Some questions about a particular system of inter-dependent linear recurrence relations.

Question

Given the following recurrence relations for $k$ terms:

$$ A_n^1 = A_{n-1}^1 + A_{n-1}^2 $$ $$ A_n^2 = A_{n-1}^1 + A_{n-1}^3 $$ $$ A_n^3 = A_{n-1}^2 + A_{n-1}^4 $$ $$ \vdots $$ $$ A_n^{k-1} = A_{n-1}^{k-2} + A_{n-1}^k $$ $$ A_n^k = A_{n-1}^{k-1} $$

Can we express each $A_n^i$ (for $i \in \{1, 2, \dots, k\}$) as a linear recurrence relation based on its own previous terms? If so, how to find the recurrence relation for each $A_n^i$ for any k, is there a general formula?

Additionally, can we prove that when each term $A_n^i$ is expressed as a recurrence relation based on its own previous terms, all of the $A_n^i$ have the same recurrence relation coefficients?

I hope i am clear.

Edit: Lets take $k=3$, then: $$ A_n^1 = A_{n-1}^1 + A_{n-1}^2 $$ $$ A_n^2 = A_{n-1}^1 + A_{n-1}^3 $$ $$ A_n^3 = A_{n-1}^2 $$

After some manipulation we get: $$ A_n^1 = A_{n-1}^1 + 2A_{n-2}^1 - A_{n-3}^1 $$ $$ A_n^2 = A_{n-1}^2 + 2A_{n-2}^2 - A_{n-3}^2 $$ $$ A_n^3 = A_{n-1}^3 + 2A_{n-2}^3 - A_{n-3}^3 $$ So we can express each $A_n^i$ using only their own previous terms and each $A_n^i$ has the same recurrence relation coefficients $[1,2,-1]$ for $k=3$.

We can express each $A_n^i$ using only their own previous terms for $k=4$ too. And in this case too each $A_n^i$ has the same recurrence relation coefficients, which is $[1,3,-2,-1]$.

It seems that for any k it is true that we can express each $A_n^i$ only their own previous terms and that each $A_n^i$ has the same recurrence relation coefficients for a particular k. Also i think the recurrence coefficient for a particular k is exactly the kth row in the following table:

Here, $T_n^i$ means $i$th column and $n$th row. $$T_n^1 = 1, \quad T_1^i = 0 \quad \text{for} \ i > 1$$ $$T_n^{2i} = T_{n-1}^{2i} + T_{n-1}^{2i-1}$$ $$T_n^{2i+1} = -T_{n-1}^{2i}$$

$$ \begin{array}{c|cccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 3 & 1 & 2 & -1 & 0 & 0 & 0 & 0 \\ 4 & 1 & 3 & -2 & -1 & 0 & 0 & 0 \\ 5 & 1 & 4 & -3 & -3 & 1 & 0 & 0 \\ 6 & 1 & 5 & -4 & -6 & 3 & 1 & 0 \\ 7 & 1 & 6 & -5 & -10 & 6 & 4 & -1 \\ \end{array} $$

I don't have a proof of any of these claims , but it seems to me that these are true, so i would like to see proofs of these claims (if these are indeed true) and a deep explanation of connection between all this.

Answer 1

The relations are equivalent to $$ A_n = \begin{pmatrix} A_n^1 \\ A_n^2 \\ \cdots \\ A_n^k \end{pmatrix} = PA_{n-1} = P \cdot\begin{pmatrix} A_{n-1}^1 \\ A_{n-1}^2 \\ \cdots \\ A_{n-1}^k \end{pmatrix}, $$ where $P$ is equal to $$ \begin{pmatrix} 1 & 1 & 0 & 0 & \ldots & 0 \\ 1 & 0 & 1 & 0 & \ldots & 0 \\ 0 & 1 & 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \ldots & 1 \\ 0 & 0 & 0 & 0 & \ldots & 0 \end{pmatrix}. $$ Though it looks like a tridiagonal matrix, your condition $A_n^k = A_{n-1}^{k-1}$ is somehow weird. I did some numerical tests and didn't find any useful properties.

You can calculate $P^{n} A_0$ to derive $A_n$.

Edit: You discovered one of the famous theorems in linear algebra. To formulate your question, we first need to prove that $P$ is invertible. Expand it along the first column and we get $$ \det(P) = \begin{pmatrix} 0 & 1 & 0 & \ldots & 0 \\ 1 & 0 & 1 & \ldots & 0 \\ 0 & 1 & 0 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots &\vdots\\ 0 & 0 & 0 & \ldots & 0 \end{pmatrix}_{k-1} - \begin{pmatrix} 1 & 0 & 0 & \ldots & 0 \\ 1 & 0 & 1 & \ldots & 0 \\ 0 & 1 & 0 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots &\vdots\\ 0 & 0 & 0 & \ldots & 0 \end{pmatrix}_{k-1}. $$ By computing these tridiagonal matrices and using induction we have $\det(P)=(-1)^{[\frac{k}2]}$. Thus $P^{-1}$ exists.

Now we can inverse the recurrence as $A_{n-1} = P^{-1}A_n$, $A_{n-2} = P^{-2}A_n$, and so on. Your question is to find $c_1, c_2, \ldots, c_m$, such that $$ A_n = c_1 A_{n-1} + c_2 A_{n-2} + \cdots + c_m A_{n-m}, $$ which means $$ A_{n} = c_1P^{-1}A_n + c_2 P^{-2} A_n + \cdots + c_m P^{-m}A_n. $$ Therefore, $$P^m -c_1P^{m-1} - c_2P^{m-2} -\cdots - c_m=0.$$ If there exists $c_1, c_2, \ldots, c_m$ that satisfy this equation, then your guess is proved.

As $P$ is a non-degenerate real symmetric matrix, Hamilton-Cayley theorem tells us that $P$ is exactly one root of its non-trivial characteristic function. Thus the coefficients of the characteristic function $c_1, c_2, \ldots, c_m$ is what you want. You may further calculate $\vert \lambda I - P \vert$ to validate your guess.

As you may have noted, Hamilton-Cayley theorem connects the characteristic root technique (in your word, express one term only using its previous terms) and the characteristic function of a matrix (the transformation matrix of $A_n$), which is the deep explanation of all these results.