Timeline for When is a matrix power nonnegative
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Dec 15, 2018 at 8:06 | comment | added | Gerry Myerson | For more discussion of index of primitivity, see math.stackexchange.com/questions/450090/… | |
| Mar 5, 2018 at 7:33 | comment | added | Jochen Glueck | All right, I see want you meant. By the way, +1 for the details on eventually positive matrices you gave in your answer, and for the excellent references. | |
| Mar 4, 2018 at 22:13 | history | edited | Pietro Paparella | CC BY-SA 3.0 |
Better exposition; references added
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| Mar 4, 2018 at 21:52 | history | edited | Pietro Paparella | CC BY-SA 3.0 |
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| Mar 4, 2018 at 21:51 | comment | added | Pietro Paparella | I see. I interpreted the statement that "all entries of $u^\top$ and $v$ have the same sign" as $\text{sgn}(u) = \text{sgn}(v)$. Thanks for pointing out my errors. | |
| Mar 4, 2018 at 20:10 | comment | added | Jochen Glueck | Is your concern about the sign of the entries of the left and right eigenvector? Robert Israel wrote that all entries of the left and the right eigenvector are non-zero and have all the same sign. That's equivalent to saying that the left and the right eigenvector are either both positive or both negative. But maybe I overlooked something or I'm misinterpreting something? | |
| Mar 4, 2018 at 18:50 | comment | added | Pietro Paparella | @JochenGlueck: see my new edit. | |
| Mar 4, 2018 at 18:49 | history | edited | Pietro Paparella | CC BY-SA 3.0 |
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| Mar 4, 2018 at 10:35 | comment | added | Jochen Glueck | I think it's very good to have all the references you gave in an answer now; the literature on eventually positive matrices is of course very relevant here. Still, I don't see how what you wrote contradicts Robert Isreal's answer. | |
| Mar 3, 2018 at 16:38 | history | edited | Pietro Paparella | CC BY-SA 3.0 |
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| Mar 3, 2018 at 14:55 | history | edited | Pietro Paparella | CC BY-SA 3.0 |
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| Mar 3, 2018 at 14:50 | comment | added | Pietro Paparella | Indeed; I always forget to leave that out! | |
| Mar 3, 2018 at 9:42 | comment | added | Jochen Glueck | The characterisation you gave at the beginning of your answer is only correct for symmetric matrices; for non-symmetric matrices you also need to assume that the left eigenvector is positive. Consider the matrix $A := \begin{pmatrix} 2 & -1 \\ 2 & -1 \end{pmatrix}$; then $A$ has spectral radius $1$, this is a simple eigenvalue and $(1,1)$ is a corresponding right eigenvector. But $A$ is a projection and thus not eventually positive. | |
| Mar 3, 2018 at 4:50 | history | edited | Pietro Paparella | CC BY-SA 3.0 |
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| Mar 3, 2018 at 4:03 | history | answered | Pietro Paparella | CC BY-SA 3.0 |