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Is it true that 2 matrices are similar if and only if they have the same Jordan form? I know that one direction is correct: if have the same Jordan form -> similar. Is the other direction correct - similar -> same Jordan form?

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Yes.${}{}{}{}{}$ –  Andres Caicedo Mar 16 at 16:30
    
Can you explain please? –  CnR Mar 16 at 16:30
    
Can you prove that similar matrices have the same eigenvalues? Not by using determinants to compute the characteristic polynomial, but by arguing directly in terms of eigenvectors and how similarity acts. If you see how to do this, the general result is just an extension of these ideas. –  Andres Caicedo Mar 16 at 16:33

1 Answer 1

If $A$ and $B$ are similar, say$$ A = QBQ^{-1} $$

The Jordan form of $B$ is $$ B = PJ_BP^{-1}\\ A = QPJ_BP^{-1}Q^{-1} = (QP)J_B(QP)^{-1} $$so $A$ and $B$ have the same Jordan form.

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