Trace of a product of gamma matrices

[BVM]

Homework Statement

A proof of equality between two traces of products of gamma matrices.

Tr(\gamma^\mu (1_4-\gamma^5) A (1_4-\gamma^5) \gamma^\nu) = 2Tr(\gamma^\mu A (1_4-\gamma^5) \gamma^\nu)

Where no special property of A is given, so we must assume it is just a random 4x4 matrix.
1_4 represents the 4x4 unity matrix.

Homework Equations

\gamma^5 := i\gamma^0 \gamma^1 \gamma^2 \gamma^3
The usual product and trace identities as found here.

The Attempt at a Solution

I have been able to prove that
Tr(\gamma^\mu (1_4-\gamma^5) A (1_4-\gamma^5) \gamma^\nu) = Tr(\gamma^\mu A (1_4-\gamma^5) \gamma^\nu) + Tr(\gamma^5 \gamma^\mu A (1_4-\gamma^5) \gamma^\nu)
using the definition of the trace and the anticommutativity of the \gamma^\mu and \gamma^5.

 

[TSny]

I have been able to prove that
Tr(\gamma^\mu (1_4-\gamma^5) A (1_4-\gamma^5) \gamma^\nu) = Tr(\gamma^\mu A (1_4-\gamma^5) \gamma^\nu) + Tr(\gamma^5 \gamma^\mu A (1_4-\gamma^5) \gamma^\nu)
using the definition of the trace and the anticommutativity of the \gamma^\mu and \gamma^5.

For your last term, try using the fact that the trace is invariant under any cyclic permutation of the matrices inside the trace. So, you can move the γ⁵ on the left to the far right. Then simplify (1₄ - γ⁵) γ$^\nu$ γ⁵.

 

[BVM]

For your last term, try using the fact that the trace is invariant under any cyclic permutation of the matrices inside the trace. So, you can move the γ⁵ on the left to the far right. Then simplify (1₄ - γ⁵) γ$^\nu$ γ⁵.

Got it! That hint really helped. Thanks.