Let $V$ and $W$ be two vector spaces such that $x_i\in V$ and $y_i\in W$. If $\sum_ix_i\otimes y_i=0$ and $\{x_i\}$ are linearly independent, then prove that $y_i=0$ for all $i$.
What I know is the fact that, $x_i\otimes y_i=0$ iff $x_i=0$ or $y_i=0$.
Possible proof
My Attempt to rephrase the proof
Let $X,Y$ are normed vectorspaces over the field $\mathbb{F}$ such that $x_i\in X$ and $y_i\in Y$. Let $X^*,Y^*$ are the algebraic dual spaces of the vector spaces $X,Y$ respectively, are defined as the set of all linear maps $f:X\to \mathbb{F}$ and $g:Y\to\mathbb{F}$.
Let's form a bilinear map $h:X\times Y\to\mathbb{F}$ such that $h(x,y)=f(x)g(x)$ for all $x\in X$ and $y\in Y$.
We can define a linear map $h_\otimes:X\otimes Y\to\mathbb{F}$ such that $h_\otimes\sum_{i,j}\Big(c_{ij}(x_i\otimes y_j)\Big)=\sum_{i,j}c_{ij}h(x_i,y_j)$ for all $x_i\in X$ and $y_i\in Y$.
$$ 0=\sum_i x_i\otimes y\implies 0=h_\otimes(0)=h_\otimes(0\otimes 0)=h_\otimes(\sum_i x_i\otimes y)=\sum_i h(x_i,y)\\ =\sum_i f(x_i)g(y_i)=\sum_i f(x_ig(y_i))=f(\sum_i x_ig(y_i)) $$ $$ 0=h_\otimes(0)=f(0)g(0)=f(\sum_i x_ig(y_i)) $$ Since $f(\sum_i x_ig(y_i))=0$ for every $f\in X^*$, say $f(x)=||x||$ then $||x||=0\implies x=0$.
We can conclude that $\sum_i x_ig(y_i)=0$
Since $x_i$ is a linearly independent set, this implies $g(y_i)=0$ for each $i$.
$g(y_i)=0$ for all $g\in Y^*$, this implies $y_i=0$ for all $i$
Is it a better explanation for the proof ?
As you say,
$$0 = h_{\otimes}(0) = f(\sum\limits_i x_ig(y_i) )$$
where $z := \sum\limits_i x_i g(y_i)$ is an element of $X$. Now $f$ represented an arbitrary element of $X^{\ast}$. Therefore $z$ is sent to $0$ by every element of $X^{\ast}$. This implies that $z$ is the zero element of $X$.