I am following a book on relativistic quantum mechanics and during some proof related to charge conjugation the author assumes that $\gamma_0$ is real (where $\gamma_0$ is one of the four 4×4 matrices that satisfy the Clifford algebra used for the Dirac equation). While it is true that the most used representations of the gamma matrices have $\gamma_0$ real, it isn't obvious to me that this is true from the properties it has by definition, namely $(\gamma_0)^2 = 1$ and $\gamma_0 = (\gamma_0)^\dagger$. Is it maybe just valid for this specific algebra?
If it helps, here are some extra details about the gamma matrices (Clifford algebra $Cl_{1,3}(R)$):
$$\{\gamma_\mu, \gamma_\nu\} = \gamma_\mu \gamma_\nu – \gamma_\nu \gamma_\mu = 2 \eta_{\mu \nu}$$
with $\mu,\nu = \overline{0,3}$ and
$$\eta = \begin{pmatrix}
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{pmatrix}$$
Best Answer
A unitary matrix has its spectrum in the unit circle. A hermitian matrix has its spectrum in the real line. Therefore, a unitary hermitian matrix has spectrum a subset of $\{-1,1\}$. If the matrix is not $\pm I$, it is a reflection which means there is a decomposition of the space $\mathbb{C}^n=V\bigoplus V^\perp$ such the matrix is the identity on $V$ and $-1$ on $V^\perp$. That said however, the matrix components don't have to be real, since $V$ may be any proper subspace of $\mathbb{C}^n$, not necessarily of a form $U\bigoplus iU$ where $U$ is a real subspace of $\mathbb{R}^n$.