# When are “square spans” not transversal?

Let $$V$$ be a finite-dimensional vector space over a field $$K$$. Given a basis $$\{v_1,\dotsc,v_n\}$$ for $$V$$, we define the "square span" of the basis to be the subspace of $$V\otimes V$$ spanned by $$v_1\otimes v_1,v_2\otimes v_2,\dotsc,v_n\otimes v_n$$.

It seems clear that, for $$n\geq 2$$, if the square spans of two bases $$v=\{v_1,\dotsc,v_n\}$$, $$v'=\{v_1',\dotsc,v_n'\}$$ have non-trivial intersection, then the two bases have proper non-trivial subsets spanning the same space. What happens given three bases $$v$$, $$v'$$ and $$v''$$ such that their square spans, taken together, span a subspace of $$V\otimes V$$ of dimension less than $$3\dim(V)$$? What can one say about $$v$$, $$v'$$ and $$v''$$ then?

• If $n=2$ then all pairs of bases have nontrivially intersecting square spans. – MTyson Dec 8 '18 at 20:04
• Let's say $n>2$. – H A Helfgott Dec 8 '18 at 21:18