# Decomposition of unitary matrix into phase and special unitary

I was reading a book on Quantum Field Theory when I came across the statement

$$U=e^{i\phi}\tilde{U}$$

where $U$ is an arbitrary $N\times N$ unitary matrix and $\tilde{U}$ is a Special Unitary $N\times N$ matrix.

I have not been able to prove this fact, nor find any mention of it anywhere else.

I agree this decomposition satisfies the required unitarity condition and satisfies $|{\rm det}U|=1$ as required, but is this decomposition unique? i.e., given $U\in U(N)$, do we have a unique $\phi\in\mathbb{R}$ and $\tilde{U}\in{\rm SU}(N)$?

## 1 Answer

Nevermind, figured it out.

$|{\rm det}U|=1$ so ${\rm det}U=e^{i\phi}$, $\phi\in\mathbb{R}$.

Define $\tilde{U}=e^{-i\phi/N}U$. Clearly this is unitary.

${\rm det}\tilde{U}=e^{-i\phi}{\rm det}U=1$

So we have $\tilde{U}\in SU(N)$.

Hence we have what we wanted to show; $U=e^{i\phi/N}\tilde{U}$, and the decomposition is unique by construction.

More generally, $$U=({\rm det}U)^{1/N}\tilde{U}$$