1
$\begingroup$

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 $D$?

Improve this question
$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Jul 13, 2020 at 5:43
  • 1
    $\begingroup$ @LSpice But an upper triangular matrix with zero diagonal entries is always Nilpotent, which is the case in the question. $\endgroup$
    – Fdost
    Commented Jul 13, 2020 at 6:04
  • 2
    $\begingroup$ 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). $\endgroup$
    – Geoff Robinson
    Commented Jul 13, 2020 at 8:53
  • $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Jul 13, 2020 at 10:33

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks for the wonderful explanation and the examples. I guess I understand it better now. Particularly, if the Jordan form respects the structure of D, then commutativity follows since each block would correspond to a scaled identity block in D and commutativity is obvious. Is the following observation true? As one can start with an arbitrarily structured D and add an arbitrary upper triangular matrix with zero diagonal entries, does this mean then, every upper triangular matrix with zero diagonal entries can be brought to every possible Jordon block form under a suitable basis? $\endgroup$
    – Fdost
    Commented Jul 13, 2020 at 9:05
  • $\begingroup$ 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. $\endgroup$
    – Federico Poloni
    Commented Jul 13, 2020 at 9:09
  • $\begingroup$ 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). $\endgroup$
    – Federico Poloni
    Commented Jul 13, 2020 at 9:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .