# The form of $2 \times 2$ unitary matrices

I've been working through "Groups and Symmetry" (Armstrong) and came across this problem in chapter 9 which I can't figure out. Any hints/help would be greatly appreciated!

Show that every $2\times 2$ unitary matrix has the form

$$\left(\begin{array}{c c} w & z \\ -e^{i \theta} z^{*} & e^{i \theta} w^{*} \end{array}\right)$$

for some $\theta\in\mathbb{R}$ and $w,z\in\mathbb{C}$. (A matrix is said to be unitary if it is invertible with its adjoint as the inverse. The symbol "*" denotes complex conjugate.)

• One more restriction is necessary, that $ww^* + zz^* =1$ Jan 4, 2013 at 4:17
• @adamW This restriction is not "necessary" here, since the OP only asked for a proof that if the matrix is unitary then it has this form for some $\theta,w,z.$ He did not ask for a proof of the converse, where that restriction would be necessary. Jul 12, 2023 at 12:43

Start with the facts you know, i.e. that you have a 2-by-2 complex matrix $\begin{pmatrix}w&z\\c&d\end{pmatrix}$, such that when you multiply it by its adjoint $\begin{pmatrix}w^*&c^*\\z^*&d^*\end{pmatrix}$ you get $\begin{pmatrix}1&0\\0&1\end{pmatrix}$. That means you have $ww^*+zz^* = 1$, $cc^*+dd^* = 1$ and $cw^*+dz^*=0$. Don't forget the other way, so you get $ww^*+cc^* = 1$, $zz^*+dd^* = 1$ and $w^*z+c^*d=0$. With these equations you should notice something immediately about both $cc^*$ and $dd^*$. You can work from there.

Adding a faster approach than the accepted answer since this post is the first result returned by Google.

For $$U := \left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$$ with $$a,b,c,d\in\mathbb{C}$$, we instead compare the matrix entries in the defining unitary equation $$U^* = U^{-1}$$.

Equating $$U^{-1} = \frac{1}{\det U} \left(\begin{smallmatrix}d&-b\\-c&a\end{smallmatrix}\right)$$ with the adjoint (conjugate ($$\overline{a}$$) transpose) of $$U$$ entrywise yields

$$d = \overline{a} \cdot \det U,\ \ a = \overline{d} \cdot \det U,\ \ \text{and}\ \ -b = \overline{c} \cdot \det U.\quad \quad \quad (1)$$

The first two equations give $$d = d \cdot \overline{\det U} \cdot \det U = d \cdot \|\det U\|^2$$ so for $$d\neq 0$$ this forces $$\|\det U\|^2=1$$, thus the scaling factor is $$\det U = e^{i\theta}$$. As $$\det U\in\mathbb{C}$$, this is true because for any complex number $$z\in\mathbb{C}$$, having unit modulus $$\|z\|=1$$ implies $$z$$ is on the unit circle, and so $$z$$ is determined by the angle $$\theta$$ such that $$z=e^{i\theta}$$. [Alternatively and more simply, use $$\det\left(U U^\dagger\right)=1$$ as suggested by @Icv.]

Now substituting $$\det U = e^{i\theta}$$ back into $$(1)$$ yields $$c = -e^{i\theta} \overline{b}\ \ \text{and}\ \ d = e^{i\theta} \overline{a},$$ hence $$U = \begin{pmatrix} a & b\\ -e^{i\theta} \overline{b} & e^{i\theta} \overline{a} \end{pmatrix}.$$

$$\rule{19cm}{0.4pt}$$

Zero-division technicality: We can't have both $$d$$ and $$c$$ equal to $$0$$ for $$U$$ unitary as this would give singular determinant $$\det U = ad-bc = 0$$. We know this because assuming otherwise leads to a contradiction: by definition of $$U$$ being unitary, we know that $$U^{-1}=U^*$$ exists, so in particular $$U$$ is invertible and in turn $$\det U \neq 0$$... contradiction. Thus if $$d=0$$ occurs, for $$U$$ unitary we know that $$c \neq 0$$, so we can argue just as we did but instead deriving $$\|\det U\|=1$$ using $$-c = \overline{b} \cdot \det U$$ and $$-b = \overline{c} \cdot \det U$$, rather than the first two equations of Eq. $$(1)$$.

• I upvoted your answer as I think this is a faster method (actually to obtain $|\det U|=1$ it's even simpler to take the det of $U^\dagger U =I$). But your expression for the inverse is wrong, and you forgot to take the transpose. The two errors cancel each other
– lcv
Jun 24, 2022 at 16:47
• Thank you, I have now updated the answer to reflect these improvements. Jun 25, 2022 at 22:44