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I'm trying to understand the proof of first theorem here. Maybe it's very simple but I would like your help because I need understand this, I have no much time and my knowledge about this subject is very limited. Some users already have helped-me here and here but I still have doubts.

The statement which I need a explanation is this: "Thus, if $\rho(A)<1$ then $|\lambda_i|<1$".

Matrices $A$ and $J$ have the same eigenvalues, because they are similar. What is relationship between this fact and scalars $\lambda_i$? Is $\lambda_i$ a eigenvalue of A? Why?

Thanks.

Community
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Pedro
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  • Are you asking why $\lambda _i$ in $J$ is an eigenvalue of $A$? That's one constructs the jordan normal form to begin with, it is supposed to have the eigenvalues on its entries. – Git Gud Jun 09 '13 at 16:15
  • I noticed you have eight questions asked and no accepted answers. Whenever you get at least a satisfying answer for a question, you should accept your favourite answer. – Git Gud Jun 09 '13 at 16:21

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As others have noted in comments, your problem seems to be with the understanding of the Jordan normal form. If $A$ has a Jordan normal form $J,$ then each entry on the main diagonal of $J$ is an eigenvalue of $J$ and hence also of $A,$ because $J$ is similar to $A.$ The multiplicity of an eigenvalue is slightly more tricky, but still routine, to calculate.

Geoff Robinson
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