$V$ is a finite vector space above the real numbers. $ f\in \operatorname{End}(v)$ with no eigenvalues. Prove that for all $v\in V$ the dimension of cylic subspace $\cal Z(v) $ is an even number.
Somehow this makes to me since every two following vector in the ordered base would be non-linearly dependent, but I can't find a formal proof for this.
Let $v\in V$ be a vector and $Z$ its associated cyclic space. Suppose that the dimension of $Z$ is odd, so that the set $\left\lbrace v, Tv, \ldots, T^kv\right\rbrace$ is a base for $Z$ for some $k$ even integer (dimension $k+1$), from this we get a polynomial $T^{k+1}v+\alpha_kT^kv+\cdots+\alpha_1 Tv+\alpha_0 v=0$, so that $T$ restricted to the cyclic space has polynomial equal to $p(x)=x^{k+1}+\alpha_kx^{k}+\cdots+\alpha_1 x+\alpha_0$ that vanishes in $T$ and which have a real root (odd degree), we have that the minimal polynomial divides $p(x)$ but if the minimal polynomial is not equal to $p(x)$ it has lower degree, but that contradicts the fact that $\left\lbrace v, Tv, \ldots, T^kv\right\rbrace$ is a base for $Z$. Therefore, $T$ have a real eigenvalue.