Existence of T-invariant complement of T-invariant subspace when T is diagonalisable Let $V$ be a complex linear space of dimension $n$. Let $T \in End(V)$ such that $T$ is diagonalisable. Prove that each $T$-invariant subspace $W$ of $V$ has a complementary $T$-invariant subspace $W'$ such that $V= W \oplus W'$.
Note: Let $\{e_1,...e_n\}$ be the set of eigenvectors together with eigenspaces $V_{\lambda_1},...V_{\lambda_n}$ of $T$. It's sufficient to show that every $T$-invariant subspace $W$ must be a direct sum of eigenspaces, then it'll be trivial to find $W'$ (just take the rest eigenspaces not in the direct sum and glue them to $W$).. But how to prove $W$ is a direct sum of eigenspaces?

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 $V$, 
$V_1\oplus V_2 \oplus \cdots \oplus V_k$.
