Coordinate-Free Definition of Trace, revisited. I have some questions about a coordinate free definition of the trace of linear operators. This questions has been asked before in this forum (see [1,2]), but I haven't found the answers of my interest to be clear enough, so I will ask some further questions about the answer which I will enumerate them as Q.i), Q.ii), ... while I ask them.
Let me begin fixing the notation. Let $V$ be a vector space, $V^*$ its dual, $V\otimes V^*$ their tensor product and $End(V)$ the vector space of linear endomorphisms in $V$ (i.e. linear operators from $V$ to $V$).
As far as I understood, the answer begins stating that the mapping $v\otimes w^*\to (u \to w^*(u)v)$ for all $v,u\in V$ and $w^*\in V^*$ is a linear isomorphism. My first question is about this mapping. Q.i) What happens to those elements of $V\otimes V^*$ that are not pure, i.e. that are of the form $\sum_i v_i\otimes w_i^*$?
Later in the answer, a scalar $w^*(v)$ is associated to each element of the form $v\otimes w^* \in V\otimes V^*$. As far as I understood, is this scalar what is identified with the trace. Q.ii) What happens when we consider non pure tensor products like $\sum_i v_i\otimes w_i^*$? Q.iii) In this case, does the scalar takes the form $\sum_i w^i(v_i)$? Q.iv) Given, that the representation $\sum_i v_i\otimes w_i^*$ is not unique, why $\sum_i w^i(v_i)$ remains invariant?
[1] Coordinate-Free Definition of Trace.
[2] Coordinate-free proof of $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$?
EDIT 1: I changed not factorizable by not pure as suggested by @TrevorGunn.
 A: Q.i) The map is first specified by its action on separable elements. Since every element in general can be written as a linear combination of separable elements, the action of the map on a general element is given by linear extension. Explicitly, elements of the form $\sum_{i}v_i \otimes w_i^*$ are taken to maps of the form
$$u \mapsto\sum_i w_i^*(u)v_i.$$
Q.ii and Q.iii) Yes, this scalar is the trace. Specifically, it is the trace of the element $v\otimes w^*$, or if you'd prefer, the linear map on $V$ that this element represents. The trace of a general (non-separable) element again has to be given through linearity. Using the same example as before, an element of the form $\sum_{i}v_i \otimes w_i^*$ has a trace of
$$\mathrm{Tr}\left(\sum_i v_i \otimes w_i^*\right)=\sum_i \mathrm{Tr}\left(v_i \otimes w_i^*\right) = \sum_i w_i^*(v_i),$$
which is exactly what you suggested.
Q.iv) This comes down to whether the map is well-defined or not. This has to do with whether or not there actually exists a linear map $\mathrm{Tr}$ such that $\mathrm{Tr}(v\otimes w^*) = w^*(v)$ for all separable elements.
In general, you have to check the consistency of the map anytime its defined on a linearly dependent set (i.e., the set of separable tensors). The easiest way to show consistency is to explicitly construct such a map, using a basis for example. Fix a basis $e_i \otimes e_j^*$ for $V\otimes V^*$ and define the trace by
$$\mathrm{Tr}(e_i\otimes e_j) = e^*_j(e_i) = \delta_{ij},$$
where $\delta_{ij}$ denotes the Kroencker delta. Note that there is no ambiguity in this definition here because we are first defining it on basis elements, and every general element has a unique decomposition in terms of basis vectors.
Given $v = \sum_{i}v_ie_i$ and $w^* = \sum_{j}w_j^*e^*_j$ (I am being a bit sloppy in distinguishing vectors and their components, but hopefully it's clear from context), we have:
$$\mathrm{Tr}(v\otimes w^*) = \sum_{i,j}v_iw_j^*\mathrm{Tr}(e_i\otimes e_j^*) = \sum_{i}v_iw_i^* = w^*(v)$$
This shows that there exists a well-defined linear map such that $\mathrm{Tr}(v \otimes w^*) = w^*(v)$ for all separable elements.
