# For an one-order Linear Recurrence of a vector sequence, does the corresponding item follow a Linear Recurrence? [closed]

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 $$k$$?

## closed as off-topic by Gerry Myerson, Anthony Quas, kodlu, David Handelman, Chris GodsilFeb 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
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• Not a question of math research, which is what this MO website is for. – Gerry Myerson Feb 11 at 3:16
• 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. – Anthony Quas Feb 11 at 3:16

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 $$X_{n+j} = A^j X_n$$