Let $V$ be a finite-dimensional complex vector space and $T:V\to V$ a linear transformation. Suppose there exists $v\in V$ such that $\{v,Tv,T^2v,\ldots,T^{n-1}v\}$ is a basis for $V$. Show that the eigenspaces of $T$ are all $1$-dimensional.
I want to use the minimal polynomial to prove the above statement. So I need to find $T^nv$. Any suggestions?
Suppose there is some eigenspace of dimension at least$~2$, say for$~\lambda$. Then $\ker(T-\lambda I)$ having dimension at least$~2$, the image $W$ of $T-\lambda I$ has dimension $d\leq n-2$ (rank-nullity). The subspace$~W$ is $T$-stable, and is annihilated by some polynomial $P[T]$ of$~T$ with $\deg(P)\leq d$ (one could take for $P$ the minimal or the characteristic polynomial of the restriction of $T$ to$~W$). Then $P(X-\lambda)$ (multiplication, not substitution!) is a polynomial of degree strictly less than$~n$ that, when evaluated in$~T$, annihilates the whole vector space: in this evaluation $P[T]\circ(T-\lambda I)$ the rightmost (first acting) factor maps $V\to W$ and the other factor vanishes on$~W$.
This shows that the minimal polynomial of$~T$ has degree strictly less than$~n$, and this contradicts the existence of a cyclic vector for$~T$. Indeed, if the minimal polynomial is $c_0+c_1X+\cdots c_mX^m$ with $m<n$, then the linear relation $0=c_0v+c_1T(v)+\cdots c_mT^m(v)$ holds for all vectors$~v\in V$, and $v,T(v),\ldots,T^{n-1}(v)$ cannot be a basis of$~V$.
