In Bjorken and Drell, Relativistic Quantum Mechanics, the following argument is constructed to show that a set of matrices derived from the $\gamma$ matrices are linearly independent. The matrices of concern (though I do not think that is is particularly important in the argument):
$$\Gamma^S = \mathrm{I},\space \space \space \space \space \space \space \ \Gamma^V_\mu = \gamma_\mu \space \space \space \space \space \space \space \Gamma^T_{\mu\nu} = \sigma_{\mu\nu}$$ $$\Gamma^P = \gamma_5 :=\gamma^5 \space \space \space \space \space \space \space \Gamma^A_\mu=\gamma_5\gamma_\mu,$$
where the $\gamma_\mu$ and $\sigma_{\mu\nu}$ have the usual meaning.
Then the following argument is made to show that the above matrices form 16 linearly independent matrices:
- $\forall\space \Gamma^n,\space (\Gamma^n)^2 = \pm \text{I}$
- $\forall \space \Gamma^n \neq\Gamma^S, \exists \space\Gamma^m \space\text{such that} \space \{\Gamma^n,\Gamma^m\} = 0$ where $\{A,B\}$ is the anticommutator, which implies Tr$(\Gamma^n) = 0$
- For $a \neq b$, $\exists \space \Gamma^n \neq\Gamma^S$ such that $\Gamma^a \Gamma^b = \Gamma^n$.
To those point I am okay with what has been said, but it is the fourth point that confuses me:
- Suppose there exists $a_n$ such that $$\sum_na_n\Gamma^n = 0$$ Then by multiplying by $\Gamma^m \neq\Gamma^S$, and taking the trace, using (3) we find that $a_m = 0$.
If someone could explain these steps in some more detail, that would be greatly appreciated.