I know this or similar questions have been asked numerous times. But the answers seem quite advanced. This problem is an exercise in a set of notes in a course where the main text is Stillwell's "Naive Lie Theory," so I would appreciate help with a solution at that level.
Show the exponential map $\mathfrak{u}(n)\rightarrow U(n)$ is surjective
where $\mathfrak{u}(n)$ is the tangent space at the identity (in the next section to be referred to as the Lie algebra) of $U(n)$, the space of unitary matrices.
Earlier it was proved that this tangent space for $U(n)$ is the set of matrices $\lbrace X\in M_n(\mathbb{C}):X+X^{*}=0\rbrace$
There is a hint that if $A$ is a unitary matrix, then there is a unitary matrix $U$ and a diagonal matrix $D$ such that $A=UDU^{*}$.
So far, I would say if $A\in \mathfrak{u}(n)$ then $e^A=e^{UDU^{*}}\in U(n)$. And $e^{UDU^{*}}=Ue^{D}U^{*}$.
Thanks
It's unclear whether you're on track with your comment, but here's a full answer:
Take any $U \in U(n)$. By the spectral theorem for normal matrices, there exists a unitary matrix $V$ such that $U = V D_UV^*$, where $D$ is diagonal with $$ D_U = \pmatrix{e^{i \theta_1} \\ & e^{i \theta_2} \\ && \ddots \\ &&& e^{i \theta_n}} $$ Notably, we have $D_U = \exp(D_X)$ where $$ D_X = i\pmatrix{\theta_1 \\ & \theta_2 \\ && \ddots \\ &&& \theta_n} $$ Now, let $X = VD_XV^*$. We have $$ \exp(X) = \exp(VD_XV^*) = V \exp(D_X)V^* = V D_U V^* = U $$ And we find that $X \in \mathfrak u(n)$ since $$ X^* = (VD_XV^*)^* = VD_X^*V^* = -VD_XV^* = -X $$ So: for any $U \in U(n)$, there is an $X \in \mathfrak u(n)$ such that $\exp(X) = U$, which is to say that the exponential map is surjective.