Elements in $U(n)$

Question

I'm following John Stillwell Naive Lie Theory book. In chapter 5, section 5.3 there's the following theorem and proof: Proof concerning the tangent space of SU(n)

In the middle of the theorem there is the following statement: ``We know that every $A \in U(n)$ is of the form $\exp(X)$ with $X+X^*= 0$.'' Why is every $A \in U(n)$ of the form $\exp(X)$? I know from the previous chapter that every $\exp(X)$ where $X+X^*=0$ complies with the above equality is in $U(n)$ but there's no proof that the other direction is also true.

Answer 1

Use the fact that every $A\in U(n)$ can be decomposed into eigenvalues and eigenvectors as $$A=\Omega\,{\rm diag}\,(e^{i\phi_1},e^{i\phi_2}\ldots e^{i\phi_n})\Omega^\ast,$$ with $\Omega\in U(n)$ and real $\phi_1,\phi_2,\ldots\phi_n$.
Define $$X=i\Omega\,{\rm diag}\,(\phi_1,\phi_2\ldots \phi_n)\Omega^\ast,$$ then, on the one hand, $X+X^\ast=0$, and on the other hand $e^X=A$.
_{for any function $f(M)$ from $\mathbb{R}^n\mapsto\mathbb{R}^n$ and any $\Omega\in U(n)$ it holds that $f(\Omega M\Omega^\ast)=\Omega f(M)\Omega^\ast$, so $e^X=\Omega\exp(i\,{\rm diag}\,(\phi_1,\phi_2,\ldots\phi_n))\Omega^\ast=\Omega\,{\rm diag}\,(e^{i\phi_1},e^{i\phi_2},\ldots e^{i\phi_n})\Omega^\ast=A$}

Answer 2

Notice that the matrices $X$ with the property $X^* + X = 0$ form precisley the Lie algebra $u(N)$. Since $U(N)$ is connected and compact, its exponential map is surjective.