$\lim_{k\to \infty} A^k v=0$ with dominant eigenvalue $1$ and generalized eigenvector Let $A$ be a $n\times n$ matrix over $\mathbb{R}$ with positive entries. Assume $v$ is a generalized eigenvector of $A$ with $(A-\lambda I)^r v=0$, where $r\geq 2$ and $|\lambda|<1$. The unique dominant eigenvalue of $A$ is $\hat{\lambda}=1$. As part of intermediate steps of a homework solution, I need to show $\lim_{k\to \infty} A^k v=0$. 
I know that if $v'$ is an ordinary eigenvector with $v'=(A-\lambda I)^{r-1} v$, then $A^k v'=\lambda^k v'$. Also, since $v$ is an generalized eigenvector, $(A^k-\lambda^k I)^r v=0$. However, the way to solve this problem is still unclear to me. Is there any hint or comment?
Also, if this statement is indeed not true, please let me know so that I can try another approach to my original homework question.
Update:
Can I do something like this (probably not):
$(A^k-\lambda^k I)^{r-1} (A^k-\lambda^k I) v=0$ implies
$(A^k-\lambda^k I)^{r-1} A^k v=(A^k-\lambda^k I)^{r-1} \lambda^k v$.
So, $A^k v=\lambda^k v$. Since $|\lambda|<1$, we get $\lim_{k\to \infty} A^k v=0$.
 A: Suppose for simplicity that $A$ is a single Jordan block. Then $e_r$ (the $r$th standard basis vector) is an $r$th-order generalized eigenvector.
By induction you can explicitly write down $A^k$, and its $r$th column is $A^k e_r$. The nonzero entries of this column are of the form $\binom{k}{l} \lambda^{k-l}$ for $l = 0, \ldots, r-1$. As $k \to \infty$, we have $|\binom{k}{l} \lambda^{k-1}| \le k^l |\lambda|^{k-l} \to 0$ because exponential growth is faster than polynomial growth. Thus $A^k e_r \to 0$.

For a general $A$, you can use its Jordan form to reduce the problem to the situation described above.
I am not implying that the Jordan form is necessary to solve this problem; I am using it just for convenience. I think you can solve the problem directly by using the definition of generalized eigenvalue to write $A^k v$ as a linear combination of $v, Av, \ldots, A^{r-1} v$, for any $k \ge r$, and show that the coefficients of this linear combination tend to $0$ as $k \to \infty$. I suspect the computation will be equivalent to the "computing the $k$th power of a Jordan block" result I was using above.
A: Well, if $v\in Ker(A-\lambda I)^r$, then there are several invariant subspaces $W_i$  such that $\oplus W_i$ contains $v$. And the restriction of $A$ on $W_i$ can be represented by a Jordan block, say $J_s(\lambda)$, once choosing basis properly. This is basically the theory of Jordan normal form. Now every entry of $J_s(\lambda)^k$ tends to $0$ as you can easily verify. So $A^kv$ tends to $0$.
