Finding the tangent space of $Z(U(n))$

Question

Recall that the center of the matrix unitary group for a given $n\in \mathbb N$ is\begin{equation*}Z(U(n))=\{\omega I:|\omega|=1\}\end{equation*} I'm trying to find the tangent space at the identity of $Z(U(n))$. so far I've been able to show that the space \begin{equation*}J=\{\alpha iI:\alpha \in \mathbb R\}\end{equation*}contained in the tangent space (via the exponential map) and is closed under Lie bracket. Is there a way to show the other way around?

Answer 1

Since $\{\omega:|\omega|=1\}\cong S^1$ is one-dimensional, your one-dimensional subspace contained in the one-dimensional identity tangent space must be the whole tangent space.