= Solution
The definition gives $T=DU$. The <matrix> $D=P^{-1}D_0P$ is invertible and <diagonalizable>. In the same basis $U$ is $I+D_0^{-1}N$, which commutes with $D_0$ because $N$ does. Conjugating back gives
$$
\boxed{T=DU=UD,\qquad D\text{ diagonalizable},\quad U\text{ unipotent}}.
$$
This is the multiplicative <Jordan decomposition in an affine algebraic group>, here applied to the general linear group. The explicit construction also handles repeated eigenvalues and nontrivial Jordan blocks.
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