 Solution

The definition gives $T=DU$. The matrix $D=P^{-1}{D}_{0}P$ 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 givesThis 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.