$O$ is not dense in $\mathbb R^n$ $A$ is a matrix of order $n$, and $x$ is an $n$-dimensional column vector. The set
$$
O=\{x,Ax,A^2x,\cdots,A^mx,\cdots\}
$$
Prove that, $O$ is not dense in $\mathbb R^n$.
This claim seems to be trivial: it just looks like $\mathbb N$ in $\mathbb R$. I recalled how we verify the denseness of $\mathbb Q$ in $\mathbb R$, but it seeems that the same method doesn't work. Can anyone help?
Suppose $O=\{A^mx:m\in\mathbb N_0\}$ is dense in $\mathbb R$. There are two possibilities:

$A^T$ has a real eigenvector $y$ corresponding to a real eigenvalue $\lambda$. Then
\begin{aligned}
&amp;\ O=\{A^mx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R^n$}\\
\Rightarrow&amp;\ y^TO=\{\lambda^m y^Tx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R$}\\
\Rightarrow&amp;\ Y=\{\lambda^m:m\in\mathbb N_0\}\text{ is dense in $\mathbb R$}.
\end{aligned}
$A^T=A^\ast$ has an eigevector $z$ corresponding to some non-real eigenvalue $\overline{\lambda}$. Then the $\mathbb R$-linear span of the entries of $z$ must be two-dimensional and hence it is equal to $\mathbb C$. Therefore,
\begin{aligned}
&amp;\ O=\{A^mx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R^n$}\\
\Rightarrow&amp;\ z^\ast O=\{\lambda^mz^\ast x:m\in\mathbb N_0\}\text{ is dense in $\mathbb C$}\\
\Rightarrow&amp;\ Z=\{\lambda^m:m\in\mathbb N_0\}\text{ is dense in $\mathbb C$}.
\end{aligned}

It remains to show that $Y$ and $Z$ aren't really dense in $\mathbb R$ and $\mathbb C$ respectively. You may continue from here.
I think the hints in the comments provide you enough help for a simple proof.
You consider two cases: the norm sequence $(||A^n x||)_{n\in\mathbb N}$ either diverges to infinity, or is eventually bounded by some real number $M$ (it could still oscillate and not converge, but we'll see that we don't care).
If it diverges, it means that any open bounded subspace of $\mathbb R^n$ (e.g., $B=\{x\in\mathbb R^n : ||x||&lt;M\}$) contains at most a finite number of elements of $O$, which means that $O$ can't be dense in $\mathbb R^n$ (because its closure doesn't even &quot;fill&quot; $B$).
If the norm is (eventually) bounded ($\leq M$), you can use the same argument for $\mathbb R^n\smallsetminus \overline{B}$ (it contains at most a finite number of elements of $O$...).
In either case, $O$ cannot be dense in $\mathbb R^n$.