Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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

share|cite|improve this question
What is the question? – Qiaochu Yuan Jun 24 '12 at 6:39
I guess his question is if there is a relation between the eigenvectors of two similar matrices. – user17762 Jun 24 '12 at 6:45
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 6:52
@Marvis Indeed, that is basically what I'm asking... – Robert S. Barnes Jun 24 '12 at 7:30
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 7:05
up vote 1 down vote accepted

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 $B$.

share|cite|improve this answer
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 7:09
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 6:27
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 6:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.