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:

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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:

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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?

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    $\begingroup$ For $SU(3)$ (and larger), consider $$\begin{pmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & -1\end{pmatrix}$$ to see that the trace need not be real then. $\endgroup$ – Daniel Fischer Sep 28 '16 at 13:17

The answer is yes for $2 \times 2$ matrices. We need some facts here:

  • The determinant is the product of all eigenvalues
  • The trace is the sum of all eigenvalues
  • The eigenvalues of a unitary matrix have magnitude $1$

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.


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