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Nov 18, 2022 at 7:35 vote accept Guido Li
Nov 16, 2022 at 16:59 answer added Jochen Glueck timeline score: 5
Nov 16, 2022 at 16:15 comment added Guido Li @CarloBeenakker indeed, but observe they are finite-dimensional arguments.
Nov 16, 2022 at 15:49 comment added David E Speyer This ought to be true because, in the finite dimensional case, if $\lambda \neq \mu$, then the generalized left $\lambda$-eigenspace is orthogonal to the right generalized $\mu$-eigenspace. So, if we were to impose in addition that $\lambda$ be algebraically simple and that the left $\lambda$-eigenspace is orthogonal to the right $\lambda$-eigenspace, then the the left $\lambda$-eigenspace is orthogonal to everything, contradiction. Not posting this as an answer because I never feel confident in the infinite dimensional world.
Nov 16, 2022 at 15:47 comment added Carlo Beenakker several proofs are at math.stackexchange.com/q/794520/87355
Nov 16, 2022 at 15:36 history edited Guido Li CC BY-SA 4.0
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Nov 16, 2022 at 15:35 comment added Guido Li algebraically simple, yes!
Nov 16, 2022 at 10:08 comment added Jochen Glueck Do I assume correctly that by "simple", you mean algebraically simple (rather than only geometrically simple)?
Nov 16, 2022 at 9:20 history asked Guido Li CC BY-SA 4.0