# Proof that U(n) is a group

1. Nov 6, 2015

### spaghetti3451

1. The problem statement, all variables and given/known data

Show that the set of all $n \times n$ unitary matrices forms a group.

2. Relevant equations

3. The attempt at a solution

For two unitary matrices $U_{1}$ and $U_{2}$, $x'^{2} = x'^{\dagger}x' = (U_{1}U_{2}x)^{\dagger}(U_{1}U_{2}x) = x^{\dagger}U_{2}^{\dagger}U_{1}^{\dagger}U_{1}U_{2}x = x^{\dagger}U_{2}^{\dagger}U_{2}x = x^{\dagger}x = x^{2}.$

So, closure is obeyed.

Matrix multiplication is associative.

The identity element is the identity matrix.

$x'^{2} = (U^{-1}x)^{\dagger}(U^{-1}x) = x^{\dagger}(U^{-1})^{\dagger}U^{-1}x = x^{\dagger}(U^{\dagger})^{-1}U^{-1}x = x^{\dagger}(UU^{\dagger})^{-1}x = x^{\dagger}x = x^{2}$.

So, the inverse of any unitary matrix is a unitary matrix.

Is my answer correct?

Last edited: Nov 6, 2015
2. Nov 6, 2015

### Staff: Mentor

yep, and if you mention that the identity is clearly unitary and the inverse exists at all, e.g. because |det(U)| = 1 it'll be perfect

3. Nov 6, 2015

### spaghetti3451

Thanks! Got it!