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Let $V=\mathbb C^2$ with standard basis $\{e_1,e_2\}$. Are there $v,w\in V$ s.t. $e_1\otimes e_1+e_2\otimes e_2\in V\otimes V$ can be written as $v\otimes w$

Is the answer no ?

for example if $v=c_1e_1+c_2e_2$ then

$v\otimes w=c_1e_1\otimes w+c_2e_2\otimes w=e_1\otimes c_1w+e_2\otimes c_2w$ is this true ?

and by uniqueness $c_iw=e_i$ then we reach a contradiction ?

Is this always valid, no matter what $V$ is ?

user257
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  • In general, this problem of determining whether a tensor can be written in separable form $\mathbf{v}\otimes \mathbf{w}$ is very difficult (NP-hard) and is the subject of a great deal of research in quantum information theory. Your particular example is indeed non-separable. In fact, in a certain sense, your example is as non-separable as it gets. – EuYu May 19 '16 at 11:54
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    Note the duplicate question has the base space of dimension $3$ rather than $2$, but this makes little difference. However, it is essential that it involves a tensor product of only two spaces; larger tensor products would be another piece of cake (in fact not a piece of cake at all). – Marc van Leeuwen May 19 '16 at 11:57

1 Answers1

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Yes, whatever $V$ is, if $(e_i)$ is a basis, then an element of the form $u=\sum e_i\otimes e_i$ with at least two terms in the sum cannot be written as a pure tensor.

You for instance can see that using the isomorphism $V\otimes V\to End(V)$ defined by $e_i\otimes e_j\mapsto (x\mapsto x_ie_j)$ (where $x_i$ is the coordinate of $x$ corresponding to $e_i$). Then the pure tensors correspond to endomorphisms of rank $1$, but the sum $u$ is sent to an endomorphism of rank the number of terms.

Captain Lama
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