How to show that the "trace" of a linear map valued in a tensor product is basis independent Suppose we have two vector spaces $C,V$, consider their tensor product $C\otimes V$ and a linear map $\rho:V\rightarrow C\otimes V$.
Let $(v_i\quad i\in I)$ be a basis of $V$.
Suppose that we can write the action of $\rho$ as follows: $\forall i\in I: \rho(v_i)=\sum_{j\in I}c_{i,j}v_j$. Now, I am asked to show that $t_V=\sum_{i\in I} c_{ii}$ is basis independent. This is a sort of trace, I think.
Now, I am trying to write the sum in a basis independent way, but the only thing I noticed is that we can write:
$$t_V=\left(C\otimes\sum_{i\in I} v_i^*\right)\rho\left(\sum_{i\in I} v_i \right)$$
where $(v_i^*\quad i\in I)$ is the dual basis of $(v_i\quad i\in I)$.
This should be intuitively basis independent, since given any basis, we can canonically write $t_V$ as above. Anyhow, I am not sure about how to build an argument.
 A: I think that unless there is a topology on these vector spaces, we must assume that elements $c_{i,j}$ are zero for all but finitely many pairs $i,j$. Otherwise, the sum $\sum_{j\in I}c_{i,j}v_j$ might be an infinite sum of non-zero elements and therefore not defined.
Under this assumption, note that $\rho$ is a linear map of finite rank. Each element $\rho \in L_{\text{FR}}(V,C \otimes V)$ can be written as a (finite) linear combination of rank-1 maps of the form
$$
\rho(w) = \alpha(w) (c \otimes v), \quad w \in V
$$
for some $c \in C, v \in V, \alpha \in V^*$.  Now, write $v = \sum_{i \in I} x_i v_i$ (where we assume that all abut finitely many $x_i$ are zero).  We find that
$$
\rho(v_i) = \alpha(v_i)\left(c \otimes \sum_{j \in I} x_j v_j\right)
= \sum_{j \in I} \alpha(v_i) x_j c \otimes v_j,
$$
so that $c_{ij} = \alpha(v_i) x_j c$. We calculate
$$
t_V(\rho) = \sum_i \alpha(v_i) x_i c = \alpha\left(\sum_i x_iv_i \right)c = \alpha(v)\,c.
$$
That is, $t_V$ is the unique linear map that takes all elements $\rho$ of the form $(c \otimes v) \alpha$ to $\alpha(v)c$.  Since this equivalent definition makes no reference to the basis $\{v_i\}$, it follows that your definition $t_V$ is indeed independent of the basis chosen.

An alternative approach, following the idea of Stefan's comment: if we write $\rho$ in the form
$$
\rho = \sum_{i,j \in I} c_{ij} \otimes (v_j v_i^*)
$$
where $(v_i^*|i \in I)$ is the dual basis, then it is easy to see that
$$
t_V = \operatorname{id}_C \otimes \operatorname{trace}_V,
$$
where $\operatorname{trace}_V : L(V,V) \to \mathbb F$ is the usual notion of the trace of a map. With that, it is clear that $t_V$ is basis independent.
