Timeline for Is this "stretched eigenvector" studied? (If so, what are its properties?)
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| May 19, 2018 at 11:55 | comment | added | user539887 | @Nathaniel There is a well-developed theory of eigenvectors for homogeneous order-preserving systems on Banach spaces, having many applications in biology. You can, for example, look up "Horst R. Thieme" on MR. | |
| May 19, 2018 at 5:01 | comment | added | user44191 | Oh, I think I see where I misunderstood what you were saying. If you care more about the "stretched spectrum" than the eigenvectors, then the normalization gives a specific "stretched eigenvalue" out of the scalings of a solution vector for equation 2; this formulation is less useful for solving for a "stretched eigenvector" given a "stretched eigenvalue". | |
| May 19, 2018 at 4:57 | comment | added | user44191 | Something that scales with $\vec{v}$. Your choice of normalization corresponds to $c = \sum_i v_i$. | |
| May 19, 2018 at 4:55 | comment | added | N. Virgo | @user44191 what determines the value of $c$ in that equation? (I think I'm missing something, because if I set it to 1 I just get my original equation 2) | |
| May 19, 2018 at 4:51 | comment | added | user44191 | I should amend my earlier comment: any solution can be scaled, by changing $\lambda$. As such, I'd look at the renormalized equation $A \vec{v}^\alpha = \lambda c^{\alpha - 1} \vec{v}$. This makes all $v$ a solution for the same $\lambda$. | |
| May 19, 2018 at 4:45 | comment | added | N. Virgo | @user44191 what I mean is, in the nonlinear case, if you change the normalisation (e.g. to $\sum_i v_i^2=1$, or $\sum_i v_i=2$) you will not in general get a multiple of the original $\mathbf v$, it will change the relative magnitudes of the entries as well. But what you say might be right also in some situations, I'm not sure. | |
| May 19, 2018 at 4:40 | comment | added | user44191 | When you say that the normalization condition is necessary because it is nonlinear, the opposite seems true to me - if it were linear, then any solution could only be unique up to scaling, and so normalization would choose that "scaling"; here, a solution might simply be unique, not even up to scaling. | |
| May 19, 2018 at 3:39 | history | asked | N. Virgo | CC BY-SA 4.0 |