# A conjecture on tensor product vectors

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 $b_1=b_2=B$

• 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. – 3-in-441 Sep 19 '16 at 21:25
• Is the $+$ a $\oplus$ (direct sum), or just a sum? – Daniel Robert-Nicoud Sep 19 '16 at 21:30
• @DanielRobert-Nicoud $+$ is the sum, not the direct sum. – jdizzle Sep 19 '16 at 21:32
• 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, – Pedro Tamaroff Sep 20 '16 at 3:24

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.

• $A$ is a subspace, not a single vector. – Daniel Robert-Nicoud Sep 19 '16 at 21:35
• No it is not. Read the OP carefully please. – KonKan Sep 19 '16 at 21:36
• Apologies, @KonKan is correct that A is a vector. The confusion was down to my poor phrasing of the question – jdizzle Sep 19 '16 at 21:38
• @ Daniel Robert-Nicoud: I think it would be reasonable to cancel your downvote. – KonKan Sep 19 '16 at 21:40
• @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 $A$?'' – jdizzle Sep 19 '16 at 21:44