We know the asymptotic behavior of the scalar geometric sequence $u_{n+1}=qu_n$ 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 $U_{n+1}=QU_n$, where $Q$ is a $2\times2$ matrix and $U_n$ is a two dimensional vector.
My attempt: $U_n=U_0Q^n$, so $|U_n|=|U_0||Q|^n$, thus if $|Q|<1$, then using scalar geometric sequences $\lim_{n}|U_n|=0$. But we should define the meaning of $|Q|$ and $|U|$.
Another issue is what happens in the other case $|Q|=1$ or $|Q|>1$.
Background of my question: The background of my question is the following probability problem:
If it rains some day, then the probability to rain the following day is (say) $0.75$. Now if it doesn't rain some day, then the probability to rain the following day is (say) $0.1$.
If we note $u_n$ the probability to rain in the $n$th day and $v_n=1-u_n$, then we would have $$u_{n+1}=0.75u_n+0.1v_n$$ $$v_{n+1}=0.25u_n+0.9v_n$$ Thus we will get the following "geometric" recurrence relation $$\begin{pmatrix} u_{n+1} \\ v_{n+1} \end{pmatrix}=\begin{pmatrix} 0.75 & 0.1\\ 0.25 & 0.9 \end{pmatrix}\begin{pmatrix} u_{n} \\ v_{n} \end{pmatrix}.$$
If matrix and vector norms are compatible in the way that $$ \|Qx\| \le \|Q\| \cdot \|x\|, $$ then $\|Q\|<1$ implies $Q^nx\to 0$ for all $x$.
If $\|Q\|=1$ then $Q$ could be the identity or a rotation matrix, with the obvious consequences on convergence of $Q^nx$.
If $\|Q\|>1$ then $Q^nx$ is not guaranteed to be unbounded. Take $Q=\pmatrix{2&0\\1&0}$ and $x=\pmatrix{1\\0}$ or $x=\pmatrix{0\\1}$.