline{x_2}=-x_3.$$ Notice that the other $$2$$ equations are just manipulations of these two. Check that left hand side is the inverse of the matrix and the left hand side is the adjoint.
Note that $$\text{SU}(2):={\{A\in M(2,\mathbb{C}): \langle Ax,Ay\rangle,\; \forall x,y\in\mathbb{C}^2,\;\text{det}(A)=1}\}$$ where $$\langle x,y\rangle:=x_1\bar{y_1}+x_2\bar{y_2}.$$ Then $$\text{SU}(2):={\{A\in M(2,\mathbb{C}): AA^*=A^*A=I_2,\;\text{det}(A)=1}\}$$ where $$A^*=\overline{(A^{t})}.$$
So for all $$A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in\text{SU}(2)$$ It is true that $$A^{-1}=A^{*}\;\text{and}\;\text{det}(A)=1$$:
$$A^{-1}=\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$$ and $$A^{*}=\begin{pmatrix}\overline{a} & \overline{c} \\ \overline{b} & \overline{d}\end{pmatrix}$$ hence $$A^{-1}=A^*$$ implies that $$d=\overline{a}\;\text{and}\; b=-\overline{c}.$$ Then $$A=\begin{pmatrix}a & -\overline{c} \\ c & \overline{a}\end{pmatrix}.$$