In a physics context I work with the SU(2) and SU(3) matrix groups. Let $U$ be such a special unitary matrix. There are expressions like $(U - \mathrm{h.c.})$ which turn out to be traceless. This means that the trace of $U$ is always real.
I played around with Mathematica to generate matrices $U$ from the generators $\sigma_i$ and the matrix exponential function. The real part of the trace of an SU(2) matrix in terms of the three algebra components nicely oscillates:
The imaginary part seems to be virtually zero, the numerical matrix exponential generates small imaginary parts of say $10^{10}$. From the color coding this seems rather zero:
I know that $U = \exp(\mathrm i \alpha_i \sigma_i)$, that $U^{-1} = U^\dagger$ and a couple of other identities. However I cannot deduce that the trace is always real.
Is it true that the trace of special unitary matrices is always real? How can one show it?
The answer is yes for $2 \times 2$ matrices. We need some facts here:
It follows that a $2 \times 2$ unitary matrix has two complex eigenvalues satisfying $\lambda_1 \lambda_2 = 1$, as well as $|\lambda_1| = |\lambda_2| = 1$.
However, this means that $\lambda_1 \overline{\lambda_1} = \lambda_1\lambda_2 = 1 \implies \overline{\lambda_1} = \lambda_2$ (the bar denotes the complex conjugate).
We then have that the trace of $U$ will be $\lambda_1 + \overline{\lambda_1} = 2 \operatorname{Re}[\lambda_1]$. So, the trace of $U$ is necessarily real.
The answer is no for $3 \times 3$ matrices. In particular, consider the matrix $U = e^{2 \pi i/3}I$. This will also fail to hold for larger matrices.