Is it true that every element of $V \otimes W$ is a simple tensor $v \otimes w$? I know that every vector in a tensor product $V \otimes W$ is a sum of simple tensors $v \otimes w$ with $v \in V$ and $w \in W$. In other words, any $u \in V \otimes W$ can be expressed in the form$$u = \sum_{i=1}^r v_i \otimes w_i$$for some vectors $v_i \in V$ and $w_i \in W$. This follows from the proof of the existence of $V \otimes W$, where one shows that $V \otimes W$ is spanned by the simple tensors $v \otimes w$; the assertion now follows from the fact that, in forming linear combinations, the scales can be absorbed in the vectors: $c(v \otimes w) = (cv) \otimes w = v\otimes (cw)$.
My question is, is it true in general that every element of $V \otimes W$ is a simple tensor $v \otimes w$?

Lets assume that $\dim V=m$ and $\dim W=n$ with $m\geq 2$ and $n\geq2$. 
Suppose that $\{v_1,v_2\}$ are linearly independent in $V$ and that $\{w_1,w_2\}$ are linearly independent in $W$. Seeking a contradiction, suppose that
$$
v_1\otimes w_1+v_2\otimes w_2=v\otimes w\tag{1}
$$
Then extend $\{v_1,v_2\}$ to a basis $\{v_1,v_2,v_3,\dotsc,v_m\}$ of $V$ and extend $\{w_1,w_2\}$ to a basis $\{w_1,w_2,\dotsc,w_n\}$ of $W$. Write
\begin{align*}
v &= \alpha_1\,v_1+\dotsb+\alpha_m\,v_m \\
w &= \beta_1\,w_2+\dotsb+\beta_n\,w_m
\end{align*}
so that
$$
v\otimes w=\sum_{i=1}^m\sum_{j=1}^n\alpha_i\beta_j\,v_i\otimes w_j\tag{2}
$$
But $\mathcal B=\{v_i\otimes v_j:1\leq i\leq m,1\leq j\leq n\}$ is a basis for $V\otimes W$ (check this!) so (1) and (2) imply
$$
\alpha_i\beta_j=
\begin{cases}
1 & i=j=1 \\
1 & i=j=2 \\
0 & \text{otherwise}
\end{cases}
$$
which clearly is equivalent to 
\begin{align*}
\alpha_i
&=
\begin{cases}
1 & i =1,2 \\
0 & i\neq 1,2
\end{cases}
&
\beta_j
&=
\begin{cases}
1 & j=1,2 \\
0 & j\neq 1,2
\end{cases}
\end{align*}
Now, we may rewrite (1) as
\begin{align*}
v_1\otimes w_1+v_2\otimes w_2
&= v\otimes w \\
&= (v_1+v_2)\otimes (w_1+w_2) \\
&= v_1\otimes w_1+v_1\otimes w_2+v_2\otimes w_1+v_2\otimes w_2
\end{align*}
which contradicts that $\mathcal B$ is a basis for $V\otimes W$.
