Say I have a space $V^{(1)}$ with basis $\{a_i \}$ and $V^{(2)}$ (with dimensions $d_1$, $d_2$ respectively) with basis $\{b_j\}$. Clearly the vectors $\{a_i\otimes b_j\}$ are a basis for $V^{(1)}\otimes V^{(2)}$. Lets call this a ``product basis''. Note that if you construct the sets given by the first and second tensor elements of this basis, you trivially recover $\{a_i\}$ and $\{b_j\}$ and these are linearly independent sets by construction. Is there a basis of $V^{(1)}\otimes V^{(2)}$ one can construct, where by the sets you get when you take the first (second) tensor elements of the vectors are linearly dependent? I.e. a set of vectors $\{s^(1)_i\otimes s^{(2)}_i \}$ where $i=1,\dots, d_1d_2$ where $\{s^{(1)}_i\}$ is a set of linearly dependent vectors, and similar for $\{s^{(2)}_i\}$?


For example, take $V^{(1)} = V^{(2)} = \Bbb R^2$, and let $\{e_1,e_2\}$ denote the canonical basis. We can take the basis $$ \{ e_1 \otimes e_1, (e_{1} + e_2) \otimes e_1, (e_1 - e_2)\otimes e_2, e_2 \otimes e_2 \} $$ By your definition, $\{s_i^{(1)}\}$ is a set of $4$ elements in $\Bbb R^2$, so it is clearly linearly dependent.

Both at the same time: $$ \{ e_1 \otimes e_1, (e_{1} + e_2) \otimes (e_1+e_2), (e_1 - e_2)\otimes (e_1 + e_2), e_2 \otimes e_2 \} $$

  • $\begingroup$ is there a choice where the sets formed by both tensor elements are linearly dependent? $\endgroup$ – jdizzle Jul 18 '16 at 20:59
  • $\begingroup$ Yep. See my edit. $\endgroup$ – Omnomnomnom Jul 18 '16 at 21:05

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