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An eigenvector is defined by $$ \lambda \mathbf{v} = A\mathbf{v}.\tag{1} $$ But suppose I change this to $$ \lambda \mathbf{v} = A\mathbf{v}^\alpha,\tag{2} $$ for real $\alpha\ne 1$, where $\mathbf{v}^\alpha$ means a vector with elements $v_i^\alpha$, along with the normalisation condition $$ \sum_i v_i = 1.\tag{3} $$ (Specifying the normalisation condition is necessary, because equation 2 is nonlinear.)

Let us call vectors that satisfy equations 2 and 3 "stretched eigenvectors", unless they have another name. I am wondering if they are studied, and if so, what their properties are. They seem somehow related to the q-exponential and q-logarithm in Tsallis statistics (with $q=\alpha$), and if that's the case it would be particularly nice to know if that connection has been worked out.

I am mostly interested in the Perron-Frobenius case, where $A$ has real, non-negative entries. Then when $\alpha=1$ there exists an eigenvector $\mathbf{v}$ with non-negative entries and the corresponding eigenvalue $\lambda$ is also real and non-negative. If the entries of $A$ are positive then this is unique.

I am interested in whether any aspects of Perron-Frobenius and its related theorems can be generalised to cases where $\alpha\ne 0$. It seems for example that the uniqueness property can go away when $\alpha>1$, but perhaps there is some sense in which it can be recovered by changing the conditions.

My motivation for this is the study of sub-exponential and super-exponential replicators, which leads to a dynamical system whose fixed points are given by equations 2 and 3.

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  • $\begingroup$ 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. $\endgroup$
    – user44191
    Commented May 19, 2018 at 4:40
  • $\begingroup$ @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. $\endgroup$
    – N. Virgo
    Commented May 19, 2018 at 4:45
  • $\begingroup$ 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 $\lambda$. $\endgroup$
    – user44191
    Commented May 19, 2018 at 4:51
  • $\begingroup$ @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) $\endgroup$
    – N. Virgo
    Commented May 19, 2018 at 4:55
  • $\begingroup$ Something that scales with $\vec{v}$. Your choice of normalization corresponds to $c = \sum_i v_i$. $\endgroup$
    – user44191
    Commented May 19, 2018 at 4:57

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