Prove that every element $\displaystyle v\in (\mathbb{C}^{2})^{\otimes 3} $ is the sum of two pure tensors $\displaystyle u_{1}\otimes u_{2}\otimes u_{3} $., where $\displaystyle u_{1},u_{2},u_{3}\in\mathbb{C}^{2} $ .

Thanks in advance.

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- Dec 6th 2009, 02:32 PMsmith09Tensor
Prove that every element $\displaystyle v\in (\mathbb{C}^{2})^{\otimes 3} $ is the sum of two pure tensors $\displaystyle u_{1}\otimes u_{2}\otimes u_{3} $., where $\displaystyle u_{1},u_{2},u_{3}\in\mathbb{C}^{2} $ .

Thanks in advance. - Dec 6th 2009, 11:26 PMNonCommAlg
you're sure that

**two**is not**three**then? the reason for asking this is that, in general, for any 2-dimensional vector space $\displaystyle V$ over a field $\displaystyle F,$ the "maximum rank" of $\displaystyle V^{\otimes 3}$ is 3 and not 2, i.e. every

element of $\displaystyle V^{\otimes 3}$ is a sum of at most 3 simple tensors. - Dec 7th 2009, 01:26 AMsmith09
Thks, but the question requires 2, not 3. They give some hints that consider $\displaystyle v $ as a linear map from $\displaystyle \mathbb{C}^2$ to $\displaystyle M_2 $ (group of 2x2 matrices) (WHY? and HOW?P) and consider two possibilities of the dimension of the image (1 or 2). But I really dont get it.

- Dec 7th 2009, 11:42 AMsmith09
Oh, sorry, $\displaystyle v$ here must be in a dense open subset.

- Dec 15th 2009, 10:24 AMsmith09
Is there any ideal then? I think the first case (1 dim) is not so hard. But the second case is ...not easy.

Thks in advance.