Suppose I have the following companion matrix ($d\times d$) The companion matrix A. $1 \geq p \geq q \geq 0$. Let $x$ ($d\times 1$) be the all one vector and my underlying problem is to analyze the first entry of $A^nx$ for some large $n$. Even if the close form doesn't exist, we know that the behavior of it will determined by the largest eigenvalue of matrix $A$. So here's my question: Is it possible to find the nice bound/approximation of the largest eigenvalue of the matrix $A$ (in terms of $d,p,q$)? Any direction or reference would be appreciated!
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1 Answer
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First of all, we can scale your matrix so there is only one parameter, so let's say $p = 1$ for simplicity.
It looks to me like the characteristic polynomial of your matrix is $$ P(\lambda) = \frac{\lambda^{n+1} - \lambda^n + q^n - q^{n+1}}{\lambda - q} $$ Thus $P(1) = q^n$ and $P(q) = (n+1) q^n - n q^{n-1}$. In particular, if $q < n/(n+1)$ we have $P(q) < 0 < P(1)$, so there is a root between $q$ and $1$.
answered Sep 9, 2018 at 23:00
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$\begingroup$ Thanks @Robert Israel ! This answer indeed give some useful direction of solving this problem. I'll try to do a more delicate analysis! $\endgroup$ Commented Sep 10, 2018 at 17:18