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I'm trying to understand the relation (if any) between the eigenvectors of similar matrices and in particular of a matrix and its diagonalization.

Given $A,D\in M^F_{n\times n}$ and invertible $P$ such that $P^{-1}AP=D$ then $AP=PD$ and the eigenvectors of A are the columns of $P$ because $AP_i=\lambda_iP_i$ and $P$ is a change of basis matrix from whatever basis $A$ is in to whatever basis $PD$ is in. $D$ itself is obviously diagonalizable, and its eigenvectors are the columns of $I$ which won't equal $P$ unless $A=D$. And that's as far as I can get at the moment.

EDIT

As Marvis noted, the heart of the question is what is the relationship ( if any ) between the eigenvectors of two similar matrices.

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    @GerryMyerson Yes of course, the equation is very clear.2012-06-25

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If $A$ has eigenvector $v$ with eigenvalue $a$, and $A$ is similar to $B$, say $B=C^{-1}AC$, then let $w=C^{-1}v$; then $Bw=C^{-1}ACC^{-1}v=C^{-1}Av=C^{-1}av=aC^{-1}v=aw$ so $w$ is an eigenvector of $B$.

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    I mean you basically get an eigenvector $w$ of $B$ from an eigenvector $v$ of $A$ by applying the change of basis matrix $C^{-1}$ to $v$. In other words, if $C^{-1}=[M]^{B_2}_{B_1}$ then $[M]^{B_2}_{B_1}[v]_{B_2} = [v]_{B_1} = w$2012-06-26