0
$\begingroup$

Let $d\geqslant 2$ be an integer and consider a convergent sequence of "companion" matrices $$A_k := \begin{pmatrix} a_{k,1} & a_{k,2} & \cdots & a_{k,d} \\\ & & & 0\\\ & I_{d-1} & & \vdots\\\ & & & 0 \end{pmatrix} \underset{k\to +\infty}{\longrightarrow} \begin{pmatrix} a_{1} & a_{2} & \cdots & a_{d} \\\ & & & 0\\\ & I_{d-1} & & \vdots\\\ & & & 0 \end{pmatrix}=: A$$ where $I_{d-1}$ is the identity matrix of order $d-1$ and $(a_{k,1})_k, \ldots, (a_{k,d})_k$ are sequences such that (EDIT) $|a_{k,j_0}|\leqslant a_k$ and which respectively converge to integers $a_1,\ldots,a_d$ such that $a_1 \geqslant \ldots \geqslant a_d\geqslant 1$. Therefore, $A$ has a dominating real eigenvalue $\alpha$ (which is also a Pisot–Vijayaraghavan number).

Can we estimate the error between $\|A_{k+1}\cdots A_{k+j_0}\|$ and $\alpha^{j_0}$ (or else $\|A^{j_0}\|$) for some matrix norm $\|\cdot\|$ and integer $j_0\geqslant 1$ ?

More precisely : does there exist a matrix norm $\|\cdot\|$, an integer $j_0\geqslant 1$ and a positive sequence $(u_{k})$ going to $0$ such that $\|A_{k+1}\cdots A_{k+j_0}\| \leqslant \alpha^{j_0}\exp(-u_{k})$ ?

This is a question to generalize the following example: if I take $d=2$, $a_{k,1}=1$ and $a_{k,2}=e^{iv_k}$ with $v_k\to 0$ i.e. $$A_k = \begin{pmatrix} 1 & e^{iv_k} \\\ 1 & 0 \end{pmatrix} \underset{k\to +\infty}{\longrightarrow} \begin{pmatrix} 1 & 1\\\ 1 & 0 \end{pmatrix} = A,$$ then $\alpha$ is the golden ratio. The spectral norm (or $2$-norm) $\|\cdot\|_2$ and $j_0=2$ work, because we have $\|A_k\|_2= \alpha$ and we can calculate/estimate by hand $$ \|A_{k+1}A_{k+2}\|^2_2 = \rho\left((A_{k+1}A_{k+2})(A_{k+1}A_{k+2})^\ast)\right) \leqslant \alpha^4-c(1-\cos(v_{k+2})) \leqslant \alpha^4-c'|e^{iv_{k+2}}-1|^2,$$ where $\rho$ is the spectral radius, and thus $\|A_{k+1}A_{k+2}\|_2 \leqslant \alpha^2\exp(-c''|e^{iv_{k+2}}-1|^2)$, where $c,c',c''>0$ are absolute constants.

The case $d=2$ seems too complex by hand for me, or else $(a_{k,j})$ has to be like $\exp(i\sum b_k)$ or $\sum \exp(ib_k)$ to expect some result... Thank you in advance, any remark/advice/help will be very appreciated.

Improve this question
$\endgroup$
5
  • $\begingroup$ In general the answer is no, imagine that there are extremely long blocks of the same matrix, whose biggest eigenvalue $\lambda$ slightly more that $\alpha$. Then the norm will be on the order of $\lambda^{j_0}$, which can be arbitrarily larger than $\alpha^{j_0}$ if $j_0$ is big enough (that is, if the block I mentioned is long enough). $\endgroup$
    – Aleksei Kulikov
    Commented Sep 13 at 20:26
  • 1
    $\begingroup$ And indeed, the claim is already false for the $d = 1$ case (you for some reason forbid it, but in my opinion forbidding the simplest cases is a bad habit, since they often make the key issues apparent). $\endgroup$
    – Aleksei Kulikov
    Commented Sep 13 at 20:28
  • $\begingroup$ If we assume $|a_{k,j_0}|\leqslant a_k$, then $\lambda$ can be less than $\alpha$. . And I don't understand your second remark for the case $d=1$ (degenerate case): I didn't forbid it, you write "already" but I have an example which works for $d=2$ and I want $d\geqslant 2$ (generalization), it doesn't make any key issue apparent. $\endgroup$
    – Kermatoni
    Commented Sep 13 at 20:57
  • $\begingroup$ Pardon me, but how are we supposed to get into your brain and guess that you had a condition like $|a_{k, j_0}| \le a_k$ in mind?! This of course changes things. As for the other thing, even if in your application the case $d\ge 2$ is needed, history teaches us again and again that it is still worthwhile to look at the simpler cases, since they could be easier to solve, and yet their solutions can shed light on the more general case. $\endgroup$
    – Aleksei Kulikov
    Commented Sep 13 at 21:01
  • $\begingroup$ I don't except you guess this, I try to ask a general question ($a_{k,j}$ have some explicit and complex expression in my case) and maybe we can add some assumption to obtain a result :) You have right but the dimension $1$ can be a strange case ^^ $\endgroup$
    – Kermatoni
    Commented Sep 13 at 21:05

0

Reset to default

You must log in to answer this question.