Proving Similarity of Non-Diagonalizable Matrices

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SUMMARY

The discussion centers on proving the similarity of a non-diagonalizable 2x2 matrix A to the Jordan form matrix L = [[Ω, 1], [0, Ω]]. Participants establish that if A cannot be diagonalized, it must have a repeated eigenvalue, which aligns with the eigenvalue structure of L. The proof hinges on demonstrating that A and L share the same eigenvalues and that there exists an invertible matrix P such that P-1AP = L. The conversation also touches on the implications of Jordan normal form and the necessity of generalized eigenvectors in constructing the proof.

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  • Concept of similarity transformations in linear algebra
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  • #31
Sorry, it's quiz time right now! :wink:

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  • #32
Lily@pie said:
But having (A-ΩI)w=0 means w is an eigenvector for A. But we know it cannot be 2 distinct eigenvectors, so w can only be a multiple of v. Which shows that v and w are linearly dependent.
Right. This is why you can't have (A-ΩI)w=0.
But we would hope to show v and w are linearly independent.

(A-ΩI)w≠0 mean (A-ΩI)w is an eigenvector for A?? So w and v are linearly independent??
don't understand ><
The only way to satisfy (A-ΩI)2w=0 is therefore (A-ΩI)w=v. Note that you've already shown that w can't be a multiple of v; therefore, w and v are independent.
Besides that writting P-1(A-ΩI)P where P=[v w] means writing the eigenvalues in the diagonal entry??
Among other things, yes, but that's not the whole story. You want to think about how you find the columns of a matrix for a linear transformation relative to the basis {v, w}.
 
  • #33
Thank you both for visiting the quiz! :)

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I won the quiz! YAY! :smile: :smile: :smile:

(Last time I finished last! :cry:)
 
  • #34
oh... okay... thanks so much for the help... ^^
 

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