Proving trace is independent of basis using tensor product This is a question that has been deleted (see here if you have sufficient reputation), but which I would like to ask again since it seems interesting.  I have an answer of my own that I will post when I have the time, but I am interested in seeing other approaches as well.

The question:

For $V,W$ vector spaces with $\dim V = n < \infty$, we let $A \in Hom(V, V \otimes W)$. Given a basis $\{v_j\}_{j=1}^{n}$ of $V$ let $w_{ij} \in W$ be defined by $Av_j = \sum_{i=1}^n v_i \otimes w_{ij}$ and let $Trace(A) = \sum_{i=1}^n w_{ii}$. I'm asked to prove that this definition is independent of choice of basis.

As is pointed out in the original post, it is notable that there is no straightforward way to compute the $w_{ij}$, let alone compute how they would change under a change of basis.
 A: One approach is as follows: note that $\operatorname{Hom}(V,V \otimes W)$ is spanned by the elements of the form $A = (v \otimes w)\alpha$ where $v \in V, w \in W,$ and $\alpha \in V^*$, the (algebraic) dual to $V$.  It therefore suffices to show that the linear map $\operatorname{Trace}:\operatorname{Hom}(V,V \otimes W) \to W$ will yield the same output for a given $A$ regardless of the basis chosen.
With that in mind: let $\{v_1, \dots ,v_n\}$ be a basis of $V$ and let $\{\alpha_1,\dots,\alpha_n\}$ denote the corresponding dual basis.  Consider any $A = (v \otimes w)\alpha$.  There exist coefficients $p_i,q_i$ such that
$$
v = \sum_i p_i v_i, \quad \alpha = \sum_{i}q_i \alpha_i.
$$
We then compute
$$
Av_j = \left[\left(\left(\sum_i p_i v_i\right) \otimes w\right) \sum_k q_k \alpha_j\right] v_j\\
= \sum_i\sum_k p_i q_k (v_i \otimes w)\alpha_k(v_j)\\
= \sum_i p_i q_j v_i \otimes w\\
= \sum_i v_i \otimes (p_i q_j w)
$$
from which it follows that
$$
\operatorname{Trace}(A) = 
\sum_{i=1}^n p_i q_i w = \left(\sum_{i=1}^n p_i q_i\right)w
$$
On the other hand, we note that
$$
\alpha(v) =  \sum_{i}q_i \alpha_i \left(\sum_j p_j v_j \right)
= \sum_{i,j} p_i q_j \alpha_i(v_j) = \sum_i p_iq_i
$$
which is to say that we may write
$$
\operatorname{Trace}(A) = \alpha(v) \cdot w.
$$
That is, $\operatorname{Trace}$ is the unique linear map that takes all elements of the form $(v \otimes w)\alpha$ to $\alpha(v) \cdot w$.  Since this equivalent definition makes no reference to the basis $\{v_i\}$, it follows that the map $\operatorname{Trace}$ is indeed independent of the basis chosen.

If you prefer the universal definition of a tensor product, we have shown that your Trace is induced by the multilinear map from $V \times W \times V^*$ to $W$ given by
$$
(v, w, \alpha) \mapsto \alpha(v)\cdot w.
$$
