# 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 $$D$$?

• It usually will not be true that the off-diagonal part of a matrix is nilpotent. Consider $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, for example. – LSpice Jul 13 at 5:43
• @LSpice But an upper triangular matrix with zero diagonal entries is always Nilpotent, which is the case in the question. – Fdost Jul 13 at 6:04
• One of the points of true Jordan decomposition (of a not necessarily nilpotent) matrix is that the diagonal part and the strictly upper triangular part commute. This will not generally be true of D and N in your question (consider the case when D has distinct diagonal entries and N is non-zero, for exmple). – Geoff Robinson Jul 13 at 8:53
• I'm sorry; I missed the "upper triangular" part. @GeoffRobinson's point is the real reason that this is not the Jordan decomposition. However, it is conjugate to the Jordan decomposition. – LSpice Jul 13 at 10:33

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.
• every upper triangular matrix with zero diagonal entries can be brought to every possible Jordon block under a suitable basis No, there are some rank constraints. You can't transform $$\begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{bmatrix}$$ into You can't transform $$\begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{bmatrix}$$, for instance. – Federico Poloni Jul 13 at 9:09
• Another comment: the way I see it, there are two "ingredients" needed in the proof: (1) write the matrix as a direct sum (block diagonal matrix) of upper triangular matrices, each one with a different eigenvalue $\lambda_i$ on the diagonal, and (2) transform those diagonal blocks into Jordan form. I don't think you considered (1) here, but that is a tricky step, because the transformations that you need must involve explicitly $\frac{1}{\lambda_i - \lambda_j}$ to clear the off-diagonal block $(i,j)$ (otherwise they would work also if the $\lambda_i = \lambda_j$, which is not the case). – Federico Poloni Jul 13 at 9:13