Can the Dirac trace of two gamma matrices always be made positive?

Question

I'm doing a calculation where I have a number of traces of just two slashed matrices. When Calculating a Dirac trace of two slashed matrices, we have the identity: $$Tr[\displaystyle{\not} a \displaystyle{\not} b] = a_\mu b_\nu Tr[\gamma^\mu\ \gamma^\nu] = 4a_\mu b_\nu \eta^{\mu\nu} = 4(a\cdot b). $$

A handful of these traces have a negative sign in them, $$Tr[- \displaystyle{\not} a \displaystyle{\not} b]. $$

Can I just ANTI-commute the two matrices to eliminate the negative sign, $$Tr[- \displaystyle{\not} a \displaystyle{\not} b] = Tr[\displaystyle{\not}b \displaystyle{\not} a],$$ and then use the cycling property to move the a matrix back to the front $$Tr[\displaystyle{\not}b \displaystyle{\not} a] = Tr[\displaystyle{\not}a \displaystyle{\not} b]? $$

Does this mean, $$Tr[-\displaystyle{\not}a \displaystyle{\not} b] = Tr[\displaystyle{\not}a \displaystyle{\not} b]? $$ Have I made a gross mistake in my understanding of commuting gamma matrices? Or is this a fun quirk of the Clifford algebra that happens when you are taking the trace of only two matrices?

Edit: I misspoke or maybe was too loose with my language when I originally posted this. I am specifically interested in the Dirac gamma matrices which follow the anti-commutation relation, $$ \gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2 \eta^{\mu\nu}.$$ Does this change anything? or is my conclusion still wrong.

Answer 1

Quite generally, you have ${\rm Tr}(-A B)=-{\rm Tr}(A B)=-{\rm Tr}(BA)$ for arbitrary square matrices $A$, $B$.