What's the easiest way to show that the covariant derivative $\nabla U^{\mu}$ is linearly independent to $U^{\mu}$, which is a vector?
I mean I'm assuming they are since I'm proving the second Bianchi identity and my proof requires this to be the case; and I'm quite confident that the Bianchi identity is true!
So I start with the Jacobi identity
$$[\nabla_a,[\nabla_b,\nabla_c]]+\text{cyclic}=0$$
I then substitute in the definition of the Riemann tensor, giving me
$$\nabla_a(R^d_{\space\space\space ebc}U^e)+\text{cyclic}=0$$
which gives
$$\nabla_a(R^d_{\space\space\space ebc})U^e+R^d_{\space\space\space ebc} \nabla_aU^e+(\text{cyclic in }a,b,c)\space\space=0.$$
So we have 6 expressions, from which if the second Bianchi identity is true, the vector and the covariant derivate of the vector must be linearly independent.
Or am I just OK to say that as the covariant derivative is (1,1) tensor and a vector is a (1,0) tensor, the result must follow straight away?