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Consider an one-order Linear Recurrence of a vector sequence, such as $${\bf x}_{n+1}={\bf A}{\bf x}_n$$ where ${\bf x}_n \in \mathbb{R}^m (\forall n)$, and ${\bf A} \in \mathbb{R}^{m\times m}$ and ${\bf x}_0$ are constants.

Denote $x_{n,i}$ be the $i$th item of ${\bf x}_n$. My question is for some $i$, whether sequence $\{x_{j,i}\}_{j \in \mathbb{N}_+}$ is always a Linear Recurrence. If it is, can we give an upper bound on the order of the Linear Recurrence? Futhermore, if the upper bound is $k$, is such order always a factor of $k$?

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closed as off-topic by Gerry Myerson, Anthony Quas, kodlu, David Handelman, Chris Godsil Feb 11 at 13:25

This question appears to be off-topic. The users who voted to close gave these specific reasons:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Anthony Quas, kodlu, Chris Godsil
  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Gerry Myerson, David Handelman
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Not a question of math research, which is what this MO website is for. $\endgroup$ – Gerry Myerson Feb 11 at 3:16
  • $\begingroup$ I think this question doesn't belong on this site. math.stackexchange.com would be more appropriate. As a hint though, you can re-write the Fibonacci recurrence as a 2*2 matrix recurrence (where you think of the entries of $\mathbf x_n$ as $(x_n,x_{n-1})$. Your last question is not precise enough to have an answer: there is no unique upper bound, after all. $\endgroup$ – Anthony Quas Feb 11 at 3:16
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each of the $m$ component sequences of your $X$ satisfies a linear recurrence of degree $m$ with characteristic polynomial given by Cayley-Hamilton. The point, I suppose, is that $$ X_{n+j} = A^j X_n $$

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