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 $v \otimes w$?
Best Answer
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 $V\otimes W$.