A question about the definition of generalized eigenspace 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 $A$


 A: Since we are proving the blue-lined statement, we write 
$$V_{\lambda_i}=\{x: (A-\lambda_i)^{m(\lambda_i)}x=0\}.$$
We prove that $\mathbb{C}^n=\oplus_{i=1}^j V_{\lambda_i}$. 
Let $v_1\in V_{\lambda_1}, v_2\in V_{\lambda_2}, \ldots, v_j\in V_{\lambda_j}$.  Suppose that there is a linear relation between these   vectors:
$$
\sum_{i=1}^j   v_i=0.
$$
For each $i$, let  $f_i(t)=\frac{p_A(t)}{(t-\lambda_i)^{m(\lambda_i)}}$. We obtain by Cayley-Hamilton theorem that 
$$
   f_i(A) v_i = 0.
$$ 
We define
$$
f(t) := \sum_{i=1}^j \frac{f_i(t)}{f_i(\lambda_i)} -1.
$$
This polynomial satisfies $f(\lambda_i)=0$ for each $i$. Then $f(t) = g(t)\prod_i (t-\lambda_i)$ for some polynomial $g(t)$. This gives
$$
f(A) = g(A)\prod_i (A-\lambda_i)=\sum_{i=1}^j \frac{f_i(A)}{f_i(\lambda_i)} -I \ \ \ (*).
$$
Applying this to $v_i$, we have 
$$
v_i = \frac{f_i(A)v_i }{f_i(\lambda_i)}-g(A)\prod_i (A-\lambda_i)v_i=-g(A)\prod_i (A-\lambda_i)v_i
$$
A finitely many iteration of above, gives $v_i=0$. 
Again by applying $(*)$ to $v$, we have for any $v\in \mathbb{C}^n$, 
$$v = \sum_{i=1}^j \frac{f_i(A)v }{f_i(\lambda_i)}-g(A)\prod_i (A-\lambda_i)v.$$
With a finitely many iteration of above, we obtain that $v\in \sum_{i=1}^j V_{\lambda_i}$. Since the sum is direct by the first part, we have $v\in \oplus_{i=1}^j V_{\lambda_i}$. 
A: 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 $\lambda$.
A: Here's another approach (the one taken in Axler's book). It avoids the Cayley-Hamilton theorem, using only the ideas of "subspace", "dimension", and "kernel" and the elementary results surrounding them.
Let $A$ be an $n \times n$ matrix. We have the following facts about powers of $A$ (try proving them and see if you like this method better):


*

*$\{0\} = \ker A^0 \subseteq \ker A^1 \subseteq \ker A^2 \subseteq \cdots$.

*If $j$ is an integer such that $\ker A^j = \ker A^{j+1}$, then the inclusions in 1 "terminate" (become equalities): $$ \{0\} \subseteq \ker A^1 \subseteq \cdots \subseteq \ker A^j = \ker A^{j+1} = \ker A^{j+2} = \cdots. $$

*$\ker A^n = \ker A^{n+1}$, i.e., the inclusions must terminate before the $(n+1)$st power: $$ \{0\} \subseteq \ker A^1 \subseteq \cdots \subseteq \ker A^n = \ker A^{n+1} = \cdots. $$


Now if $\lambda$ is an eigenvalue of $A$, consider the above inclusions with $A$ replaced by $A - \lambda I$. Let $j$ be the least integer for which the inclusions terminate, so the inclusions before the power $j$ are strict. We get
$$
(*) \quad \{0\} \subsetneq \ker(A - \lambda I)^1 \subsetneq \cdots \subsetneq \ker(A - \lambda I)^j = \cdots = \ker(A - \lambda I)^n.
$$
We know that $j$ exists because fact 3 says $j \leq n$.
Now, the algebraic multiplicity $m$ of $\lambda$ is defined to be $m = \dim \ker (A - \lambda I)^n$. Since the $\subsetneq$'s in $(*)$ are strict inlcusions of subspaces, the dimension must increase by at least $1$ each time (this is how we prove fact 3 above!); hence $\dim \ker (A - \lambda I)^j \geq j$. Thus by taking dimensions in $(*)$ we have 
$$
j \leq \dim \ker (A - \lambda I)^j = \cdots = \dim \ker (A - \lambda I)^n = m.
$$
Using $j \leq m$  and $(*)$ once more, we see that 
$$
\ker (A - \lambda I)^j = \cdots = \ker (A - \lambda I)^m = \cdots = \ker (A - \lambda I)^n,
$$
and the last equality is what you wanted.
