# Show the exponential map $\mathfrak{u}(n)\rightarrow U(n)$ is surjective

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

• Look at the eigenvalues of a unitary matrix $U$—does the polar form of non-zero complex numbers suggest anything to you? – Branimir Ćaćić May 11 '17 at 0:21
• @BranimirĆaćić The eigenvalues of a unitary matrix have modulus equal to $1$. So the polar forms would $e^{i\theta_n}$. And thus the conjugate transpose would have eigenvalues $e^{-i\theta_n}$. And since $U$ is diagonalizable and $D$ is diagonal, then $UDU^{*}$ will just be $D$. Hope this is a correct step in the right direction? Thanks, – user12802 May 11 '17 at 1:23

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.
• It is notable that in general, the exponential map of $G$ is surjective whenever $G$ is a compact, connected Lie Group. $U(n)$ is indeed both compact and connected. – Omnomnomnom May 11 '17 at 1:39