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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.

Robert S. Barnes
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    What is the question? – Qiaochu Yuan Jun 24 '12 at 06:39
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    I guess his question is if there is a relation between the eigenvectors of two similar matrices. –  Jun 24 '12 at 06:45
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    So the relation is not between the eigenvectors of $A$ and the eigenvectors of the diagonalization of $A$, it's between the eigenvectors of $A$ and the matrix that is used to perform the diagonalization of $A$. – Robert Israel Jun 24 '12 at 06:52
  • @Marvis Indeed, that is basically what I'm asking... – Robert S. Barnes Jun 24 '12 at 07:30
  • The relation between the eigenvectors of two similar matrices is given in my answer, isn't it? If $A$ has eigenvector $v$, and $B$ is similar to $A$, then $B$ has eigenvector $w=C^{-1}v$, where $C$ is the matrix that makes $A$ and $B$ similar --- $B=C^{-1}AC$. – Gerry Myerson Jun 25 '12 at 07:05
  • @GerryMyerson Yes of course, the equation is very clear. – Robert S. Barnes Jun 25 '12 at 09:41

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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 $B$.

Gerry Myerson
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  • I understand the equation itself, but not so much the reasoning behind it. Why is $C^{-1}v$ an eigenvector of B? – Robert S. Barnes Jun 25 '12 at 07:09
  • Because, as the displayed equations show, when you multiply $C^{-1}v$ by $B$, you get the scalar $a$ times $C^{-1}v$. That's what being an eigenvector of $B$ means, right? – Gerry Myerson Jun 25 '12 at 13:03
  • OK, I think I figured out what was bugging me. $A$ and $B$ both represent the same linear transformation $T$ according to different bases, say $A=[T]{B_1}$ and $B=[T]{B_2}$. $C$ is the change of basis matrix from $B_1$ to $B_2$. So each of $B$'s eigenvectors is a linear combination of the columns of the change of basis matrix with the combination being given by an eigenvector of $A$. So each eigenvector of $B$ is the coordinate vector of an eigenvector of $A$ according to the base $B_1$. $A$ and $B$ have the same eigenvectors in different bases. Does that sound right? – Robert S. Barnes Jun 25 '12 at 20:22
  • Everything sounded right, except I don't follow the very last thing. $A$ and $B4 don't have the same eigenvectors. I don't know what "the same eigenvectors in different bases" means to you, but whatever it means to you I don't think it's a good way to express it. Still, it sounds like you have a good idea of what's going on. – Gerry Myerson Jun 26 '12 at 06:27
  • 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 $[M]^{B_2}{B_1}[v]{B_2} = [v]{B_1} = w$ – Robert S. Barnes Jun 26 '12 at 06:56