Timeline for Is this lemma in elementary linear algebra new?
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| Nov 16, 2019 at 0:23 | history | edited | Libli |
This question is intimately related to the projective geometry of $\mathbb{P}(A) \times \mathbb{P}(B) \subset \mathbb{P}(A \otimes B)$. This is why I added two tags : algebraic-geometry and projective geometry
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| Nov 16, 2019 at 0:21 | answer | added | Libli | timeline score: 3 | |
| Sep 6, 2013 at 20:24 | vote | accept | George Bergman | ||
| Sep 6, 2013 at 16:50 | answer | added | Clément de Seguins Pazzis | timeline score: 20 | |
| Sep 6, 2013 at 15:04 | comment | added | Suvrit | Somehow it feels that the following "dual" result is very closely related, and could for some fields also yield your inequality (by splitting the tensor product into "separable" and "joint" spaces, whose dimensions add up to $d_Ad_B$): On the maximal dimension of a completely entangled subspace..." by K. Parathasarathy; ias.ac.in/mathsci/vol114/nov2004/Pm2342.pdf --- in particular, your subspaces have the "separable" state property, while the cited paper considers "full entangled" subspaces. | |
| Sep 6, 2013 at 2:30 | comment | added | darij grinberg | Probably also related: Flanders' theorem (§8.3 in Prasolov's Linear Algebra book www2.math.su.se/~mleites/Prasolov/prasLinAlg/pr-linAlg-main.dvi ). | |
| Sep 6, 2013 at 1:23 | comment | added | Martin Brandenburg | This feels like a statement from projective geometry. $\mathbb{P}(V) \subseteq \mathbb{P}(A \otimes B)$ somehow "intersects enough" $\mathbb{P}(A) \times \mathbb{P}(B)$ so that $\dim(\mathbb{P}(V)) \geq \dim(\mathbb{P}(A) \times \mathbb{P}(B))$. | |
| Sep 6, 2013 at 0:50 | comment | added | Jack Huizenga | I don't have time to think about this right now, but it seems strikingly familiar to the following theorem of Hopf. If $f:A \otimes B\to C$ is a linear map which is injective on each factor separately, then $\dim f(A\otimes B) \geq \dim A + \dim B - 1.$ However, this theorem is true over $\mathbb{C}$ but false over $\mathbb{R}$ (the proof is given by algebraic topology), so maybe it is only a superficial observation. | |
| Sep 5, 2013 at 23:59 | comment | added | darij grinberg | In characteristic $\neq 2$, the space spanned by diagonal bilinear forms actually contains all symmetric bilinear forms. | |
| Sep 5, 2013 at 23:55 | comment | added | Michael | Could you clarify my confusion? Take $A=B$, identify their tensor product with bilinear forms, and take $V$ be the space spanned by diagonal bilinear forms. What am I missing here? | |
| Sep 5, 2013 at 21:09 | comment | added | user6976 | Welcome to Mathoverflow! | |
| Sep 5, 2013 at 20:47 | review | First posts | |||
| Sep 5, 2013 at 20:59 | |||||
| Sep 5, 2013 at 20:30 | history | asked | George Bergman | CC BY-SA 3.0 |