again thank you for the answer! This was really enlightening and I feel happy to understand a little more by having such contact with experts on the subject!

@Abdelmalek hahaha that's true.

I will take a look at both references! Thanks!

Pedro, thanks for the answer. Although I'm not much familiar with $C^{*}$-algebras, I got the idea and appreciate the details and quality of your exposure. I know Bratelli & Robinson's book, so I might need to learn a little bit of $C^{*}$-algebra to fully understand your reasoning. In any case, your answer will be an important guide for me. Also, this discussion led me to think about another topic, and I should post another question soon.

Thanks Carlo, this seems like a very useful reference!

@IgorKhavkine it seems that the easiest way to prove the irreducibility of $\rho$ is to prove that the family of four matrices $\gamma^{\mu}$ is irreducible, no? These are only four matrices and if there is no jointly invariant subspace for them, there will be no jointy invariant subspace for $\rho(x)$, $x \in \mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)$.

@IgorKhavkine yes, I thought so but how does the isomorphism implies irreducibility? I was trying to use it on my edit but this is where I got confused. Just to clarify, when you write $\rho: \mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi) \to M_{4\times 4}(\mathbb{C})$ you are considering both algebras over $\mathbb{R}$ right?

@IgorKhavkine this is very helpful. Is it easy to prove this irreducibility? I thought I knew how to prove but now I'm not sure.

@CalvinMcPhail-Snyder I added some new thoughts. Maybe this is a more natural way of thinking on the problem.