yasusar0

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 $\underset{n}{lim}{u}_{n}=0$, $q=1$ 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.

Tamoni5e

Expert

Step 1
If matrix and vector norms are compatible in the way that $‖Qx‖\le ‖Q‖\cdot ‖x‖,$ then $‖Q‖<1$ implies ${Q}^{n}x\to 0$ 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 $Q=\left(\begin{array}{cc}2& 0\\ 1& 0\end{array}\right)$ and $x=\left(\begin{array}{c}1\\ 0\end{array}\right)$ or $x=\left(\begin{array}{c}0\\ 1\end{array}\right)$

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