TITLE: Showing that $T:V\to V$ has a cyclic vector if its eigenspaces all have dimension one.
QUESTION [8 upvotes]: Let $V$ be an $n$-dimensional complex vector space and $T:V\to V$. Suppose that $$\{v\in V: Tv = \lambda v\}$$ has dimension $1$ or $0$ for all $\lambda\in \mathbb{C}$. Show that there exists some $w\in V$ such that $\{w,Tw,\dots, T^{n-1}w\}$ is linearly independent.
I tried using the canonical forms to answer this question but I don't think that's the right way to go.

REPLY [3 votes]: Hints:

It suffices to show that the minimal polynomial of $m(x)$ of $T$ is equal to the characteristic polynomial $p(x)$ of $T$, which has degree $n$.
If $\lambda$ is an eigenvalue of $T$, then the multiplicity of $\lambda$ in $m$ is equal to the size of the largest Jordan block corresponding to $\lambda$.
If $\lambda$ is an eigenvalue of $T$, then the algebraic multiplicity of $\lambda$, i.e. the multiplicity of $\lambda$ in $p$, is equal to the sum of the sizes of the Jordan blocks corresponding to $\lambda$.
If $\lambda$ is an eigenvalue of $T$, then the number of Jordan blocks corresponding to $\lambda$ is equal to the geometric multiplicity of $\lambda$, which in your problem is $1$ for all eigenvalues $\lambda$.