TITLE: Existence of T-invariant complement of T-invariant subspace when T is diagonalisable
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: Let $\{e_1,\dots,e_n\}$ be the set of eigenvectors and define $W'$ as the complement of $W$ with respect to $V$. Without losing generality, let $W\ne\{0\}$ and $W\ne V$ (these cases are trivial). For each $e_i$, it should happen that $e_i\in W$ or $e_i\in W'$. Since we assume that $W\ne\{0\}$ and $W\ne V$, this implies that $W'\ne\{0\}$ and $W'\ne V$. If $w\in W$, then $w=a_1e_1+\ldots+a_ne_n$, and this implies $a_{k+1}e_{k+1}+\ldots+a_ne_n=w-a_1e_1-\ldots-a_ke_k\in W$, then $a_{k+1}=\ldots=a_n=0$, so $span\{e_1,...e_k\}=W$ and $span\{e_{k+1},...e_n\}=W'$, for some $k\ge1$. This proves that $V= W \oplus W'$ since $\operatorname{span}\{e_1,\dots,e_k\} \cap \operatorname{span}\{e_{k+1},\dots,e_n\}=\varnothing$. They are invariant under $T$ since if $e_i\in W$ (or $e_i\in W^{\mathrm c}$) then $T(e_i)=\lambda_ie_i\in W$ (or $\lambda_ie_i\in W^{\mathrm c}$).