Timeline for Are Diagonally dominant Tridiagonal matrices diagonalizable?
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Nov 12, 2023 at 22:50 | comment | added | jdods | What if the matrix is irreducible? The counterexamples here seem to be reducible. I know that doesn't changes things for a transition matrix, since an irreducible, aperiodic one can still be non-diagonalizable. I can't find any diagonalizable info on generators, or generally matrices with negative diagonal. | |
| Dec 18, 2016 at 11:33 | history | edited | Federico Poloni | CC BY-SA 3.0 |
alternate idea
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| Dec 18, 2016 at 11:18 | comment | added | Federico Poloni | @EmilioPisanty Done! | |
| Dec 18, 2016 at 11:16 | history | edited | Federico Poloni | CC BY-SA 3.0 |
Added how to construct it, on request
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| Dec 18, 2016 at 10:52 | comment | added | Emilio Pisanty | Ah, that's a nice fact to have around. Thanks for the insight! (and consider writing it into the answer). | |
| Dec 18, 2016 at 8:16 | comment | added | Federico Poloni | @EmilioPisanty My plan was "let's construct a Jordan block setting some $c_i$ to 0 and then pad it". It is a known result that if $T_{11}$ and $T_{22}$ have no eigenvalues in common, then $\begin{bmatrix}T_{11}&T_{12}\\0&T_{22}\end{bmatrix}$ and $\begin{bmatrix}T_{11}&0\\0&T_{22}\end{bmatrix}$ are similar, so I just made $T_{11}$ a Jordan block, chose $T_{22}$ with no eigenvalues in common with it, and the rest followed. | |
| Dec 18, 2016 at 8:13 | comment | added | Federico Poloni | @PatDevlin Sorry for the scooping! | |
| Dec 18, 2016 at 3:58 | comment | added | Pat Devlin | @EmilioPisanty When I came up with my example (I've been scooped!) the thought process was (1) try to make it obviously not diagonalizable [e.g., in this case, the Jordan block in the top left does the trick], and (2) make it otherwise as simple as possible. Counterexamples are easy to come by, I'm sure. Another thought here is that you don't want to check "random" or "generic" things because those are diagonalizable. Instead, you want to think of small cases (2x2 and 3x3 don't yield counterexamples without thinking too long about it, so try 4x4). | |
| Dec 18, 2016 at 1:23 | comment | added | Emilio Pisanty | Is there some deeper underlying reason for this? Or are the OP's conditions simply not specific enough for any strong results to follow? That is, what was the thought process that generated this counterexample? | |
| Dec 17, 2016 at 19:05 | vote | accept | KNN | ||
| Dec 17, 2016 at 19:01 | history | edited | Federico Poloni | CC BY-SA 3.0 |
added 109 characters in body
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| Dec 17, 2016 at 18:53 | comment | added | KNN | Interesting ! Thank you for your time ! | |
| Dec 17, 2016 at 18:51 | comment | added | Pat Devlin | I'm very glad to see this counterexample. Could you also provide the characteristic polynomial just to make it easier to verify? | |
| Dec 17, 2016 at 18:50 | comment | added | Federico Poloni | @DuyNguyen It did before you edited it. :) Anyway, there are counterexamples with $a_i\neq 0$ as well, I have modified the answer to show one. | |
| Dec 17, 2016 at 18:49 | history | edited | Federico Poloni | CC BY-SA 3.0 |
added 31 characters in body
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| Dec 17, 2016 at 18:48 | review | Low quality posts | |||
| Dec 17, 2016 at 18:52 | |||||
| Dec 17, 2016 at 18:33 | comment | added | Pat Devlin | Ah. I didn't see that. | |
| Dec 17, 2016 at 18:32 | comment | added | KNN | I don't think this answer my question, it requires $a_i<0$ | |
| Dec 17, 2016 at 18:31 | comment | added | Pat Devlin | You beat me to it! I was just typing this now. | |
| Dec 17, 2016 at 18:30 | history | answered | Federico Poloni | CC BY-SA 3.0 |