Timeline for If there are eigenvectors with largest components $i$ resp. $j$, then is there an eigenvector with two largest components $i$ and $j$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 6, 2021 at 22:17 | vote | accept | M. Winter | ||
| Jan 6, 2021 at 11:22 | answer | added | M. Winter | timeline score: 0 | |
| Jan 5, 2021 at 10:30 | comment | added | M. Winter | @dodd For an arbitrary subspace there is no meaning in "an edge $ij\in E$". But you could consider the smallest eigenvalue (that is, the corresponding eigenspace) of the 5-cycle graph. Then the $u_i$ exist for all vertices, but $u_{ij}$ exists for no edge. | |
| Jan 5, 2021 at 2:38 | comment | added | M. Winter | @dodd This subspace is not just any subspace, but it is a very special eigenspace of an irreducible symmetric 01-matrix, and I ask for the existence of $u_{ij}$ only if the $(i,j)$-entry of that matrix is one. I consider the formulation in terms of graphs more natural than just talking about this matrix (and I do not see how this can be reasonably formulated just in terms of subspaces). Of course, I am happy with an answer in any language, whether graphs, matrices, subspaces, etc. | |
| Jan 5, 2021 at 2:18 | comment | added | M. Winter | @dodd I restricted the question to connected graphs now. | |
| Jan 5, 2021 at 2:18 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Jan 5, 2021 at 2:16 | comment | added | markvs | Try the graph with two vertices and no edges. It has one eigenvalue of multiplicity 2. | |
| Jan 5, 2021 at 2:02 | comment | added | M. Winter | @dodd Correct me if I am wrong, but I think this follows from the Theorem of Perron-Frobenius, at least for connected graphs. If it makes a difference, I should restrict my question to connected graphs, but I do not think so. | |
| Jan 4, 2021 at 22:02 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Jan 4, 2021 at 21:57 | history | asked | M. Winter | CC BY-SA 4.0 |