How does $\{\gamma^\mu , \gamma^\nu\}= 2g^{\mu\nu}I$ imply that $\gamma^\mu$ is traceless?

Question

How does $\{\gamma^\mu , \gamma^\nu\}= 2g^{\mu\nu}I$ imply that $\gamma^\mu$ is traceless?

where I represents the identity matrix

I know that $$\{\gamma^\mu , \gamma^\nu\}=\gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu \tag{1}$$ and that

$$g^{\mu\nu}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix} \tag{2}$$

I know the real values of the gamma matrices but I suppose I am meant to ignore the fact that I know those values and simply use $g^{\mu\nu}$ as the reason behind their tracelessness.

I have attempted using $(1)$ by only "paying attention" (for the lack of better words) to the diagonal terms, as follows:

$$\gamma^\mu= \begin{pmatrix} a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & 0 & 0 & d \\ \end{pmatrix}$$

and multiplying this by a similar matrix with diagonal terms e,f,g,h.

This led me nowhere.

Answer 1

Using the relation, we know that $ \frac{\gamma^b \gamma^b}{2g^{bb}} = 1 $. Thus, assuming $a\ne b$ we write \begin{align} Tr(\gamma^a) &= \frac{1}{2g^{bb}}Tr(\gamma^a \gamma^b \gamma^b) \end{align} This can be manipulated in two ways:

1. Using a cyclic permutation property of the trace, we obtain $$ Tr(\gamma^a) =\frac{1}{2g^{bb}}Tr(\gamma^b \gamma^a \gamma^b) $$
2. Using the (anti-)commutation relation $\gamma^a \gamma^b = − \gamma^b \gamma^a$ for $a\ne b$. $$ Tr(\gamma^a) = -\frac{1}{2g^{bb}}Tr(\gamma^b \gamma^a \gamma^b) $$

Both conditions must be satisfied. This is only possible if the trace is zero.