Linear recurrent sequences and matrices.

Question

Let $k$ be a field, let $d$ be an integer greater than $1$, let $(v,x)\in k^d\times k^d$ and let $A\in k^{d\times d}$ be invertible.

For all $n\in\mathbb{N}$, let define the following element of $k$: $$u_n:={}^tvA^nx.$$

I would like to show that:

$(u_n)_{n\in\mathbb{N}}$ is a recurrent linear sequence.

This fact is well-known when $A$ is a companion matrix (or its transpose, that depends on your definition).

Surely I can use Frobenius decomposition theorem and get the result, but that might be overkill.

Any hint will be appreciated!

Using the Frobenius decomposition theorem seems like a perfect way to get the result — Ben Grossmann, May 31 '16 at 09:23
Thank you for your input! I guess I should be satisfied with my current proof. — C. Falcon, May 31 '16 at 09:30
You might want to post your current proof (in a new question) and get feedback on it that way — Ben Grossmann, May 31 '16 at 09:32

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

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This should be the same as the linear systems we often get, just $y_n = A^n x.$ Here the matrix $A$ is square, you are calling it $d.$

The Cayley-Hamilton Theorem says that $A$ satisfies a polynomial, degree no larger than $d.$ The same equation is satisfied by the column vectors $y_{k+d}, y_{k+d-1}, \ldots, y_k.$ That is why the $d$ entries of $y$ each obey the same linear recursion.

Finally, your $u_n$ obeys the same recursion, coefficients are those of the characteristic polynomial of $A,$ or the minimal polynomial if that has smaller degree.

Here is a recent one How does one solve this recurrence relation?

I have answered many questions the same way...

edited Apr 13 '17 at 12:21

Community

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answered Jun 25 '16 at 01:16

Will Jagy

139,541

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That is a really neat way of seeing it! I am not sure Cayley-Hamilton theorem is easier than the Frobenius decomposition though, but both theorems suit me. To give some background I wanted to show the equivalence between Skolem-Mahler-Lech theorem and this statement: if $k/\mathbb{Q}$ is a field extension, $\sigma$ is a linear automorphism, $x\in k^d$ and $H$ is an hyperplane, then ${n\in\mathbb{N}\textrm{ s.t. }\sigma^n(x)\in H}$ is the union of a finite set and of a finite number of arithmetic progressions. – C. Falcon Jun 25 '16 at 01:29
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@C.Falcon This method comes up often for me because of integer quadratic forms; if $H$ is the symmetric matrix of second partial derivates of a quadratic form, given by $h(x) = (1/2) x^T H x,$ then a matrix $A$ that solves $A^T H A = H$ automatically gives $h(Ax) = h(x).$ The most frequent example on this site is the ordinary Pell equation $x^2 - n y^2 = 1.$ The outcome is that $x$ and $y$ obey identical linear recurrences, degree $2$ because we have two variables. – Will Jagy Jun 25 '16 at 01:52
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This year I actually studied (in an arithmetic course) the link between integer quadratic forms, Pell equations and units of the ring of integers of a quadratic field extension. It is nice to have here a quick reminder, it helped me clarify some of my ideas. – C. Falcon Jun 25 '16 at 02:02

Linear recurrent sequences and matrices.

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