Long run matrix equilibrium

gerry87

How do approximate the long run state of a matrix

eg.

|.9 .1|
|.2 .8| = A

so

Z = I + A + A^2 + A^3 + ... + A^n
AZ = A + A^2 + A^3 + ... + A^n

Z - AZ = I
Z(I - A) = I
Z = (I - A)^-1

so (I - A)^-1 is the same as I + A + A^2 + A^3 + ... + A^n. Is this right?

so if i wanted to do this for matrix A above it would be

(I - A) =
|.1 .1|
|.2 .2|

(I - A)^-1 = ?

The determinant is 0 so that means it's a non-singular, right? So can it be done? I got the answer by just multiplying it out using a matrix calculator, but am i doing something wrong?

Thanks

MathsManiac

You haven't stated what exactly it is you're looking for.

Are you trying to determine whether A^n converges to a finite matrix as n tends to infinity? If so, the answer is yes: it converges to the matrix with [2/3 1/3] as both top and bottom rows.

If this is a right stochastic matrix (as it looks like it might be), the long term behaviour of the system is that it will be in state 1 with probability 2/3 and in state 2 with probability 1/3.

Once you know that a matrix is a right stochastic matrix, you know that 1 is an eigenvalue. You can find a corresponding eigenvector (ie a stable state vector) by letting the row vector v = [p (1-p)] and then solve the matrix equation vA=v. In your case, you get p=2/3, which tells you that the state vector [2/3 1/3] is invariant under A. You also know that this also forms each row of the limit of A^n.

Not sure, though, if this is what you were asking about.