Schur's decomposition says any matrix $A$ is similar to a upper triangular matrix $U$ i.e., there exists unitary $Q$ such that $A = Q^{-1}UQ$. If we split $U$ as $D+N$ where $D$ is the diagonal part and $N$ is the off-diagonal part, then we know $N$ is nilpotent. Any Nilpotent matrix can be brought to Jordan form using a basis $P$ i.e., there exists $P$ such that $N = P^{-1}J_NP$ where $J_N$ is the Jordan form of $N$. Thus we have,

\begin{align*} QAQ^{-1} & = & U \\ & = & D + N \\ & = & D + P^{-1}J_NP \\ \end{align*} This implies $$(PQ)A(PQ)^{-1} = PDP^{-1} + J_N$$

If $P$ commutes with $D$, then we get that in the basis given by $PQ$, the matrix splits into diagonal + nilpotent parts. Is this the same as the Jordan decomposition? If so, why should $P$ commute with $D$?

conjugateto the Jordan decomposition. $\endgroup$ – LSpice Jul 13 at 10:33