Naive question about the group $SU(n)$? As usual, let $SU(n)$ represent the set of all the $n\times n$ unitary matrices with determinant $1$. 
It's easy to show that any matrix $U$ takes the form $U=e^{iA}$ ($A$ is a $n\times n$ traceless hermitian matrix) must belong to $SU(n)$. Vice verse, for any $U\in SU(n)$, can $U$ be written as the form $U=e^{iA}$? How to prove it? Thank you very much.
 A: This follows from the description of the Lie algebra ${\frak{su}}(n)$. The exponential function $\exp\colon {\frak{su}}(n)\rightarrow SU(n)$ is surjective, since the group $SU(n)$ is compact, i.e., every $U$ in $SU(n)$ is of the form $\exp(iH)=e^{iH}$.
A: Here's an alternative answer that's a bit more explicit in constructing $A$:
The crucial fact we will use it that because $U$ is unitary, it is unitarily diagonalizable, meaning there exists a diagonal matrix $D$ and a unitary matrix $V$ (with $V^{-1} = V^\ast$) such that
$$U = V D V^{-1}.$$
Because $U$ is unitary with determinant $1$, its eigenvalues, which are the diagonal entries of $D$, are complex numbers of modulus $1$, and the product of the eigenvalues is $1$. That means $D$ is of the form
$$D = \text{diag}(e^{i\theta_1}, \dots, e^{i\theta_n})$$
for some real numbers $\theta_1, \dots, \theta_n$. (Note that each of the $\theta_i$'s is unique up to adding integer multiples of $2\pi$.) Since the product of the eigenvalues is $1$, the sum $\theta_1 + \dots + \theta_n$ is an integer multiple of $2\pi$. By adjusting, say, $\theta_1$ by that multiple of $2\pi$ (which doesn't change $e^{i\theta_1}$, and doesn't change $D$), we can ensure that $\theta_1 + \dots + \theta_n = 0$.
Now define $A_0$ by
$$A_0 = \text{diag}(\theta_1, \dots, \theta_n),$$
so that $D = e^{iA_0}$. The "adjustment" we just made ensures that $A_0$ is traceless. (Note that the adjustment was not unique, illustrating that the choice of $A_0$, and therefore $A$, is not unique.)
Finally, define $A$ by
$$A = V A_0 V^{-1},$$
so that $e^{iA} = V e^{iA_0} V^{-1} = V D V^{-1} = U$, as desired. Note that $A$ is traceless since $A_0$ is traceless (using the cyclic property of the trace), and $A$ is hermitian since $A_0$ is hermitian and $V$ is unitary.
