Solving simultaneous linear difference equations

[Ad123q]

Hi

Was wondering if anyone could help me with this topic, which relates to the solution of simultaneous linear difference equations and diagonizable matrices.

Sorry for the lack of latex, but as such I will denote an exponent by "^" and a subscript by "_".

Say we have a system of simultaneous linear difference equations

x_n+1 = ax_n + by_n
y_n+1 = cx_n + dy_n

where a, b, c, d are numbers, and "x_n+1" etc is "x subscript n+1".

Denote v_n by the column vector (x_n , y_n) and let the coefficients a, b, c, d form a matrix
A= a b
c d

Then we can write v_n+1 = Av_n.

The solution of the system is then meant to be, in general, v_n = A^nv_0,
where A^n is matrix A raised to power n, calculated by considering that P^-1AP = D (diagonal) and so A^n = PD^nP^-1,

where P^-1 is the inverse of the matrix P, viz the matrix formed by eigenvectors.

Thanks very much for any help in advance.

 

[Ad123q]

Sorry, just realized that I didn't in fact highlight what I don't understand!
Basically I don't understand the whole reason why we can go from v_n+1 = Av_n to v_n = A^nv_0 (ie the process to do this); is it induction of some kind?

This type of problem must be fairly easy but I can't see it.

Best wishes.

 

[Ad123q]

Anyone?

 

[Cantab Morgan]

I'm not sure I understand what you're asking, but I see it this way... $v_1 = A v_0$ right? Just plug in n=0 into the difference equation. Then multiply both sides of the equation by A, and what do you get? $A v_1 = A A v_0 = A^2 v_0$. But we already know what $A v_1$ is by the difference equation again. It's $v_2$. So, $v_2 = A^2 v_0$. Keep multiplying by A, and see that $v_n = A^n v_0$.