I'm trying to calculate the covariant derivative of my connection $\mathbf{\mathcal{A}} $ (in this case $\mathbf{\mathcal{A}} $ is also a vector valued matrix but nevermind, I don't think is relevant for the purpose of the question...). Of course $\mathbf{\mathcal{A}} $ is also a traceless Hermitian matrix.
I'm up to evaluate then the non-Abelian curvature $\mathbf{\mathcal{F}} $ that results from the covariant derivative of $\mathbf{\mathcal{A}} $, i.e., in differential form language, the term \begin{equation} D\mathbf{\mathcal{A}} = d\mathbf{\mathcal{A}}+\mathbf{\mathcal{A}} \wedge\mathbf{\mathcal{A}} \end{equation}
In particular, I don't understand why the last term of the previous equation should be different from zero (in principle, the non-commutativity of the algebra underlying the system should manifest itself in this term) and how to carry out the calculations analytically.
Here is an example that explains why the wedge of a 1-form with itself need not be zero. Let our 1-forms be valued in the Lie algebra $(\Bbb R^3, \times)$. Consider $\omega = \hat idx + \hat j dy$. Then by definition, we have $\omega \wedge \omega = [\hat i, \hat j] dx \wedge dy = 2\hat k dx \wedge dy$. This would of course be zero if our Lie algebra was commutative.