Distinct eigenvalues gives rise to linearly independent vectors which can be generated by the matrix

Question

Let $A$ be a complex n dimensional square matrix with $n$ distinct eigenvalues. Then, does there exist a vector $u\in \mathbb{C}^n$ such that $u, Au, A^2u, \ldots, A^{n-1}u$ be linearly independent?

I think yes, and the proof somehow hinges on the diagonalizability of the matrix. Any hints. Thanks beforehand.

Answer 1

This is true for any field $k$, and you can prove it using the $k[X]$-module structure associated to the matrix/endomorphism of $k^n$ associated to $A$, defined by $$P(X)\cdot u=P(A)\,u,\quad P\in k[X].$$ The existence of a vector $u$ such that $\,(u, Au,A^2u,\dots,A^{n-1}u)\,$ be linearly independent means the $k[X]$-module $(k^n,u)$ is cyclic, and it is a known result that this module is cyclic if and only if its minimal polynomial has degree $n$.

Now, if there are $n$ distinct eigenvalues $\lambda_1,\dots, \lambda_n$, the minimal polynomial is equal to the characteristic polynomial $$\chi_u=(X-\lambda_1)\dotsm(X-\lambda_n),$$ which has degree $n$.