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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?

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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 $A^{-1}$?

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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
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