Finding eigenvectors of similar matrices

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Discussion Overview

The discussion revolves around finding eigenvectors of similar matrices, specifically how to derive an eigenvector of the matrix PAP-1 from an eigenvector of matrix A using an invertible matrix P. The scope includes theoretical understanding and conceptual clarification of matrix similarity and eigenvalues.

Discussion Character

  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant inquires about the method to find an eigenvector w of PAP-1 given an eigenvector v of matrix A, suggesting that similarity might be relevant.
  • Another participant proposes using the transformation w = Pv, asserting that PAP-1(Pv) = PAv = Pλv = λPv, indicating that Pv is an eigenvector of PAP-1.
  • A participant expresses a sentiment about the simplicity of the problem once understood, indicating a level of frustration with the apparent ease of the solution.
  • One participant provides an intuitive explanation, suggesting that an invertible matrix P represents a "symmetry" of Rn and that conjugation by P reflects this symmetry in the context of nxn matrices.
  • Another participant adds that two matrices are similar if they represent the same linear operator in different bases, emphasizing that they share the same eigenvalues despite differing eigenvectors due to the change of basis.

Areas of Agreement / Disagreement

Participants generally agree on the relationship between eigenvectors and similar matrices, with multiple perspectives on the interpretation of similarity and its implications for eigenvalues and eigenvectors. No consensus is reached on a singular method or deeper implications beyond the stated relationships.

Contextual Notes

The discussion does not delve into specific mathematical proofs or assumptions regarding the properties of eigenvectors and matrices, leaving some aspects of the transformation and implications of similarity unexamined.

Alupsaiu
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If v is in Rn and is an eigenvector of matrix A, and P is an invertible matrix, how would you go about finding an eigenvector w of PAP-1?
I'm thinking you have to use a fact about similarity?
 
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Just use Pv. Then PAP-1(Pv) = PAv = P [itex]\lambda[/itex] v = [itex]\lambda[/itex] Pv.
 
Man I hate how painfully simple problems like these are when you see them done haha...thanks a bunch
 
No problem. To give some intuition for this, you should think of an invertible matrix P as giving a "symmetry" of Rn. Then conjugation by P is the corresponding symmetry of the space of nxn matrices. That's really why people care about similar matrices. If two matrices are similar, then there is some "symmetry" that can transform one matrix into the other.
 
Another way to think about it: two matrices are similar if and only if they represent the same linear operator, written in different bases (P is the "change of basis" matrix). They necessarily have the same eigenvalues since changing bases would only change vectors, not scalars.
 

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