TITLE: A question about the definition of generalized eigenspace
QUESTION [8 upvotes]: Given an $n\times n$ matrix $A$, its generalized eigenspace pertaining to an eigenvalue $\lambda_i$ is defined as

$V_{\lambda_i}=\{x:(A-\lambda_i I)^n x=0\}$

The question is to prove

$V_{\lambda_i}=\{x:(A-\lambda_i I)^n x=0\} = \{x:(A-\lambda_i I)^{m(\lambda_i)} x=0\}$ where $m(\lambda_i)$ is the algebraic multiplicity of $\lambda_i$.

It is obvious that $\{x:(A-\lambda_i I)^{m(\lambda_i)} x=0\} \subseteq \{x:(A-\lambda_i I)^n x=0\}$, since $(A-\lambda_i I)^{m(\lambda_i)} x=0\Rightarrow (A-\lambda_i I)^n x=0$. The problem is the other direction.
The solution hints that using Hamilton-Cayley theorem, i.e.

$\prod\limits_{{\lambda } \in \sigma (A)} {{{(A - {\lambda}I)}^{m({\lambda })}}}  = O$ where ${\sigma (A)}$ is the spectrum of $A$

REPLY [3 votes]: To simplify notation, take $\lambda$ to be $\lambda_r$, the last one. Write $m_j$ instead of $m(\lambda_j)$. 
Suppose that there exists $x$ with $(A-\lambda_r I)^{m_r}x\ne0$ and $$\tag{*}(A-\lambda_r I)^{m_r+1}x=0.$$ We have
$$\tag{**}
\prod_{j=1}^{r-1}(A-\lambda_j I)^{m_j}\,(A-\lambda_r I)^{m_r}x=0.
$$
Write $$v=(A-\lambda_1 I)^{m_1-1}\prod_{j=2}^{r-1}(A-\lambda_j I)^{m_j}\,(A-\lambda_r I)^{m_r}x.$$
Then by $(**)$ we have $(A-\lambda_1 I)v=0$. We also have, using that all terms commute, that $(A-\lambda_r I)v=0$. Thus $\lambda_1 v=\lambda_r v$, which implies that $v=0$. In other words, we were able to reduce the power in $(**)$ by $1$. 
If we keep repeating this procedure many times, we'll eventually get to
$$
(A-\lambda_{r-1}I)\,(A-\lambda_r I)^{m_r}x=0.
$$
Applying the above reasoning one again, we get that $(A-\lambda_r I)^{m_r}x=0$, a contradiction. 
It follows that $$V_{\lambda}=\{x:(A-\lambda I)^n x=0\} = \{x:(A-\lambda I)^{m(\lambda)} x=0\}.$$ The procedure can be repeated for any $\lambda$.