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 $\underset{n}{lim}{u}_{n}=0$, $q=1$ 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 $2\times 2$ matrix and ${U}_{n}$ is a two dimensional vector.

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 $2\times 2$ matrix and ${U}_{n}$ is a two dimensional vector.

Answer & Explanation

Tamoni5e

Expert

2022-07-22Added 14 answers

Step 1

If matrix and vector norms are compatible in the way that $\Vert Qx\Vert \le \Vert Q\Vert \cdot \Vert x\Vert ,$ then $\Vert Q\Vert <1$ implies ${Q}^{n}x\to 0$ for all x.

Step 2

If $\Vert Q\Vert >1$ then Q could be the identity or a rotation matrix, with the obvious consequences on convergence of ${Q}^{n}x$.

If $\Vert Q\Vert >1$ 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 $x=\left(\begin{array}{c}0\\ 1\end{array}\right)$

If matrix and vector norms are compatible in the way that $\Vert Qx\Vert \le \Vert Q\Vert \cdot \Vert x\Vert ,$ then $\Vert Q\Vert <1$ implies ${Q}^{n}x\to 0$ for all x.

Step 2

If $\Vert Q\Vert >1$ then Q could be the identity or a rotation matrix, with the obvious consequences on convergence of ${Q}^{n}x$.

If $\Vert Q\Vert >1$ 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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