A proof uses the very general following result (known in France as the lemme des noyaux (= kernels lemma):
Lemma: Let $E$ a $K$ vector space (not necessarily finite dimensional), $f$ an endomorphism of $E$, $P_1, \dots, P_n$ pairwise coprime polynomials in $K[x]$. Then
$$\ker\bigl(P_1(f)\bigr)\oplus\dots\oplus \ker\bigl(P_n(f)\bigr)=\ker\bigl((P_1\dots P_n)(f)\bigr)$$
Apply this lemma to the characteristic polynomial $\,\chi_f(x)=(x-\lambda_1)^{m_1}\dots(x-\lambda_r)^{m_r}$. The generalised eigenspaces are precisely the $\,\ker(f-\lambda_i)^{m_i}$s $\,(i=1,\dots,r)$ and $\,\ker\chi_f(x)=\ker0=V$ by Hamilton-Cayley.
Proof of the lemma (sketch):
By induction of the number of factors: we have to prove that if $P$ and $Q$ are coprime polynomials, $\ker P(f)\oplus\ker Q(f)=\ker (P\circ Q)(f) $.
As $(P\circ Q)(f)=(Q\circ P)(f)$, each of $\ker P(f),\ker Q(f)$ is contained in $(P\circ Q)(f)$, hence $\ker P(f)+\ker Q(f)\subset (P\circ Q)(f)$.
Conversely, since $P$ and $Q$ are coprime, we have a Bézout identity: $U(x)P(x)+ V(x)Q(x)=1$ for some polynomials $U(x), V(x)$. Substituting $f$ for $x$, this translates as:
$$U(f)\circ P(f)+ V(f)\circ Q(f)= \operatorname{Id}_V.$$
Now let $x$ be in $\ker(P\circ Q)(f)$. We have
$$x=U(f)\circ P(f)(x)+ V(f)\circ Q(f)(x)\tag{1}$$
Let's denote $x_2=U(f)\circ P(f)(x)$, $\,x_1=V(f)\circ Q(f)(x)$. It is easy to check $\,P(x_1)=0$ and $\,Q(x_2)=0$. Thus $x=x_1+x_2$ lies in $\ker P(f)+\ker Q(f)$.
Finally, the sum is direct, if $x\in\ker P(f)\cap\ker Q(f)$, we have $x=0$ from $\,(1)$.
The inductive step is straightforward (left as an exercise! :o) —.
Best Answer
The minimal polynomial is of the form \begin{equation} p=(x-c_1)(x-c_2)\cdots (x-c_k), \end{equation} where $c_1,c_2,\ldots,c_k$ are the distinct eigenvalues of $T$.
By primary decomposition \begin{equation} V=W_1 \oplus W_2 \oplus \cdots \oplus W_k, \end{equation} where $W_i$ is the eigenspace corresponding to $c_i$, $1\leq i \leq k$.
From Hoffman & Kunze, Page 226, Exercise 10, one should be able to see that \begin{equation} W=(W\cap W_1) \oplus (W\cap W_2) \oplus \cdots \oplus (W\cap W_k). \end{equation} Clearly, $W\cap W_i$ is $T$-invariant, $1\leq i \leq k$.
Let $\{\alpha_1,\alpha_2,\ldots,\alpha_{r_i} \}$ be an ordered basis for $W\cap W_i$. Since $W\cap W_i$ is a subspace of the eigenspace $W_i$, $\{\alpha_1,\alpha_2,\ldots,\alpha_{r_i} \}$ can be extended to $\{\alpha_1,\alpha_2,\ldots,\alpha_{r_i},\alpha_{r_i+1},\ldots,\alpha_{s_i} \}$, a basis for $W_i$. Let $V_i$ be the subspace spanned by $\{\alpha_{r_i+1},\ldots,\alpha_{s_i} \}$. Then $W_i=(W\cap W_i)\oplus V_i$.
Hence \begin{equation} V=(W\cap W_1)\oplus V_1 \oplus (W\cap W_2)\oplus V_2 \oplus \cdots \oplus (W\cap W_k)\oplus V_k, \end{equation} i.e., $W$ has $T$-invariant complementary subspace of $V$, $V_1\oplus V_2 \oplus \cdots \oplus V_k$.