# Tensor product and linear dependence of vectors

Let $V_1, \ldots, V_k$ be complex vector spaces. Given $k$ vectors $v_1 \in V_1, \ldots, v_k \in V_k$, we define that the tensor product $v_1 \otimes \ldots \otimes v_k$ has rank 1. For any tensor $T \in V_1 \otimes \ldots \otimes V_k$, the rank of $T$ is the minimum $r \in \mathbb{N}$ such that $T$ can be written as a sum of $r$ rank 1 tensors. In this case, there are vectors $v_{1,1}, \ldots, v_{r,1} \in V_1, \ldots v_{1,k}, \ldots, v_{r,k} \in V_k$ such that

$$T = \sum_{i=1}^r v_{i,1} \otimes \ldots \otimes v_{i,k}.$$

I have two questions about the relation between this decomposition and the linear dependency of the vectors.

1) Suppose we have linearly independent vectors $v_{1,j}, \ldots,v_{r,j} \in V_j$, for each $j=1 \ldots k$, and construct the tensor $T = \sum_{i=1}^r v_{i,1} \otimes \ldots \otimes v_{i,k}$. Is it right write to say the rank of $T$ is $r$? If not, what conditions should be considered instead just independence?

2) On the other hand, suppose we know $T$ has rank $r$ and can be written as $T = \sum_{i=1}^r v_{i,1} \otimes \ldots \otimes v_{i,k}$. Is it right to say the vectors $v_{1,j}, \ldots,v_{r,j} \in V_j$, for each $j=1 \ldots k$, are linearly independent?

I'm aware that tensors are not so simple and probably these relations doesn't hold. In this case I'm also accepting suggestions in the following sense:

1) What properties the vectors should have in order to construct a tensor of rank $r$.

2) In the case we already have a tensor of rank $r$ (together with its decomposition), what properties the vectors forming it should have?

Thank you.

• crap there's no answer. This is exactly what I wanted to know! – user3391229 Feb 25 '18 at 20:24

Lemma 3.5. For $\ell=1 \ldots k$, let $x_1^{(\ell)}, \ldots, x_r^{(\ell)} \in \mathbb{R}^{d_\ell}$ be linearly independent. Then the tensor defined by $$\sum_{j=1}^r x_j^{(1)} \otimes x_j^{(2)} \otimes \ldots \otimes x_j^{(k)}$$ has rank $r$.
Proposition 4.6. Let $x_1, y_1 \in \mathbb{R}^{d_1}, x_2,y_2 \in \mathbb{R}^{d_2}$, and $x_3, y_3 \in \mathbb{R}^{d_3}$ be vectors such that each pair $x_i,y_i$ is linearly independent. Then the tensor $$x_1 \otimes x_2 \otimes y_3 + x_1 \otimes y_2 \otimes x_3 + y_1 \otimes x_2 \otimes x_3$$ has rank 3.
The tensor has rank 3 but in the decomposition we use $x_1$ as the first factor two times, so it's not necessary to have linearly independent factors in $\mathbb{R}^{d_1}$. The same goes for the other positions.