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Let $PJP^{-1}$ denote the Jordan decomposition of $M$. The matrix $J$ is a direct sum of Jordan blocks; it is unique up to permutation of the Jordan blocks. The matrix $P$ is not unique.

There are two questions here, which are similar enough that they deserve to be asked together:

  1. Can the Jordan decomposition be computed in a backwards stable way?
  2. The Jordan decomposition is often avoided in numerical linear algebra, even though it is a very convenient decomposition for developing a lot of theory. My question is: Why?

One unconvincing answer to 2 is that it's discontinuous: The matrices $P$ and $J$ vary discontinuously as $M$ varies. But this is a bad explanation because it also applies to the SVD (note that the factors $U$ and $V$ vary discontinuously in the SVD), but the SVD still gets used in numerical linear algebra.

[EDIT: Please do not tell me that in the SVD, the map $M \mapsto \Sigma$ is Lipschitz continuous (and not just $\frac1 2$-Hölder as one answer claims) because note that people in numerics often want to compute the whole SVD, including the matrices $U$ and $V$. And that's discontinuous.]

I somehow suspect that the problem might lie with $P$ being non-unitary.

wlad
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  • To me, the Jordan decomposition of a matrix $M$ is a pair of commuting, matrices, one semisimple, one nilpotent (or unipotent if you take the multiplicative decomposition), that sum (or multiply) to $M$. When you say "Let $P J P^{-1}$ be the Jordan decomposition of $M$", do you mean that, in the sense in which I refer to it, $P J P^{-1}$ is the nilpotent (or perhaps unipotent) part of the Jordan decomposition? – LSpice Dec 24 '22 at 16:34
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    @LSpice I mean this: Jordan Normal Form. I'm not comfortable saying JNF because I'm considering $P$ and $J$ together, and not just $J$ alone, and I think the terminology might be ambiguous. It's $P$ and $J$ together that describe the whole decomposition. – wlad Dec 24 '22 at 16:39
  • @LSpice The matrix $J$ can be written as $D + N$, where $D$ is diagonal (semisimple) and $N$ is nilpotent. Alternatively, $M$ itself can be written as such a sum, which would then be $PDP^{-1} + PNP^{-1}$. So it's related to the decomposition you mention, but not the same thing. – wlad Dec 24 '22 at 16:44
  • Since every matrix is arbitrarily close in norm to a diagonalizable one, what would backward stability of the Jordan decomposition even mean? Technically, the standard backwards stable algorithms for computing the eigendecomposition of a nonsymmetric matrix are backward stable algorithms for the Jordan decomposition which happen to always compute a diagonal $J$. – James Dec 24 '22 at 20:14
  • @James You need the $P$ matrix as well. If you always perturb $M$ so that $J$ is diagonal, $P$ might end up ill-conditioned. Perturbing $M$ so that $J$ is non-diagonal might perhaps result in a lower condition number for $P$. – wlad Dec 24 '22 at 20:41
  • @James To be clear, I'm asking for $P$ to be computed as well as $J$. Your comment reads like you think I'm only interested in computing the matrix $J$. – wlad Dec 24 '22 at 21:06
  • @wlad I understood you want $P$ too, and I understand your concerns about $P$ being ill-conditioned (which is exactly the answer to your second question, really). But that's a separate concern from backward stability, i.e., not part of your first question. Can you make your question more precise? – James Dec 24 '22 at 21:44
  • @James If it's exactly the answer to my 2nd question, I'd like to see why. I posted another question about that here – wlad Dec 25 '22 at 07:12

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