Consider the following recurrence relations for the sequences $a_n$ and $b_n$:

$ (\gamma_1+ \alpha_1 + n \gamma) a_n = \lambda_1 a_{n-1} + \alpha_2 b_n $

$ (\gamma_2+ \alpha_2 + n \gamma) b_n = \lambda_2 b_{n-1} + \alpha_1 a_n $

subject to the initial condition, say, $a_0 = k_1 $ and $b_0=k_2$.

I want to reexpress the above system in the matrix form, such as

$ A_{n+1} = B A_n $,

where $A_n$ is the vector $(a_n, b_n)^T$ and $B$ is the coefficient matrix. Note that the coefficient matrix commutes with itself for all $n$. Also, if possible, I want to go further and have both a diagonal matrix and a coefficient matrix on the right-hand side of the equality.

Any help would be appreciated!


I am looking for a matrix-based generalisation of the solution to the above system. That is, What if I had one more sequence, say $c_n$, and therefore the following system:

$ (\gamma_1+ \alpha_1 + \beta_1 + n \gamma) a_n = \lambda_1 a_{n-1} + \alpha_2 b_n + \alpha_3 c_n$

$ (\gamma_2+ \alpha_2 + \beta_2 + n \gamma) b_n = \lambda_2 b_{n-1} + \alpha_1 a_n + \beta_3 c_n$

$ (\gamma_3+ \alpha_3 + \beta_3 + n \gamma) c_n = \lambda_3 c_{n-1} + \beta_1 a_n + \beta_3 b_n$

  • $\begingroup$ This may be more complicated than it appears - since $n$ appears in the recurrence relationship as a variable, your matrix $B$ will have to be non-constant, a function of $n$. $\endgroup$
    – FShrike
    Commented Jul 27, 2021 at 9:04
  • $\begingroup$ Also, isn't the recurrence relationship a bit circular? $a_n$ is defined in terms of $b_n$, which is defined in terms of $a_n$... $\endgroup$
    – FShrike
    Commented Jul 27, 2021 at 9:05
  • $\begingroup$ Yes, it is. However, it is solvable too. For the small values of $n$, I can get the terms of the sequences solving by hand. The matrix $B$ should be manipulated so that one can get something like this: $n \hat{B}$, where $\hat{B}$ contains only constants. $\endgroup$
    – Lynx
    Commented Jul 27, 2021 at 10:29
  • $\begingroup$ What is the pattern here? I don't see the general way in which the $RHS$ of these equations are built from the $RHS$ of your original question $\endgroup$
    – FShrike
    Commented Jul 27, 2021 at 14:29
  • 1
    $\begingroup$ Well, I can't help you any further without hearing the problem name / further context. Good luck with your problem, hopefully my methodology works for the rest of your problem $\endgroup$
    – FShrike
    Commented Jul 27, 2021 at 16:04

1 Answer 1


It's an ugly mess, but I got through it:

$$\begin{align*}&(\gamma_1+\alpha_1+n\gamma)a_n=\lambda_1a_{n-1}+\alpha_2b_n \\&(\gamma_2+\alpha_2+n \gamma)b_n=\lambda_2b_{n-1}+\alpha_1a_n \\&\implies a_n=\frac{\lambda_1a_{n-1}+\alpha_2b_n}{\gamma_1+\alpha_1+n\gamma} \\&\implies b_n=\frac{\lambda_2b_{n-1}+\alpha_1a_n}{\gamma_2+\alpha_2+n\gamma} \\&\implies a_n(\gamma_1+\alpha_1+n\gamma)=\lambda_1a_{n-1}+\frac{\lambda_2\alpha_2b_{n-1}+\alpha_1\alpha_2a_n}{\gamma_2+\alpha_2+n\gamma} \\&\implies b_n(\gamma_2+\alpha_2+n \gamma)=\lambda_2b_{n-1}+\frac{\lambda_1\alpha_1a_{n-1}+\alpha_1\alpha_2b_n}{\gamma_1+\alpha_1+n\gamma}\end{align*}$$

For convenience, let:


Now if we skip over an ugly in-between step,

$$a_n\cdot\omega=a_{n-1}(\lambda_1\cdot\tau)+b_{n-1}(\lambda_2\cdot\alpha_2) \\b_n\cdot\omega=a_{n-1}(\lambda_1\cdot\alpha_1)+b_{n-1}(\lambda_2\cdot\sigma)$$

And with a pretty sort of symmetry, we get:


For factorisation, this is the best I can do:

Let $\sigma'=\gamma_1+\alpha_1$ and $\tau'=\gamma_2+\alpha_2$:


And by the distributivity of matrix multiplication:

$$\begin{bmatrix}a_n\\b_n\end{bmatrix}=\frac{1}{\omega}\begin{bmatrix}\lambda_1\cdot\tau'&\lambda_2\cdot\alpha_2\\\lambda_1\cdot\alpha_1&\lambda_2\cdot\sigma'\end{bmatrix}\begin{bmatrix}a_{n-1}\\b_{n-1}\end{bmatrix}+\frac{n\gamma}{\omega}\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\begin{bmatrix}a_{n-1}\\b_{n-1}\end{bmatrix} $$

And if you're running a computer analysis or something that requires you to extrapolate future $a_n$ maybe this helps, since the $1/\omega$ terms can be easily computed and collected as functions of $n$, and the left matrix is constant, and the right matrix is only very trivially a function of $n$.

  • $\begingroup$ Thank you for the idea! I am trying to extend the above system to, say, K system of recurrence relations. So, at each step, I get one more equation, e.g. an equation of $c_n$, and a term like $\alpha_3 c_n$ on the RHS of the former two equations. Then, a question arises: how would I solve such a system if I wanted to generalise the problem? $\endgroup$
    – Lynx
    Commented Jul 27, 2021 at 14:12
  • $\begingroup$ I'm sorry @lynx I don't really understand - could you formalise your question and edit it into your original question please? Then I'll have a look $\endgroup$
    – FShrike
    Commented Jul 27, 2021 at 14:13
  • $\begingroup$ @lynx Do you have some way of constructing recurrence relations similar to the one I just solved? $\endgroup$
    – FShrike
    Commented Jul 27, 2021 at 14:22
  • $\begingroup$ Thanks for your interest, @FShrike. I have explained it in the OP. $\endgroup$
    – Lynx
    Commented Jul 27, 2021 at 14:26

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