Rank of generalized eigenvector for matrix power

Question

$A$ is a square matrix with real entries. Let $v$ be a generalized eigenvector of $A$ with rank $r$, i.e., $(A-\lambda I)^rv=0$ and $(A-\lambda I)^{r-1} v\neq0$, where $v$ is non-null and $\lambda$ is an eigenvalue of $A$. We know that $v$ is also a generalized eigenvector of $A^n$ with rank $s\leq r$, because $(A^n-\lambda^n I)^rv=Q(A, \lambda)(A-\lambda I)^rv=Q(A, \lambda)0=0$. If $\lambda=0$, I think $ s=\lfloor r/n \rfloor$. Question is: if $\lambda\neq0$, what can we say about $s$? Is $s=r$? Feel free to throw in additional hypotheses that may help to solve the problem.

Answer 1

First of all, you need to realise that $v$ will be a generalised eigenvector of$~A^n$ not for$~\lambda$, but for the eigenvalue$~\lambda^n$. Now you know that it takes $r$ applications of $A-\lambda I$ to reduce $v$ to $0$, and you want to know how many applications of $A^n-\lambda^nI$ it will take to also reduce $v$ to $0$. Now the polynomial $X^n-\lambda^n$ is divisible by $X-\lambda$ with quotient $Q=X^{n-1}+\lambda X^{n-2} +\cdots+ \lambda^{n-2}X+\lambda^{n-1}$. When $\lambda\neq0$, the polynomial $Q$ does not have$~\lambda$ as a root, and is therefore relatively prime to $X-\lambda$. Hence $Q[A]$ acts invertibly of the generalised eigenspace for$~\lambda$ of$~A$. It follows that $$(A^n-\lambda^nI)^k(v) = ((A-\lambda I)Q[A])^k(v)=(Q[A])^k((X-\lambda I)^k(v)) $$ which is $0$ if and only if $k\geq r$. So indeed, if $\lambda\neq0$, the rank of $v$ with respect to $A^n$ equals$~r$.