Timeline for answer to What are some counter-intuitive results in mathematics that involve only finite objects? by Borbei
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| when toggle format | what | by | license | comment | |
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| Jul 26, 2019 at 20:12 | comment | added | user584285 | “It is my experience that proofs involving matrices can be shortened by 50% if one throws the matrices out.” -- E. Artin (Geometric Algebra, p. 14) | |
| Feb 14, 2019 at 17:01 | comment | added | leftaroundabout | @JyrkiLahtonen I only gave SVD as one alternative lin-map representation example. The point of my comment was that linear mappings should foremostly be understood as linear functions, for which matrices are also merely one representation. | |
| Feb 14, 2019 at 14:24 | comment | added | Jyrki Lahtonen | @leftaroundabout I don't find the use singular value decompositions very illuminating here because those only exist for real or complex matrices whereas the theorem holds for any field. | |
| Dec 16, 2016 at 14:41 | audit | Low quality posts | |||
| Dec 16, 2016 at 14:41 | |||||
| Dec 11, 2016 at 11:02 | comment | added | Mozibur Ullah | @leftaroundabout: well, I don't see what we're arguing about in this case; I already mentioned 'linear mappings' in my comment; the use of the word abstract, in my mind, is simply an added qualifier to distinguish between matrices as a concrete representation of linear maps from linear maps axiomatically (ie abstractly treated) treated. | |
| Dec 11, 2016 at 10:57 | comment | added | leftaroundabout | @MoziburUllah there is no such difference, I just said abstract to emphasize that there's no need to write out the linear mapping in any particular notation (like a matrix). It's sufficient to describe it as a generic function $\varphi : V\to W$ between two vector spaces which has the linearity property $\varphi(\mu\cdot u + v) = \mu\cdot\varphi(u) + \varphi(v)$. Sure enough, any such function (if the spaces are finite-dimensional) can be written in matrix notation, but my point is that this isn't always such a good idea since it obscures interesting properties (like well-defined rank). | |
| Dec 11, 2016 at 10:12 | comment | added | Mozibur Ullah | @leftaroundabout: I fail to see the difference between abstract linear mappings and just linear mappings - can you explain the difference? | |
| Dec 11, 2016 at 10:09 | comment | added | leftaroundabout | @MoziburUllah there are plenty of representations. One representation that makes it particularly clear that the rank doesn't depend on the space dimension is the singular value decomposition (which can be defined without ever talking about matrices, only abstract linear mappings). | |
| Dec 11, 2016 at 6:51 | comment | added | Mozibur Ullah | @leftaroundabout: matrices are the only representation I know of linear mappings, what else did you have in mind? | |
| Dec 6, 2016 at 23:33 | comment | added | Mike Jones | “Young man, in mathematics you don't understand things. You just get used to them.” -- John von Neumann | |
| Dec 3, 2016 at 21:17 | comment | added | leftaroundabout | Arguably, the reason for the counterintuitive matrix rank is that matrices are just a not-really-optimal representation of linear mappings. It's fairly intuitive that the image of a linear mapping has the same dimension as the image of its adjoint mapping. | |
| Dec 3, 2016 at 19:58 | history | answered | Borbei | CC BY-SA 3.0 |