# Trace properties of the gauge potential in non-Abelian gauge theory

I want to proof equation 69.18 in Srednicki's book "Quantum field theory", which reads: $$$$A_\mu^a(x)=2\text{Tr}[A_\mu(x)T^a].\tag{69.18}$$$$ $$A_\mu(x)$$ is the non-Abelian gauge field and $$T^a$$ are the generator matrices, which can be used to expand $$A_\mu(x)$$ as $$$$A_\mu(x)=A_\mu^a(x)T^a.\tag{69.17}$$$$

Srednicki uses the normalisation $$$$\text{Tr}[T^aT^b]=\frac{1}{2}\delta^{ab}.\tag{69.8}$$$$

A simple proof seems to be: \begin{aligned} 2\text{Tr}[A_\mu(x)T^a]&=2\text{Tr}[A_\mu^b(x)T^bT^a] \\ &=2\text{Tr}[T^bT^a] A_\mu^b(x) \\ &=2 \left( \frac{1}{2}\delta^{ab} \right) A_\mu^b(x) \\ &=A_\mu^a(x) . \end{aligned} If it is correct, then why is the following equality true? (second equality above)

$$$$2\text{Tr}[A_\mu^b(x)T^bT^a]=2\text{Tr}[T^bT^a] A_\mu^b(x).$$$$ The the elements of the gaugefield are matrices, so it is not obvious why it should be true. Note that the elements of the gaugefield are traceless and hermitian, maybe this plays a role in the resolution to my question.

$$\mathsf A^\mu = A^\mu_a \mathsf T^a$$ is a matrix but $$A^a_\mu$$ is not.