Let $V$ is a finite dimensional vector space over a field $\mathbb F.$ Let $\rm dim (V)=n>0$ and $\mathcal {B}=\{v_1,\ldots,v_n\}$ be a basis of $V.$ Now we know dimension of $V \otimes_\mathbb {F}V$ is $n^2$ as $V \otimes_\mathbb {F}V \cong \mathbb {F^{n^2}}.$ Now since the set $\mathcal {A}=\{v_i\otimes v_j:1 \leq i,j\leq n\}$ spans $V \otimes_\mathbb {F}V$ and the number of elements in $\mathcal {A}$ is $n^2$, $\mathcal {A}$ forms a basis for $V \otimes_\mathbb {F}V$ as a vector space over $\mathbb F.$ In this way every element in $\mathcal {A}$ is non zero.
Now my question is if I don't want to use the above arguments how can I show that for any $i$ and $j$ the element $v_i\otimes v_j$ is non zero in $V \otimes_\mathbb {F}V$. By the construction of tensor product if it can be shown then it will help me.
Help me. Many Thanks.