Trace is real for special unitary group It seems to be very simple, but I do not understand why trace of element of special unitary group is real and is it right at all or not?
 A: Let $\omega=e^{2\pi i/3}$ and $A=\operatorname{diag}(\omega,-\omega,-\omega)$. Then $A\in SU(3)$ but $\operatorname{trace}(A)=-\omega\ne\mathbb R$.
A: I don't think it is for $SU(3)$.  
Every Unitary matrix is diagonalizable and its eigenvalues lie on the unit circle.
If it is special, then the product of the eigenvalues is $1$.
In $SU(2)$, you have two eigenvalues $\lambda_1,\lambda_2$ whose product is $1$ ,
so the second is the reciprocal of the first. Since they lie on the unit circle the reciprocal is the complex conjugate. Since the trace is the sum of the eigenvalues it is $\lambda_1+\overline{\lambda}_1$ which is real. So the statement is true for $SU(2)$.
Here is what is true in general. Let $s_k:SU(n)\rightarrow \mathbb{C}$ be $(-1)^{n-k}$ times coefficient of $\lambda^{n-k}$ in the characteristic polynomial, that is $$det(A-\lambda Id).$$  These are a complete set of conugacy invariants. Notice $s_1(A)$ is the trace, and $s_n(A)=1$ because it's the determinant of $A$ and we are in the special unitary group.
If the eigenvalues of $A$ are $\lambda_i$ where $i$ ranges from $1$ to $n$ then $s_k$ is the $k$th elementary symmetric function evaluated on the eigenvalues. 
That is
$$s_k(A)=\sum_{i_1<i_2<\ldots <i_k}\lambda_{i_1}\lambda_{i_2}\ldots \lambda_{i_k}.$$
Since $1=det(A)=\prod_{i=1}^k\lambda_i$, if $\{j_1,\ldots,j_{n-k}\}$ is the complement of $\{i_1,\ldots i_k\}$ in $\{1,2,\ldots,n\}$ then that product of the $\lambda_{i_m}$ is the complex conjugate of the product of the $\lambda_{j_n}$.  Hence  $s_k(A)=\overline{s}_{n-k}(A)$.
Therefore, for all $n$, $s_k(A)+s_{n-k}(A)$ is real.  
In the case of $SU(2)$, this means $s_1(A)+s_1(A)$ is real, meaning the trace of a $2\times 2$ special unitary matrix is real.
For some laughs you should plot the image of the trace on $SU(3)$ in $\mathbb{C}$.
