Proving Similarity of Non-Diagonalizable Matrices

[I like Serena]

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[vela]

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}.

 

[I like Serena]

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[Lily@pie]

oh... okay... thanks so much for the help... ^^