Showing that $T:V\to V$ has a cyclic vector if its eigenspaces all have dimension one. 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.
 A: 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 $\lambda$.

A: I undeleted my answer because I like my explicit approach. Clearly this is not a complete answer, but only a special case.
If you know that the matrix is diagonalizable then all diagonal entries are different. Now test the vector $(1,\dots,1)$ against it in this new basis and you see that the column matrix of this set of vecors is nothing but the Vandermonde matrix, which is invertible.
For the general case see the above answer
