Because of the Fundamental Theorem of Algebra, operators on complex vector spaces have special properties. In particular, the minimal polynomial of an operator on a complex vector space has the form
\begin{equation*}
p(x)=(x-\lambda_1)^{k_1}(x-\lambda_2)^{k_2}\ldots(x-\lambda_1)^{k_m}\text{.}
\end{equation*}
In this case, we claim that \(V\) is a direct sum of generalized eigenspaces.
Proof.
We will use induction on the dimension of \(V\text{.}\) To establish the base case, we assume that \(\dim V = 1\text{.}\) In this case, \(p(x) = x-\lambda\) so \(T-\lambda I = 0\) or \(T=\lambda I\text{.}\) Then \(V=G_\lambda\text{.}\)
For the inductive step, we will assume the result is true for all vector spaces of dimension less than \(\dim V\text{.}\) We will choose an eigenvalue \(\lambda\) and write the minimal polynomial \(p\) as
\begin{equation*}
p(x) = q(x)(x-\lambda)^k
\end{equation*}
where \(q(\lambda)\neq 0\text{.}\)
Notice that
\begin{equation*}
\nul((T-\lambda I)^k) \cap \range((T-\lambda)^k) = \{0\}\text{.}
\end{equation*}
To see this, suppose that \(\vvec\) is in this intersection. Then \(\vvec = (T-\lambda)^k\uvec\) for some vector \(\uvec\text{.}\) Moreover,
\begin{equation*}
0 = (T-\lambda I)^k\vvec = (T-\lambda I)^{2k}\uvec\text{.}
\end{equation*}
This implies that \(p_\uvec\) divides \((x-\lambda)^{2k}\) so that the only factors of \(p_\uvec\) are \(x-\lambda\text{.}\) Since \(p_\uvec\) also divides the minimal polynomial \(p\text{,}\) we also know that \(p_\uvec\) divides \((x-\lambda)^k\text{.}\) Therefore,
\begin{equation*}
\vvec = (T-\lambda I)^k\uvec = 0\text{.}
\end{equation*}
\begin{equation*}
\dim\nul((T-\lambda I)^k) +\dim \range((T-\lambda I)^k) =
\dim V\text{,}
\end{equation*}
which says that
\begin{equation*}
V = \nul((T-\lambda I)^k) \oplus \range((T-\lambda I)^k)\text{.}
\end{equation*}
If we define \(U=\range((T-\lambda I)^k)\text{,}\) then we also have
\begin{equation*}
V = G_\lambda \oplus U\text{.}
\end{equation*}
Since we have written the minimal polynomial \(p(x)=q(x)(x-\lambda)^k\text{,}\) we can see that the minimal polynomial of \(T|_U\) is \(q\text{.}\) For instance, if \(\uvec\) is in \(U\text{,}\) then \(\uvec=(T-\lambda
I)^k\vvec\) for some vector \(V\text{.}\) Then
\begin{equation*}
q(T)\uvec = q(T)(T-\lambda I)^k \vvec = p(T)\vvec = 0\text{.}
\end{equation*}
This shows that the minimal polynomial \(p_U\) of \(T|_U\) divides \(q\text{.}\) However, if the minimal polynomial of \(T|_U\) has a smaller degree than \(q\text{,}\) this would contradcit the fact that the minimal polynomial of \(T\) has the smallest possible degree.
By the inductive hypothesis, \(U\) may be written as a direct sum of its generalized eigenspaces. All that remains is to show that, if \(\mu\) is an eigenvalue distinct from \(\lambda\text{,}\) the generalized eigenspace of \(T\) associated to \(\mu\) is the same as the generalized eigenspace of \(T|_U\) associated to \(\mu\text{.}\) To this end, suppose that \(\vvec\) satisfies \((T-\mu
I)^l\vvec\) for some \(l\text{.}\) Because \(V=G_\lambda
\oplus U\text{,}\) we can write
\begin{equation*}
\vvec = \nvec + \uvec
\end{equation*}
where \(\nvec\in G_\lambda\) and \(\uvec \in U\text{.}\) We have
\begin{equation*}
0 = (T-\mu I)^l\vvec = (T-\mu I)^l\nvec + (T-\mu I)^l\uvec\text{.}
\end{equation*}
Because these subspaces are \(T\)-invariant, we have
\begin{align*}
(T-\mu I)^l\nvec \amp = 0\text{.}
\end{align*}
This means that
\(\nvec\in G_\lambda\cap G_\mu = \{0\}\) by
Proposition 2.4.4 so that
\(\nvec = 0\text{.}\) Therefore,
\(\vvec = \uvec\in U\) and
\begin{equation*}
(T-\mu I)^l\uvec = (T|_U-\mu I)^l \uvec = 0\text{.}
\end{equation*}
