Is it true that every element of $V \otimes W$ is a simple tensor $v \otimes w$?

Question

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 $v \otimes w$?

Answer 1

This is in general not true. One easy way to see this, is the following: Assume $V$, $W$ are $n$ and $m$ dimensional vector fields over the complex numbers, then it is fairly easy to show that $V\otimes W$ is isomorphic to $\mathbb{C}^{n\times m}$ with the following isomorphism $\phi: V\otimes W \rightarrow \mathbb{C}^{m\times n}$, which is defined as

$\phi(u\otimes w) := uw^{H}$ for elementary tensors and extended by linearity for all $x \in V\otimes W$.

Now due to ismorphism between both spaces, we can study the same question in $\mathbb{C}^{n\times m}$: Notice that elementary tensors $u\otimes w$ correspond to matrices $uw^{H}$, hence only of rank one. A general matrix $M\in\mathbb{C}^{n\times m}$ can be decomposed into a linear combination of rank one matrices (using SVD for example) but obviously is not corresponding to a rank one matrix. Hence transferring this property to $V\otimes W$ using the isomorphism, we see that not every element in $V\otimes W$ is an elementary tensor.

Answer 2

It is necessarily possible to write all tensors as a simple tensor when either $V$ or $W$ has dimension $1$ (or $0$). I will leave this to you to confirm, the proof is fairly trivial.

On the other hand, suppose that $V$ and $W$ have dimensions $2$ or greater, and that we are given a basis of vectors $v_i$ and $w_i$ for the respective spaces. Then I am fairly certain that we can say: $$ v_1 \otimes w_2 + v_2 \otimes w_1 \neq v \otimes w \qquad \forall v \in V, w \in W $$ In the context of quantum mechanics, this amounts to the statement "the pure state $v_1 \otimes w_2 + v_2 \otimes w_1$ is entangled" (QM assumes, however, that $V$ and $W$ have an inner product).

In fact, the states $v_1 \otimes w_1 \pm v_2 \otimes w_2$ and $v_1 \otimes w_2 \pm v_2 \otimes w_1$ are referred to as the Bell States, since they are the "canonical example" of entanglement (i.e. none of these vectors can be written as simple tensors).

Answer 3

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 $V\otimes W$.

Answer 4

As has been pointed out, $V\otimes W$ is isomorphic to $Hom(V, W)$ once you pick a basis, with simple elements going to rank one matrices. Since (over any coefficients) as soon as $V$ and $W$ both have dimension at least 2 there is always a linear map from $V$ to $W$ with at least 2D image, we know that not every matrix is rank 1 and so not every tensor in $V\otimes W$ is simple.

I want to note that there is another proof that works for real or complex coefficients, or coefficients in a finite field. It works simply by comparing dimensions.

In the real and complex cases, the canonical map $V\times W/scalars \to V\otimes W$ is not linear, but it is smooth. And there is no surjective smooth map from a smaller dimensional space to a larger dimensional space, by say, Sard's theorem. This means there are non-simple elements in $V\otimes W$.

If $V$ and $W$ are real vector spaces the dimensions or direct sum and tensor product are $\dim V+\dim W$ and $\dim V\dim W$, respectively. (For complex vector spaces the real dimensions are $2\dim V+2\dim W$ and $2(\dim V\dim W)$.) So as long as $\dim V+\dim W-1< \dim V \dim W$ i.e. $(1-\dim V)(1-\dim W)<0$ i.e. one of $V$ and $W$ is at least 2-dimensional, we are done.

For coefficients in finite fields, we can see that there is no surjective map from a smaller dimensional space to a larger dimensional space simply by counting elements, and then the proof is the same.

Answer 5

Is every real $m\times n$ matrix of the form $u \cdot v^T$ for some vectors $u\in \mathbb R^m$ and $v\in \mathbb R^n$?