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.)

  • 8
    $\begingroup$ One more restriction is necessary, that $ww^* + zz^* =1$ $\endgroup$
    – adam W
    Jan 4, 2013 at 4:17
  • $\begingroup$ @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. $\endgroup$ Jul 12, 2023 at 12:43

2 Answers 2


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 $$ U = \begin{pmatrix} a & b\\ -e^{i\theta} \overline{b} & e^{i\theta} \overline{a} \end{pmatrix}. $$


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. $(1)$.

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    $\begingroup$ 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 $\endgroup$
    – lcv
    Jun 24, 2022 at 16:47
  • $\begingroup$ Thank you, I have now updated the answer to reflect these improvements. $\endgroup$
    – foam78
    Jun 25, 2022 at 22:44

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