Vector spaces and Kronecker product

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.