Asymptotic behavior of the geometric sequence U n + 1 = Q U n in...

yasusar0

yasusar0

Answered

2022-07-21

Asymptotic behavior of the geometric sequence U n + 1 = Q U n in a probability problem
We know the asymptotic behavior of the scalar geometric sequence u n + 1 = q u n with respect to the values of q. That is, | q | < 1 implies lim n u n = 0, q = 1 implies u n is constant, etc.
Now I am just asking myself the same question about the geometric sequence U n + 1 = Q U n , where Q is a 2 × 2 matrix and U n is a two dimensional vector.

Answer & Explanation

Tamoni5e

Tamoni5e

Expert

2022-07-22Added 14 answers

Step 1
If matrix and vector norms are compatible in the way that Q x Q x , then Q < 1 implies Q n x 0 for all x.
Step 2
If Q > 1 then Q could be the identity or a rotation matrix, with the obvious consequences on convergence of Q n x.
If Q > 1 then Q n x is not guaranteed to be unbounded. Take Q = ( 2 0 1 0 ) and x = ( 1 0 ) or x = ( 0 1 )

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?