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Let $1\leq a\leq b$ with $a,b\in\mathbb{N}$. Moreover, let $c:=a+b$.

Consider the polynomial $$ f_{c }(x)=x^{c+1}-x^{c}-1. $$ By $x_{c}$ denote the largest positive root of $f_{c}$.

It can be shown that, for large $c$, $$ x_c\sim1+\frac{\log c}{c}~~~(*) $$ and hence $\lim_{c\to\infty}x_{c }=1$.

Note $f_{c}$ is the characteristic polynomial of the $(c+1)\times (c+1)$-matrix $A_{c}$ with entries $$ a_{1,1}=1, \quad a_{c+1,1}=1 $$ and $a_{i,j}=1$ on the right side-diagonal and 0 elsewhere.

For example, if $c=2$, then $$ A_{2}=\begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{pmatrix}. $$

Question

I wonder: Does there exists a „scaling exponent“ $\alpha=\alpha(c)$ such that

(1) $x_{c}^{\alpha(c)}$ is the largest root of the characteristic polynomial of the matrix $A_{c}^{\alpha(c)}$ and

(2) the limit $\lim_{c\to\infty}x_c^{\alpha(c)}\neq 1$ exists?

At least for exponents $\alpha(c)\in\mathbb{N}$, ad (1) should be satisfied.

Using $(*)$, one should have that, for large $c$, $$ x_c^{\alpha(c)}\sim\left(1+\frac{\log c}{c}\right)^{\alpha(c)} $$ So, maybe it is useful to consider $$ \lim_{c\to\infty}\left(1+\frac{\log c}{c}\right)^{\alpha(c)} $$ Here, I dont know how to continue.

mathfemi
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1 Answers1

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If $c$ is even, then $A_c$ has no $\leq 0$ eigenvalues. Then, for any real $\alpha(c)$, you can define ${A_c}^{\alpha(c)}$ as a real matrix with $\rho({A_c}^{\alpha(c)})={x_c}^{\alpha(c)}$.

It suffices to choose $\alpha(c)=\dfrac{c}{\log(c)}$. Indeed $\log({x_c}^{\alpha(c)})\sim \alpha(c)\dfrac{\log(c)}{c}=1$ and ${x_c}^{\alpha(c)}$ tends to $e$.

Yet, you obtain the same result if you choose $\alpha(c)=Int(\dfrac{c}{\log(c)})$ ("Int" denotes the integer part). Moreover, for such an $\alpha(c)$, the result works also for odd $c$.

EDIT. I forgot to say that $x_c$ is the root of $f_c$ with maximum modulus.

  • I think another option would be to choose $\alpha(c)=\frac{1}{\log x_c}$ which also should give that $x_c^{\alpha(c)}$ tends to $e$? – mathfemi Jan 21 '18 at 11:43
  • I don't see where is the interest. You don't know $x_c$ ! –  Jan 21 '18 at 11:50
  • But we know that $x_c\sim 1+\frac{\log c}{c}$ and hence can compute that $\lim_{c\to\infty}x_c^{1/\log x_c}=e$. – mathfemi Jan 21 '18 at 11:56
  • Do as you want. I did the job for free. Now, it's your business. –  Jan 21 '18 at 11:58
  • Please, I was not complaining about your answer, on the contrary. I just wanted to mention that there is another option maybe. - - By the way, what do you mean by "you can define $A_c^{\alpha(c)}$ as a real matrix with...". In which way define? – mathfemi Jan 21 '18 at 12:00
  • @mathfemi but $b^{1/\log b} = e$ for all $b > 0$, no limits involved. – Antonio Vargas Jan 21 '18 at 20:34
  • @mathfemi , ${A_c}^{\alpha(c)}$ is defined as $\exp(\alpha(c)\log(A_c))$, where $\log$ is the principal logarithm; this $\log$ is defined for the matrices that have no $\leq 0$ eigenvalues and sends every real matrix in a real matrix. Yet, you do not need the previous notion because we can be content with entire powers of $A_c$. –  Jan 21 '18 at 23:11