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A popular form of the Perron-Frobenius theorem states the following result: if $A$ is a $d \times d$ real matrix all of whose entries are positive, then the spectral radius of $A$ is a simple eigenvalue of $A$, and every other eigenvalue of $A$ is smaller in modulus. If we let $\lambda_1(A),\ldots,\lambda_d(A)$ be a listing of the eigenvalues of $A$ in order of decreasing modulus (allowing repetition in the event of multiplicity) then the Perron-Frobenius theorem tells us in particular that if $A$ is positive, then $|\lambda_1(A)|>|\lambda_2(A)|$.

If we write $\sigma_1(A),\ldots,\sigma_d(A)$ for the singular values of $A$, then using the well-known formula $$|\lambda_k(A)|=\lim_{n \to \infty} \sigma_k(A^n)^{\frac{1}{n}}$$ we can restate the above result as follows: if $A$ is positive, then there exist $C>0$ and $\tau \in (0,1)$ such that $$\sigma_2(A^n) \leq C\tau^n\sigma_1(A^n)$$ for all $n \geq 1$. I am interested in generalisations of this property to sets of matrices (which in my situation happen to be invertible). In particular I need to use the following result: if $A_1,\ldots,A_r$ are positive $d \times d$ matrices, then there exist $C>0$ and $\tau \in (0,1)$ such that $$\sigma_2(A_{x_n}\cdots A_{x_1}) \leq C\tau^n\sigma_1(A_{x_n}\cdots A_{x_1})$$ for every $x_1,\ldots,x_n \in \{1,\ldots,r\}$ and every $n \geq 1$. I can see that this result follows from a much stronger theorem due to J. Bochi and N. Gourmelon (Some characterizations of domination, Mathematische Zeitschrift 263 (2009) 221--231). However, given that the Perron-Frobenius theorem is over 100 years old, I would be a little surprised if the oldest reference for the above result is as recent as 2009. Is anyone able to point out to me an older reference?

Thanks in advance!

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  • $\begingroup$ So I guess you could get something by showing the family of matrices uniformly contract the positive orthant in the Hilbert metric. Probably not the kind of answer you had in mind though. $\endgroup$ – Anthony Quas Mar 18 '15 at 14:08
  • $\begingroup$ I'm sure that it can be proved in a page or two using the Hilbert metric (without the invertibility requirement), but my main question is whether someone has already done so and published it. $\endgroup$ – Ian Morris Mar 18 '15 at 14:14

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