Let $\mathbb V$ be an $n$ dimensional complex vector space. If $T \in L(V,V),$ prove that there exists a diagonalizable operator $D$ and a nilpotent operator $N$ such that $T = D + N$ and $DN = ND$
and show that these mappings are uniquely determined.
Attempt: If $V$ be a vector space with $\dim V=n$, Then, the "kernel lemma" states that $$V = \bigoplus_{i=1}^n \ker (T- \lambda_i \mathrm{I})^{n}$$ and each of the subspaces $G(\lambda_i,T) = \ker (T- \lambda_i \mathrm{I})^{n}$ is invariant under $T$.
Now on each $G(\lambda_i,T)$, we have have $T = \lambda_i \mathrm{I}+ (T - \lambda_i \mathrm{I})$ and the second term is nilpotent.
For a vector $v$ we can write it as $v=v_1+...+v_n$ with $v_i \in G(λ_i,T).$ Then $Nv=∑_iNv_i=∑_i(T−λ_iI)|_{G(λi,T)}v_i$
This produces a sum of nilpotent operators. The sum of nilpotent operators might not be nilpotent. So,How do I use this to define a general $T$? Thanks a lot for the help