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Consider a real vector space $V^{(1)}\otimes V^{(2)}$ where $\otimes$ is the tensor product. Product vectors are of the form $v_1\otimes v_2$ where $v_1\in V^{(1)}$, $v_2\in V^{(2)}$, anything else is a non-product vector. I conjecture the following

I have a set of product vectors $a_i\otimes b_i$, $i=1,\dots,n$ that sum to make a product vector $A\otimes B$, $A\otimes B=\sum_i \alpha_i a_i \otimes b_i$ where $\alpha_i$ are real, non-zero constants and there are no repetitions of $a_i$. All of these product vectors must have $b_i= \lambda \, B$ $\forall$ $i$, where $\lambda$ is some non-zero real.

I believe this is true, but am not sure how to proceed. For example, if $n=2$, then $A\otimes B=a_1\otimes b_1 + a_2 \otimes b_2$ is not a product vector unless $a_1=a_2$ or $b_1 =b_2$, and then it is clear that either $a_1=a_2=A$ or $b_1=b_2=B$

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  • $\begingroup$ You can show your vector space has basis $v \otimes w$ where $v$ runs over the basis of $V^1$ and $w$ over the basis of $V^2$. That's probably how it was defined actually, since you have this question tagged with linear algebra. $\endgroup$ – 3-in-441 Sep 19 '16 at 21:25
  • $\begingroup$ Is the $+$ a $\oplus$ (direct sum), or just a sum? $\endgroup$ – Daniel Robert-Nicoud Sep 19 '16 at 21:30
  • $\begingroup$ @DanielRobert-Nicoud $+$ is the sum, not the direct sum. $\endgroup$ – jdizzle Sep 19 '16 at 21:32
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    $\begingroup$ @jdizzle: I have already put enough effort in answering your original question, keeping up with your comments and trying to help you reformulating. Now you are asking me to delete my answer (which referred to the initial form of your question and has been already accepted by you). I think this is unfair for me and disruptive for the site itself. This is why I advised you to ask ask a new question. Letting alone that the new form of your question is rather vague (again). you are leaving me no other choice than to flag your post (this means: asking moderators intervention to handle it). $\endgroup$ – KonKan Sep 19 '16 at 23:14
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    $\begingroup$ If someone has taken the time to answer your question, do not modify it in such a way it deems the answers useless. Please, post another question instead. I will revert to the original question. Regards, $\endgroup$ – Pedro Tamaroff Sep 20 '16 at 3:24
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Since, $X\otimes Y = P\otimes Q + A$ then yor statement:

$A$ is a linear sum of terms of this form ...

is valid (by definition). But the rest are not. Take for example $\mathbb{R}^4$, and a basis $\mathbf{e_1,e_2,e_3,e_4}$. Let $$ A=\mathbf{e_1\otimes e_2+e_3\otimes e_4} $$ Then $A$ has none of the forms: $P\otimes (\dots )$ or $(\dots )\otimes Q$ or $X\otimes (\dots )$ or $(\dots )\otimes Y$

Thus, your conjecture is not true in general.

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  • $\begingroup$ $A$ is a subspace, not a single vector. $\endgroup$ – Daniel Robert-Nicoud Sep 19 '16 at 21:35
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    $\begingroup$ No it is not. Read the OP carefully please. $\endgroup$ – KonKan Sep 19 '16 at 21:36
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    $\begingroup$ Apologies, @KonKan is correct that A is a vector. The confusion was down to my poor phrasing of the question $\endgroup$ – jdizzle Sep 19 '16 at 21:38
  • $\begingroup$ @ Daniel Robert-Nicoud: I think it would be reasonable to cancel your downvote. $\endgroup$ – KonKan Sep 19 '16 at 21:40
  • $\begingroup$ @KonKan could you elaborate on this please? Is the a counter example or do you agree to some extent with the conjecture? If $A=e_1\otimes e_2+e_3\otimes e_4$, and we add a product vector $B$ to it to make another product vector, then the conjecture would imply that $B$ must be something like $B=e_1\otimes x - e_3 \otimes e_4$, etc. The equation can also be phrased as ``I have a general vector $A$, and ai add a product vector to it $B$. The result is a product vector $C$. Knowing $B$ and $C$, how does this constrain $A$?'' $\endgroup$ – jdizzle Sep 19 '16 at 21:44

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