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I've seen two definitions of generalized eigenvectors. One is $(A−λ⋅I)^kv=0$ and one where it is $Av = \lambda B v$. I get that they are the same thing but I'm curious. What is the point of both formulations and how does the first lead to the second?

Request for more info:

https://www.netlib.org/lapack/lug/node54.html

Thanks!

user1357015
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  • Could you define the second definition more precisely? What is $B$? If it's arbitrary, what's to stop it from being $B = \lambda^{-1} A$, thus making any $\lambda \neq 0$ a "generalised eigenvalue"? – Theo Bendit Oct 01 '18 at 02:05
  • Also, if $\lambda = 0$, then $Av = \lambda B v$ becomes $Av = 0$, which produces the (non-generalised) eigenvectors corresponding to the eigenvalue $0$. I don't think they're equivalent at all. – Theo Bendit Oct 01 '18 at 02:07
  • @TheoBendit: I added additional information. I'm getting it form the generalized eigenvalue package in python and C – user1357015 Oct 01 '18 at 02:15
  • This appears to be an unrelated problem. For one, notice that $A$ is assumed to be self-adjoint (symmetric or Hermitian), which implies that all the eigenvectors of $A$ are just plain old eigenvectors (i.e. $A$ is diagonalisable). – Theo Bendit Oct 01 '18 at 02:19
  • @TheoBendit: Ok -- so in that case, any clarity why even define the second version as they do? – user1357015 Oct 01 '18 at 02:22
  • As you might be able to guess from my previous comments, I'm not really familiar with generalised eigenproblems. It seems like one of those problems that arose in some application, but I'm not familiar with these applications. – Theo Bendit Oct 01 '18 at 02:26
  • The way to understand this is to take square matrices they give you and construct $P$ so that $P^{-1} A P = J$ is the Jordan form. See, for example, https://math.stackexchange.com/questions/2937108/finding-a-jordan-basis-after-finding-the-jordan-canonical-form/2937155#2937155 from today – Will Jagy Oct 01 '18 at 02:36

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