Let $T: V \to V$ be a linear operator and $V$a finite dimensional vector space over $F$. Suppose that $m_T = (x- \lambda_1)^{m_1} \cdots (x - \lambda_k)^{m_k}$. Show that there is a diagonalizable operator $D:V \to V$ and a nilpotent operator $N: V\to V$ s.t $T = D+N$ and $DN= ND$. Futhermore, prove that $D$ and $N$ are polynomials in $T$.
I know that some variations of this question has asked before here:
Diagonalizable and nilpotent operators
However, a tried to solve my way, so I have questions about my solution:
Since $m_t$ has that form in $F[x]$ then $p_T$ does, that is, $T$ admit a jordan canonical form, call $J$.
Since each block of $J$ is sum of a diagonalizable and a nilpotent operator we have $J = D' + N'$, where $D$ is a diagonal matrix and $N$ is nilpotent.
Therefore, for each basis $B$ of $V$ there exists $P$ s.t $[T]_{B} = P^{-1} J P = P^{-1}(D'+N')P = P^{-1}D' P + P^{-1}N' P$.
Now take $D:V \to V$, where $[D]_{B}= [P^{-1}D'P]$ and $N: V \to V$, where $[N]_{B} = [P^{-1}N' P]$. So we have $D$ diagonalizable and $N$ nilpotent.
But $ND = P^{-1} N' D' P \neq P^{-1}D'N'P = DN$ (it's easy to show a counter example). Where is my wrong?
Also, How can I prove that $D$ and $N$ are polynomials in $T$?