# How can I show that $v_i\otimes v_j$ is non zero in $V \otimes_\mathbb {F}V$

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.

Whatever your definition of tensor is, it should be true that any bilinear function $$V\times V\longrightarrow \mathbb{F}$$ must factor uniquely through $$V\times V\longrightarrow V\otimes_\mathbb{F} V\longrightarrow \mathbb{F}$$ Now take any linear functional $\ell :V\to\mathbb{F}$ such that $\ell(v_i)\neq0$ and $\ell(v_j)\neq 0$. Define a bilinear function $f:V\times V\to \mathbb{F}$ as $$f(u,v) = \ell(u)\ell(v)$$ Since $f$ is clearly bilinear, $f$ must factor through a homomorphism $g:V\otimes_\mathbb{F} V\to \mathbb{F}$, i.e. $g(u\otimes v) = f(u,v)$. Since $g(v_i\otimes v_j) = f(v_i,v_j) = \ell(v_i)\ell(v_j)\neq 0$, $v_i\otimes v_j$ is nonzero.
If $(v_i,v_j)$ are non-zero there is a bilinear application $\beta : V \times V \to k$ with $\beta(v_i,v_j) = 1$ and by the universal property of tensor product we get a map $b : V \otimes V \to k$ with $b(v_i \otimes v_j) = 1$ so it cannot be zero.