I am trying to understand why a matrix $M$ that commutes with all gamma matrices $\gamma^\mu $ is proportional to the Identity matrix. I am following example 3.18 in Voja Radovanovic's book on solved QFT problems. considering the set of 16 linearly independent Gamma matrices: $$\Gamma^a={\{I,\gamma^\mu,\sigma^{\mu\nu},\gamma_5,\gamma_5\gamma^\mu\}}$$ I understand most of the procedure. I begin with the fact that I can write any Matrix as a sum of the Gamma matrices and separate one of them. (with $\Gamma^b\neq I$): $$M=\sum_{a}c_a\Gamma^a=c_b\Gamma^b+\sum_{a\neq b}c_a\Gamma^a$$ Now I multiply from the left by $\Gamma_d$ and from the right by $\Gamma^d$, such that $\Gamma^d$ and $\Gamma^b$ anticommute: $$\Gamma_d M\Gamma^d=c_b\Gamma_d\Gamma^b\Gamma^d+\sum_{a\neq b}c_a\Gamma_d\Gamma^a\Gamma^d$$
Since $M$ by definition commutes with all $\gamma^\mu$, and all $\Gamma^d$ are comprised of said $\gamma^\mu$, $M$ commutes with any $\Gamma^d$. Also, $\Gamma^b$ will anticommute with $\Gamma^d$, and $\Gamma_d\Gamma^d=1$: $$M=-c_b\Gamma^b+\sum_{a\neq b}c_a\Gamma_d\Gamma^a\Gamma^d$$ But I know that the product of any two $\Gamma$ simply give another $\Gamma$ up to a factor $\eta=+-1 $ or $\eta=+-i$ $$M=-c_b\Gamma^b+\sum_{a\neq b}c_a\eta\Gamma^a$$ Here, I can multiply the first equation as well as the previous one by $\Gamma_b$ and take the trace, giving the following two expressions: $$Tr(M\Gamma_b)=Tr(c_b\Gamma^b\Gamma_b)+Tr(\sum_{a\neq b}c_a\Gamma^a\Gamma_b)$$ $$Tr(M\Gamma_b)=Tr(-c_b\Gamma^b\Gamma_b)+Tr(\sum_{a\neq b}c_a\eta\Gamma^a\Gamma_b)$$ Clearly, $c_b$ has to be 0 for the first two expressions to be equivalent. But the book now says that it is "obvious" that all $c_a$ terms are $0$ except for the one multiplying the identity matrix, but I do not see how I could draw that conclusion. How can we know that?
