Is it possible to prove the Jordan decomposition starting from Schur's decomposition? 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 $D$?
 A: I am not sure I get what you mean by "brought to Jordan form", but if you don't consider the structure of $D$ while changing basis for $N$ then it won't work. Example:
$$
U = 
\begin{bmatrix}
1 & 1& 0 & 0\\
0 & 1& 0 & 0\\
0 & 0 & 2 & 1\\
0 & 0 & 0 & 2
\end{bmatrix}
$$
has
$$
N = 
\begin{bmatrix}
0 & 1& 0 & 0\\
0 & 0& 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{bmatrix}
$$
which will presumably be transformed (without further information from $D$) into
$$
J_N = 
\begin{bmatrix}
0 & 1& 0 & 0\\
0 & 0& 1 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}
$$
and the transformation that does it does not commute with $D$.
Other example:
$$
U = 
\begin{bmatrix}
1 & 1& 0 & 0\\
0 & 1& 1 & 0\\
0 & 0 & 2 & 1\\
0 & 0 & 0 & 2
\end{bmatrix}
$$
has
$$
N = 
\begin{bmatrix}
0 & 1& 0 & 0\\
0 & 0& 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{bmatrix}
$$
which will presumably be left unchanged, but that doesn't produce a Jordan decomposition.
On the other hand, if you mean something else that uses the structure of $D$ by "brought to Jordan form", then sure, there is a change of basis that turns $U$ into Jordan form, but that's just obvious by following the proof of the Jordan decomposition theorem.
