Timeline for The rank of a "triangle-free" matrix
Current License: CC BY-SA 3.0
12 events
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Nov 16, 2017 at 22:51 | history | edited | Seva | CC BY-SA 3.0 |
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Jul 15, 2017 at 7:49 | history | edited | Seva | CC BY-SA 3.0 |
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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Apr 6, 2017 at 19:21 | comment | added | Wolfgang | @FedericoPoloni Yes don't worry, I agree. I just wanted to point out that it's not THE adjacency matrix. :) And to note that without the -1's, the problem might be much easier. | |
Apr 6, 2017 at 18:32 | comment | added | Seva | "Triangle-free" is just the term that came to my mind. Let's say, the support of $A$ is free of some triangles. Anyway, the linguistics is not that much of an issue here, the math is... | |
Apr 6, 2017 at 18:26 | comment | added | Federico Poloni | @Wolfgang What you write is true, but still "triangle-free" seems like the perfect name to describe this concept to me. | |
Apr 6, 2017 at 16:22 | comment | added | Tony Huynh | @Wolfgang Thanks you're right! Deleted my silly comment. | |
Apr 6, 2017 at 15:30 | comment | added | Wolfgang | @TonyHuynh The indices in $a_{ij}a_{jk}a_{ki}$ are cyclic. It is rather triangles which are not right ones - at least not with sides parallel to the axis - and only some of those. Try $\{i,j,k\}=\{1,2,3\}$. | |
Apr 6, 2017 at 15:21 | comment | added | Wolfgang | @FedericoPoloni If we have two matrices, one with (for a certain pair i,j) $a_{ij}=1,a_{ji}=0$ and one with $a_{ij}=0,a_{ji}=-1$, both describe the same directed graph, don't they? But in terms of ranks, they may have different ranks. I suppose this possibility is intended by the OP, otherwise it would be stated "all elements in $\{0,1\}$". | |
Apr 6, 2017 at 12:47 | comment | added | Federico Poloni | @AmirSagiv Because it is the adjacency matrix of a triangle-free directed graph. | |
Apr 6, 2017 at 11:15 | comment | added | Amir Sagiv | Just out of curiousity, why do you call it "triangle free"? | |
Apr 6, 2017 at 11:05 | history | asked | Seva | CC BY-SA 3.0 |