Vector spaces and Kronecker product Can we express any vector  $z \in \mathbb{R}^{nm}$ as the Kronecker product of elements of $x\in\mathbb{R}^n$ and $y\in\mathbb{R}^m$?
I was working a little example where $n=m=2$ and we can write the standard/natural basis of $\mathbb{R}^4$ as the Kronecker product of the standard basis for $\mathbb{R}^2$,
$$ \left[
\begin{array} 
 \\ 1   
 \\ 0  
\end{array} \right] \otimes \left[
\begin{array} 
 \\ 1   
 \\ 0  
\end{array} \right] = \begin{array}[llll] \\ [1 & 0 & 0 & 0]^\top \end{array} 
$$
$$ \left[
\begin{array} 
 \\ 1   
 \\ 0  
\end{array} \right] \otimes \left[
\begin{array} 
 \\ 0  
 \\ 1  
\end{array} \right] = \begin{array}[llll] \\ [0 & 1 & 0 & 0]^\top \end{array} 
$$
and so on.
I would appreciate some insights.
 A: The answer is no. Consider for example the case of a vector
$$V=(u,v,w,x) \in \mathbb {R^4}$$
Let us show that it isn't in general the Kronecker product of two vectors
$$(a,b), \ (c,d)\in \mathbb {R^2}$$
Indeed, identifying
$$(a,b) \otimes (c,d)=(ac,ad,bc,bd)$$
with $U$ gives the 4 conditions:
$$\begin{cases}ac&=&u\\ad&=&v\\bc&=&w\\bd&=&x\\\end{cases}$$
but this would imply that the following products are identical (to $abcd$)
$$ux=vw$$
which has no reason to hold in general.
A: No, we have $$\mathbb{R}^{nm} \cong \mathbb{R}^n \otimes \mathbb{R}^m$$, so this is equivalent to asking whether every tensor is pure, and this is false (at least for either $n$ or $m$ greater than $1$). To see this, note that $\otimes: V \times W \to V \otimes W$ cannot be surjective unless $m + n = \dim(V \times W) \geq \dim(V \otimes W) = mn$ by a dimension argument.
A: The answer is no. But you already knew that.
But quite many times we can find very close low-rank approximations of tensors.
Singular Value Decomposition ( SVD ) compression is a special case of this for matrices.
In fact this is quite often used both in data compression and machine learning and was a hot research subject at least 5-10 years ago and maybe still is today.
