Linear system of recurrence equations

xt = a11xt-1 + a12yt-1

yt = a21xt-1 + a22yt-1

In matrix form the above can be written as xt = Axt-1

let xt = Put and sub this into above to Put = APut-1

Providing P^-1 exists, multiply P^-1 to the other side to ut = P^-1APut-1

P^-1AP becomes diagonal matrix D so... ut = Dut-1

ut is a vertical vector with arbitrary terms ut and vt that are used to solve the problem, the ut and vt have to be converted back to xt and yt!!

u0 = P-1x0 can be used to find u0 and v0 in terms of x0 and y0
 * Question solving
 * Find P, P^-1 and D of A
 * Find xt = Put then u0 = P^-1x0
 * ut here is the vertical vector of eigenvalues^t multiplied by u0 and v0 respectively which is D multiplied by the u0 vector
 * x0 will be given
 * We now know what u0 is
 * Then we sub ut back into xt = Put then we multiply the vectors to get the solution

The n x n system Markov processes
 * For a more than 2x2 system, we use a different approach. You can use powers of a matrix to solve the recurrence equations.
 * xt = A^tx0
 * A^t = PD^tP^-1
 * therefore = xt = PD^tP^t-1x0
 * We use this method to write the solution as a linear combination of the eigenvectors of A
 * A Markov process is a closed process where the population changes to one state from another in time intervals and is observed as scheduled times
 * The probabilities of a state changing to one state to another is listed in a matrix
 * In a Markov process the matrix A has non-zero values and the sum of the columns is equal to 1
 * We need to predict the vector xt for which we know the answer to which is xt = PD^tP^t-1x0