How to prove that a restriction $(T − \lambda_j I)|_{U_j}$ is nilpotent. Suppose V is a complex vector space and $T \in L(V).$ Let $\lambda_1, . . . , \lambda_m$ be the distinct eigenvalues of $T$, and let $U_1, . . . , U_m$ be
the corresponding subspaces of generalized eigenvectors. Then:
(a) $V = U_1 \oplus \cdots\oplus U_m$;
(b) each $U_j$ is invariant under $T$;
(c) each $(T − \lambda_j I)|_{U_j}$ is nilpotent.
This appears in Axler "Linear algebra done right" and I already proved a and b, but I have problemos doing c
I have established that the restriction $N_j = (T − \lambda_j I_d): U_j → U_j$ is nilpotent.
 A: Recall that the generalised eigenspace can be written as
\begin{equation}
U_j = \bigcup \limits_{r=1}^{\infty} \ker(T-\lambda_jI)^r.
\end{equation}
Now, let $x \in U_j$ be arbitrary (so it lies in one of the kernels). This means there is an $r \in \Bbb{N}$ such that $(T- \lambda_jI)^r(x) = 0$. I hope you know that restriction to $U_j$ commutes with polynomials in $T$; so this justifies the first step:
\begin{align}
\left((T- \lambda_jI) \bigg|_{U_j} \right)^r(x) &= \left((T- \lambda_jI)^r \bigg|_{U_j} \right)(x) \\
&:= (T- \lambda_jI)^r(x) \tag{since $x \in U_j$} \\
&= 0.
\end{align}
By the way you should have mentioned that $V$ is finite dimensional, so that each $U_j$ will be as well. Now, make use of the following lemma, with $X = U_j$ and $S = (T- \lambda_jI)|_{U_j}$.

Lemma. Let $X$ be a finite dimensional (complex) vector space, and let $S:X \to X$ be a linear operator. Suppose that for every $x \in X$, there is an $r \in \mathbb{N}$ such that $S^r(x) = 0$. Then, $S$ is nilpotent.

The proof is simple: if $\beta = \{x_1, \dots, x_k\}$ is a basis of $X$, then for $i \in \{1, \dots, k\}$ you can find an $r_i$ such that $S^{r_i}(x_i) = 0$. Let $r = \max\{r_1, \dots, r_k\}$. Then, $S^r = 0$.
