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If you have a system of difference equations like this:

$x_{n+1} = a_{11}x_n + a_{12}y_n$

$y_{n+1} = a_{21}x_n + a_{21}y_n$

And you know that there is only one unique eigenvalue (ie: $\lambda_1=\lambda_2=\lambda$), how do you show that the general solution is $x_n=C_1\lambda^n + C_2n\lambda^n$?

My work so far: I've combined the two equations into a single equation $x_{n+2} - (a_{11} + a_{22})x_{n+1} + (a_{22}a_{11} - a_{12}a_{21})x_n = 0$

I know that this type of difference equation typically has the form $C\lambda^n$, but I have no idea how to come up with or show the bit with $Cn\lambda^n$.

tempestuousfool
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1 Answers1

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Let $A=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$ have equal (algebraic) eigenvalues: $\lambda_1=\lambda_2=\lambda$. Then either $a_{12}=a_{21}=0$ (in which case the situation is very simple) or the Jordan normal form of $A$ is

$$J=\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}.$$

In the latter case, you can show that

$$J^n=\begin{bmatrix} \lambda^n & n \lambda^{n-1} \\ 0 & \lambda^n \end{bmatrix}$$

by induction. Your result follows from there, since $A^n=V J^n V^{-1}$, where $V$ is the generalized eigenvector matrix.

Alternately, you can just guess-and-check, again using induction.

Ian
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  • I don't want to edit myself because I'm not certain, but from my study of analogous differential equations I believe the upper-right $\lambda$ in the Jordan form should be a $1$? – user10478 May 26 '20 at 02:08
  • @user10478 Thanks, I edited. – Ian May 26 '20 at 02:26