Timeline for Spectral Radius and Spectral Norm for Markov Operators
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 17, 2023 at 17:29 | vote | accept | Sam OT | ||
| Mar 16, 2023 at 17:49 | comment | added | Sam OT | So, to be clear, my chain is nothing like yours. You have shown, very clearly, that a generic statement that I'd want doesn't hold. I guess I need to dig more into the details of my chain. I wanted to avoid that! 🙁 | |
| Mar 16, 2023 at 17:44 | comment | added | Sam OT | I think it is, unfortunately! I was thinking that any matrix $P$ with eigenvalues $1$ (constant vector) and $0$ (multiplicity $n-1$) must have all rows equal. But, that's only true if you require $P$ to be self-adjoint! (and so has a spectral decomposition). I guess I should just check that the spectral gap of the additive symmetrisation does indeed tend to $0$, but I guess this should be easy... | |
| Mar 16, 2023 at 14:39 | comment | added | DRJ | Indeed, thanks, this is now corrected. Is this not a counter-example to what you asked ? | |
| Mar 16, 2023 at 14:39 | history | edited | DRJ | CC BY-SA 4.0 |
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| Mar 16, 2023 at 10:53 | comment | added | Sam OT | Thanks, @DRJ, for your answer. I'm a bit confused/sceptical of a couple of things. Before diving into those, perhaps you can correct a typo in your example? For you, $x \in \{0,1\}^n$, but then the two potential states are both in $\{0,1\}^{n-1}$. Were they supposed to be $(x_2, ..., x_n, 0/1) \in \mathbb R^n$? In words, "Kill first bit and append a uniform bit." | |
| Mar 15, 2023 at 16:18 | history | edited | DRJ | CC BY-SA 4.0 |
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| Mar 15, 2023 at 16:10 | history | edited | DRJ | CC BY-SA 4.0 |
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| Mar 15, 2023 at 9:34 | history | edited | DRJ | CC BY-SA 4.0 |
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| Mar 15, 2023 at 9:29 | history | answered | DRJ | CC BY-SA 4.0 |