A question about the definition of generalized eigenspace

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 $v\in \oplus_{i=1}^j V_{\lambda_i}$.

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 $\lambda$.