Linear map dense trajectory Is there a finite-dimensional $\mathbb{R}$-normed space $V$ with $x_0 \in V$ and linear map $T \colon V \to V$ such that $\{T^nx_0\}_{n=1}^{\infty}$ is dense in $V$?
I got this question of Bollobas' Introductory Linear Analysis. My intuition leans towards no, but I can't come up with a convincing argument without assuming things. There is a hint given and it is to first prove the result over a $\mathbb{C}$-normed space.
 A: Assume there exist such $V,T,x_0$.
Let $V_{\mathbb C} = V\oplus V$ be the complexification of $V$, where we cononically embed $V$ in $V_\mathbb C$, and $T_{\mathbb C}:V_{\mathbb C}\to V_{\mathbb C}$ be the complex-linear extension of $T$. $V_{\mathbb C}$ and then aswell ${V_\mathbb C}'$ is finite dimensional, therefore ${T_{\mathbb C}}'$ must have have an eigenvector $x'$ with corresponding eigenvalue $\lambda$. Since $V$ is an $\mathbb R$-linear subspace of $V_{\mathbb C}$ and $x'$ is linear, $x'(V)$ is an $\mathbb R$-linear subspace of $\mathbb C$. It can't be the trivial subspace because then $x'$ would vanish on all of $V_\mathbb C$, which contradicts $x'\neq 0$. We claim that $\{x'({T_\mathbb C}^nx_0): ~n\in\mathbb N\}$ is dense in $x'(V)$, in particular, it contains arbitrary small and arbitrary large values (in terms of modulus). In order to see that, let $y\in x'(V)$ and $x\in V$ with $x'(x) = y$. Then, by assumption, there is a sequence of natural numbers $(n_k)_{k\in\mathbb N}$ with ${T_\mathbb C}^{n_k}x_0 = T^{n_k}x_0 \to x$. By continuity of $x'$, we have $x'({T_\mathbb C}^{n_k}x_0) \to y$.
On the other hand, $x'({T_\mathbb C}^nx_0) = [({T_\mathbb C}')^nx']x_0 = \lambda^n x'(x_0)$. But if $|\lambda|>1$, this doesn't assume arbitrary small values and if $|\lambda|\leq 1$ then it doesn't assume arbitrary large values.
