# Condition for a tensor to be decomposable

Let $V$ be a vector space of dimension 3 with basis $e_1,e_2,e_3$. Let $W$ be a vector space of dimension 2 with basis $f_1,f_2$. Is $e_1\otimes f_1+e_2\otimes f_2$ decomposable? What about $e_1\otimes f_2+e_1\otimes f_3-e_2\otimes f_2-e_2\otimes f_3$? What is the condition for a tensor $$v=\sum_{i=1}^3 \sum_{j=1}^2 a_{i,j} e_i \otimes f_j$$ to be decomposable, that is, has the form $v\otimes w$ for $v\in V, w\in W$. As a related question, if I'm given any two vector spaces with dimensions $m,n$ and given bases, what is the condition for a tensor in $V\otimes W$ to be decomposable?

I think this has something to do with the linear independence but I'm not very comfortable with the tensor product, so I'm not really sure where to begin.

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Basically the nonzero tensor associated to the matrix $A=[a_{ij}]$ is decomposable when $A=vw^T$ for some column vectors $v,w$; in other words rank 1. –  anon Apr 5 '12 at 5:02

$\newcommand\P{\mathbb{P}}$Let $V$ and $W$ be complex vector spaces, and let $\P(V)$, $\P(W)$ and $\P(V\otimes W)$ be the projective spaces attached to $V$, $W$ and $V\otimes W$, respectively. If $v\in V$ is non-zero, I'll write $[v]$ the point of $\P(V)$ corresponding to it; it is the equivalence class of $v$ in $V\setminus0$ for an the equivalence relation of linear dependence.

Since decomposability of a tensor in does not change when we multiply it by a non-zero scalar, we can talk about the indecomposable elements of $\P(V\otimes W)$. Your question is therefore more or less equivalent to

how can we describe the set of indecomposable elements of $\P(V\otimes W)$?

Now, there is a map $f:\P(V)\times\P(W)\to\P(V\otimes W)$ which maps $([v],[w])$ to $[v\otimes w]$. This is a map of projective varieties (in the sense of algebraic geometry) and its image is precisely the set of indecomposable tensors. The image is in fact a subvariety of $\P(V\otimes W)$, which means that it is the common zero set of a finite set of polynomials. Finding these polynomials explicitely in a classical problem, solved long ago. If you want information about this, the keyword is Segre embedding; Wikipage has a page on it, and most introductions to algebraic geometry will say something.

In the special case where $\dim V=3$ and $\dim W=2$, with bases $\{x_1,x_2,x_3\}$ and $\{y_1,y_2\}$, we want conditions on the coefficients of a tensor $$\sum_{\substack{1\leq i\leq 3\\1\leq j\leq2}}f_{i,j}x_i\otimes y_j$$ to be equal to a product $$\Bigl(\sum_{1\leq i\leq 3}v_ix_i\Bigr)\otimes\Bigl(\sum_{1\leq j\leq 2}w_iy_i\Bigr).$$

It is easy to see that we must have $$f_{i,j}f_{k,l}=f_{k,l}f_{i,l}$$ for all $i,k\in\{1,2,3\}$ and all $j,l\in\{1,2\}$ for that to happen, and some work will show that this conditions are in fact sufficient. We can express all these conditions by saying that the matrix $$\begin{pmatrix}f_{1,1}&f_{1,2}\\f_{2,1}&f_{2,2}\\f_{3,1}&f_{3,2}\end{pmatrix}$$ has rank $1$. Proving this is «just» linear algebra.

The answer in the general case where the dimensions are arbitrary is of the same spirit.

N.B.: it is interesting to know that the question «which tensors have rank $k$?» for $k\geq2$ and mora than two factors is much, much harder, and very important —and I think this is unsolved in general. Someone who knows algebraic geometry might be able to tell us.

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