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

This question pop'd up when I was studying graph. I am thinking about the relation between principal eigenvector of adjacency matrix $A$ and its inverse $A^{-1}$, do they have any relation?

share|cite|improve this question
up vote 3 down vote accepted

Suppose that $v$ is an eigenvector of $A$ with eigenvalue $\lambda.$ Then $v$ is also an eigenvector of $A^{-1}$, but with eigenvalue $1/\lambda:$

$$v = 1.v = (A^{-1}A)v = A^{-1}(\lambda v) = \lambda (A^{-1}v) \quad \Rightarrow \quad A^{-1}v = \lambda^{-1} v.$$

Now let $\{ \lambda_i \}_{i=1}^n$ be the spectrum of $A$, and let the $\lambda_i$ be ordered: $\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_n.$ What can you say about the spectrum of $A^{-1}$?

share|cite|improve this answer
This is all true, but I don't think it says anything about the eigenvectors - does it? – Chris Taylor Oct 18 '11 at 10:46
@Chris, Yes it does. The eigenvalues of $A^{-1}$ are ordered in reverse ordered, but they correspond to the same eigenvectors. So, .... – Tapu Oct 18 '11 at 11:33
@Chris: to make it explicit, let's suppose that all eigenvalues are strictly positive. If $v_1$ is a principal eigenvector of $A$, it corresponds to smallest (positive) eigenvalue of $A^{-1};$ similarly, if $v_n$ corresponds to the smallest (positive) eigenvalue of $A$, it corresponds to the largest eigenvalue of $A^{-1}.$ (Maybe convince yourself with an example.) – Gerben Oct 18 '11 at 16:49
Sure, that much is fine. But knowing the principal eigenvector of $A$ tells you nothing about the principal eigenvector of $A^{-1}$, and that's what is being asked. – Chris Taylor Oct 18 '11 at 23:51
In that case: to know the principal eigenvector of $A^{-1}$, you need to know the 'least important' eigenvector of $A$ - but I guess in statistical applications you rarely have access to this information. – Gerben Oct 19 '11 at 5:01

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.