# Prove a linear combination of tensor product can not be written as a single tensor product

Prove that in $\mathbb{C}^{3}\otimes\mathbb{C}^{3}$, the state vector $$\mathbf{h}=\frac{1}{\sqrt{8}}=e_{1}\otimes e_{1}+e_{2}\otimes e_{2}+e_{1}\otimes e_{2}+e_{2}\otimes e_{1}+e_{1}\otimes e_{3}+e_{3}\otimes e_{1}+e_{2}\otimes e_{3}+e_{3}\otimes e_{2}$$

cannot be written in the form $h_{1}\otimes h_{2}.$

========= My proof goes as follows:

Without loss of generality, let $e_{1}=\left(\begin{array}{c} 1\\ 0\\ 0\end{array}\right), e_{2}=\left(\begin{array}{c} 0\\ 1\\ 0\end{array}\right), e_{3}=\left(\begin{array}{c} 0\\ 0\\ 1\end{array}\right),$

then $e_{1}\otimes e_{1}=\left(\begin{array}{c} 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right), e_{2}\otimes e_{2}=\left(\begin{array}{c} 0\\ 0\\ 0\\ 0\\ 1\\ 0\\ 0\\ 0\\ 0\end{array}\right), e_{1}\otimes e_{2}=\left(\begin{array}{c} 0\\ 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right),$ $e_{2}\otimes e_{1}=\left(\begin{array}{c} 0\\ 0\\ 0\\ 1\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right), e_{1}\otimes e_{3}=\left(\begin{array}{c} 0\\ 0\\ 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right), e_{3}\otimes e_{1}=\left(\begin{array}{c} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 1\\ 0\\ 0\end{array}\right), e_{2}\otimes e_{3}=\left(\begin{array}{c} 0\\ 0\\ 0\\ 0\\ 0\\ 1\\ 0\\ 0\\ 0\end{array}\right),$

$e_{3}\otimes e_{2}=\left(\begin{array}{c} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 1\\ 0\end{array}\right),$ and $\mathbf{h=\frac{1}{\sqrt{8}}}\left(\begin{array}{c} 1\\ 1\\ 1\\ 1\\ 1\\ 1\\ 1\\ 1\\ 0\end{array}\right)$

And eventually we will get some contradiction. But can this method be really applied "without loss of generality"? What is the way to do this without writing all these big vectors?

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## 2 Answers

Consider ${\mathbb C}^3 \otimes {\mathbb C}^3$ as corresponding to $3 \times 3$ matrices with complex entries. $u \otimes v$ corresponds to the matrix $u v^T$, which has rank 1 (if $u$ and $v$ are nonzero column vectors). So what is the rank of the matrix corresponding to $\bf h$?

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I believe h must have a rank of 8, but unsure about the precise steps of showing that. – Xiaowen Li Dec 2 '11 at 20:04
You believe a $3 \times 3$ matrix can have a rank of 8? I think you need to refresh your memory of the meaning of rank. – Robert Israel Dec 4 '11 at 7:56

write $h_1, h_2$ as linear combinations of $e_1, e_2, e_3$ and compute the tensor product in this representation. Compare the coefficients.

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to Thomas: that's exactly the where I am not sure how to do. Do we just simply follow the normal rules (like linearly) about tensor products? – Xiaowen Li Dec 2 '11 at 19:57
this comment (regarding rank) relates to Roberts answer ;-) – user20266 Dec 2 '11 at 19:58
look at the coefficient of $e_3 \otimes e_3$. What do you mean by 'normal rules'. You need to know, of course, how to multiply linear combinations of tensors. Look at the definitions/theorems you were shown. – user20266 Dec 2 '11 at 19:59
oops that was a wrong reply. Thanks for point it out. – Xiaowen Li Dec 2 '11 at 20:04