Prove there is a basis-independent isomorphism between $T^1_1(V)$ and the space of linear maps $V \rightarrow V$ 
Let $V$ be a finite-dimensional vector space. Prove there is a basis-independent isomorphism between $T^1_1(V)$ and the space of linear maps $V \rightarrow V$.

I want to show there is a bijective between elements in $T^1_1(V)$ and elements in $V \rightarrow V$, and them show there is a homomorphism. Mission failed.
Also, I read this question and do not understand: Dimension of Hom(U, V)
However, my question is more primitive - so regardless the complex discussion related to the two paper in the question and accepted answer, if the dimension of $\text{Hom}(U,V)$ just dim$U \times$ dim$V$?
Thanks!
 A: Let $a \in V$ and $\alpha \in V^*$.  Let $A \in T^1_1$, so that $A(a, \alpha) \in K$, where $K$ is the base field of $V$.
There exists some derivative operator $\partial^{(\alpha)}$ that can be written in terms of a basis of vectors $e_1, e_2, \ldots$ and components of $\alpha$ labeled $\alpha_1, \alpha_2, \ldots$ as
$$\partial^{(\alpha)} = \sum_{i=1}^n e_i \frac{\partial}{\partial \alpha_i}$$
where $n$ is the dimension of $V$ (and $V^*$).  While I have written this derivative operator in terms of a basis, the operator is itself basis independent; for instance, this operator could be defined using an integral (especially in the presence of a volume form or metric), or it could merely be shown to have the proper transformation laws of a vector.
Once this derivative is in place, we can construct a linear map $f: V \to V$ by
$$f(a) = \partial^{(\alpha)} A(a, \alpha)$$
Again, explicitly using a basis, this is
$$f(a) = \sum_{i=1}^n e_i A(a, e^i)$$
where $e^i$ are basis covectors.
Conversely, if there is a linear map $k: V \to V$, we can construct some $B \in T_1^1$ via the following.  Use the identity
$$\partial^{(\alpha)} \alpha(a) = a$$
(This, again, can be proven in a basis, but the result is basis independent.)  It's clear then that
$$\partial^{(\alpha)} (\alpha \circ k)(a) = k(a)$$
And thus, setting $B(a, \alpha) = (\alpha \circ k)(a)$ is consistent with the construction given earlier.
So far, we have proven the existence of these corresponding maps, but what about uniqueness?  That comes from linearity.  Suppose there were a tensor $C(a, \alpha)$ such that $\partial^{(\alpha)} C(a, \alpha) = f(a)$, then we would have
$$0 = \partial^{(\alpha)} A(a, \alpha) - \partial^{(\alpha)} C(a, \alpha)$$
Through linearity, we can rewrite this as
$$0 = \partial^{(\alpha)} (A-C)(a, \alpha)$$
This readily implies that $A = C$.  Either they are both zero maps, or they are identical.  A similar argument works the other way around.
To be honest, I do worry that this argument may not be basis-independent enough for you. I guess that will depend on how readily you can accept $\partial^{(\alpha)}$ being basis independent.  This, I admit, is rather tricky; it's much much easier to construct it when you have a little more structure than just that of a vector space.
