I'm currently going through Sheldon's Axler's Linear Algebra Done Right and am struggling to understand his proof of Theorem 8.23.
Suppose we are given a complex vector space $V$ and a linear transformation $T \in L(V)$. Let $\lambda_1, \dots, \lambda_m$ be the distinct eigenvalues of $T$, and let $U_1, \dots, U_m$ be the corresponding subspaces of the generalized eigenvectors. Theorem 8.23 claims it must be the case that $V = U_1 \oplus\dots\oplus U_m$.
The full proof can be found here, listed as Theorem 8.23 on page 174.
In his proof, Axler defines $U = U_1 + \dots + U_m$ and a function $S = T|_U$. He then claims that $S$ has the same eigenvalues, with the same multiplicities as $T$ because all the generalized eigenvectors of $T$ are in $U$, the domain of $S$.
I don't understand why this is the case. Consider a generalized eigenvector $v$ of $T$, such that $$ v \in \operatorname{null}(T - \lambda_i I)^{\dim V}. $$ To be a generalized eigenvector of $S$, that would mean $v \in \operatorname{null}(T - \lambda_i I)^{\dim U}$. I don't understand how this follows from the previous statement.
Thanks in advance.