Ricci identity/Riemann curvature tensor and covectors Can somebody please explain to me how the following statement is true?

The Riemann curvature tensor $R^c_{dab}$ is given by the Ricci identity $$(\nabla_a\nabla_b-\nabla_b\nabla_a)V^c\equiv R^c_{dab}V^d$$ where $\nabla_a$ denotes the covariant derivative. It is linear in $V^c$, hence may be shown by the Quotient theorem to be a tensor.

Now, I can see that the $R^c_{dab}$ is a tensor by construction -- based on the LHS of the Ricci identity. However, I don't understand how the linearity in $V^d$ comes to play.

Also, it is  given that for covectors, the Ricci identity takes the form 

$$(\nabla_a\nabla_b-\nabla_b\nabla_a)V_c\equiv -R^d_{cab}V_d$$ 

How does this follow from the Ricci identity for (contravariant) vectors? 
If I write $$(\nabla_a\nabla_b-\nabla_b\nabla_a)V_c=(\nabla_a\nabla_b-\nabla_b\nabla_a)(g_{cd}V^d)$$ and in GR, the Levi-Civita connection has that the metric is covariantly constant, we have 
$$(\nabla_a\nabla_b-\nabla_b\nabla_a)(g_{cd}V^d)=g_{cd}(\nabla_a\nabla_b-\nabla_b\nabla_a)V^d\\=g_{cd}R^d_{eab}V^e=R_{ceab}V^e=R^d_{cab}V_d$$ 
 Where has my minus sign gone?
I have read that you can  the Ricci identity for covectors by arguing using the fact that the Levi-Civita connection is symmetric, but I don't know how they mean. 
Thanks in advance for any help!
 A: You got it right up to
$$
...=R_{ceab}V^e=-R_{ecab}V^e=-R^d_{\cdot cab}V_d.
$$
There is a difference between the first index raised ($R^c_{\cdot dab}$) and the second idex raised ($R^{\,d}_{c\cdot ab}$).
As I understand it, the statement is that from the LHS of the Ricci identity you only know that $T^c_{ab}(V)=R^c_{dab}V^d$ is a tensor for any $V$. And then you use the Quotient theorem to deduce that $R^c_{dab}$ is a tensor.
A: It might be simple to get the minus sign difference between first index raised and second index raised curvature tensor from components definition.
For first index raised curvature, $R^{\kappa}{}_{\lambda\mu\nu}$ is defined as
$$\langle dx^\kappa, R(e_\mu,e_\nu)e_\lambda \rangle = \langle dx^\kappa, \nabla_\mu\nabla_\nu e_\lambda-\nabla_\nu\nabla_\mu e_\lambda\rangle$$
where (3,1)-type curvature tensor is defined as $R:\mathfrak{X}(M)\times \mathfrak{X}(M)\times \mathfrak{X}(M) \to \mathfrak{X}(M)$ with $R(X,Y,Z)=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z$. Plug in definition of $\nabla_\mu$, you get
$$R^{\kappa}{}_{\lambda\mu\nu}=\partial_\mu \Gamma_{\nu \lambda}^\kappa-\partial_\nu \Gamma_{\mu \lambda}^\kappa+\Gamma_{\mu\eta}^\kappa \Gamma_{\nu \lambda}^\eta-\Gamma_{\nu\eta}^\kappa \Gamma_{\mu \lambda}^\eta$$
Now the fourth(second) index raised curvature tensor is defined as
$$R_{ijk}{}^l\omega_l=[\nabla_i,\nabla_j]\omega_k$$
again straightforward calculation yields
$$R_{ijk}{}^l = \partial_j \Gamma_{ik}^l-\partial_i \Gamma_{jk}^l+\Gamma_{ik}^s\Gamma_{js}^l-\Gamma_{jk}^s\Gamma_{is}^l$$
Note $\{\mu\}\leftrightarrow \{i\}$, $\{\nu\}\leftrightarrow \{j\}$, $\{\lambda\}\leftrightarrow \{k\}$ while $\{\kappa\}\leftrightarrow \{l\}$ are both covariant indices. Use metric tensor to bring down all indices and move the covariant index to the fourth slot, you will see where the minus sign comes from
$$R_{\kappa\lambda\mu\nu}=R_{\mu\nu\kappa\lambda}=-R_{\mu\nu\lambda\kappa}$$
