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?