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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 $v \otimes w$?

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  • $\begingroup$ I am fairly sure that this will be true if and only if $V$ or $W$ has dimension $1$. I don't remember a proof off-hand. Unfortunately, proving that objects can't be reduced tends to be difficult when using any definition except the universal property. $\endgroup$ – Omnomnomnom Mar 16 '15 at 5:38
  • $\begingroup$ Intuitively, if we could make this simplification, then $V \otimes W$ would "look like" $V\oplus W$, which is "too small". $\endgroup$ – Omnomnomnom Mar 16 '15 at 5:39
  • $\begingroup$ Worth noting: This fact appeared on someone's math overflow post about "common false beliefs in mathematics" mathoverflow.net/a/23552/33377 $\endgroup$ – Brian Fitzpatrick Mar 16 '15 at 6:04
  • $\begingroup$ This kind of elements of the tensor or exterior algebra are usually called decomposable. $\endgroup$ – hjhjhj57 Mar 16 '15 at 6:05
  • $\begingroup$ Another common term for the same is separable. $\endgroup$ – Omnomnomnom Mar 16 '15 at 6:06
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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.

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  • $\begingroup$ I like this answer! This is the natural approach in the context of multi-linear algebra, wherein the tensor product of linear maps is taken to be a multilinear map. $\endgroup$ – Omnomnomnom Mar 16 '15 at 6:38
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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 $V\otimes W$.

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    $\begingroup$ $$ \alpha_i\beta_j= \begin{cases} 1 & i=j=1 \\ 1 & i=j=2 \\ 0 & \text{otherwise} \end{cases} $$ Does it not just imply that for example $\alpha_1$ and $\beta_1$ need to be inverses of each other, and not necessarily that $\alpha_1=1$ and $\beta_1=1$? $\endgroup$ – Pawel Jul 27 '16 at 2:00
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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).

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    $\begingroup$ I thought of a proof! I think. I'll give it a shot. $\endgroup$ – Omnomnomnom Mar 16 '15 at 5:52
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    $\begingroup$ Never mind, Brian has it covered. $\endgroup$ – Omnomnomnom Mar 16 '15 at 5:55
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    $\begingroup$ Is mine the same proof you were thinking of? What I ended up writing ended up sloppier than what I was thinking in my head! $\endgroup$ – Brian Fitzpatrick Mar 16 '15 at 5:56
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    $\begingroup$ @BrianFitzpatrick it's essentially the same; basically, we expand the product end up with a system of equations that has no solution. Maybe there's a way to make this nice with matrices. $\endgroup$ – Omnomnomnom Mar 16 '15 at 6:01
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This is equivalent to asking, does every multivariable polynomial factor into polynomials of one variable? No.

Consider the polynomial $x^2 + y$. If it could be factored into $P(x)Q(y)$ Then for some value of $y$, $Q(y) = 0$, and thus $x^2 + y = 0$ no matter the value of $x$. This is obviously false.


To be precise, let $\{1, x, x^2\}$ be a basis of $X$ and $\{1, y\}$ be a basis of $Y$. Then the basis of $X \otimes Y$ is just

$$\{x^2 \otimes y, x^2 \otimes 1, x \otimes y, x \otimes 1, 1 \otimes y, 1 \otimes 1 \}$$

The element $x^2 \otimes 1 + 1 \otimes y$ cannot be factored into a pure tensor.

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