Element which is independent Let $V$ be a complex vector space with a product satisfying $xy=-yx$ for all $x,y\in V$.
Chosse a basis $v_i$ of $V$ and a basis $w_i$ such that $v_iw_j=\delta_{ij}$.
Why is the element $\sum_i w_i\otimes v_i$ independent of the basis $v_i$?
 A: From the question before the last edit: "I have chosen another basis $v_i'$, hence I get another basis $w_i'$ such that $(v_i',w_j')=\delta_{ij}$. If I write $v_i'=\sum_k\lambda_{k,i}v_k$  resp. $w_i'=\sum_l\mu_{l,i}w_l$ and plug in into $(v_i',w_j')=\delta_{ij}$ I get $$\delta_{ij}=\sum_k\lambda_{k,i}\mu_{k,j}.$$
Further I get $$\sum_i w_i'\otimes v_i'=\sum_{i,l,k}\mu_{l,i}\lambda_{k,i}w_l\otimes v_k.$$
Why is this equal to $\sum_i w_i\otimes v_i?$"
The relations $\delta_{ij}=\sum_k\lambda_{k,i}\mu_{k,j}$ for all $i,j$ mean that
the matrices $L=(\lambda_{i,j})$ and $M=(\mu_{i,j})$ satisfy $L^T \,M=I$, where $L^T$ means
the transpose of $L$ and $I$ is the identity matrix. Therefore $M=(L^T)^{-1}$ and we also have
$M\,L^T=I$. In the entries of $L,M$ this means that
$$\delta_{lk}=\sum_i\mu_{l,i}\lambda_{k,i}$$
for all $l,k$. Then your sum
$$\sum_{i,l,k}\mu_{l,i}\lambda_{k,i}w_l\otimes v_k=\sum_{l,k}\delta_{lk}w_l\otimes v_k=
\sum_k w_k\otimes v_k$$
as wanted.
A: Here is another proof with much less calculation. If $V$ is finite dimensional then $V\otimes V$ can be identified with the space of endomorphisms of $V$ such that $$(w\otimes v)\,h=(v\,h)\, w$$ for all $v,w,h$. For this we need the existence of bases $v_i$, $w_j$ such that $v_iw_j=\delta_{ij}$ for all $i,j$. Then $\sum_i w_i\otimes v_i$ is the identity because $\sum_i(w_i\otimes v_i) w_j=\sum_i\delta_{ij}w_i=w_j$ for all $j$ and $w_j$ form a basis. For the same reason $\sum_i w_i'\otimes v_i'$ is the identity for any other bases $v_i'$, $w_j'$ satisfying $v_i'w_j'=\delta_{ij}.$ Therefore $\sum_i(w_i\otimes v_i)=\sum_i(w_i'\otimes v_i')$.
