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I am working on a problem where I have an ($n \times n $) matrix A and an eigenvalue of A, $\lambda$, where $\lambda$ has geometric multiplicity 1. The right and left eigenvectors of A corresponding to $\lambda$ are component-wise positive. How can I show that there are no other component-wise non-negative eigenvectors?, with the exception of scalar multiples of these?

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That's not true in general. It is true in some cases: see en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem . –  Qiaochu Yuan Sep 15 '11 at 17:55
    
Can I ask something really silly? Doesn't geometric multiplicity $1$ mean that the dimension of the eigenspace is $1$ and hence if you find any eigenvector, then this is a basis for the eigenspace and hence all other eigenvectors are scalar multiples? So if you throw out multiples, then there are no other eigenvectors at all, you don't have to worry about what the components are. –  Matt Sep 15 '11 at 18:10
    
Oh. Nevermind. I see. You don't mean other eigenvectors associated to the eigenvalue $\lambda$, you mean other eigenvectors in general. Sorry. –  Matt Sep 15 '11 at 18:14
    
@Matt: which are you asking about? –  Qiaochu Yuan Sep 15 '11 at 18:47
    
@Qiaochu: I figured it out. I was just misreading the question. –  Matt Sep 15 '11 at 20:04

2 Answers 2

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Start by observing that eigenvectors corresponding to different eigenvalues left eigen-vector corresponding to one eigenvalue is orthogonal to right-eigenvector corresponding to a different eigenvalue should be orthogonal.

Let $u_{\lambda}$ denote right eigenvector, i.e. $A u_{\lambda} = \lambda u_{\lambda}$, and $v_{\lambda}$ denote left eigenvector, i.e. $v_{\lambda} A = \lambda v_{\lambda}$.

Let $\mu \not= \lambda$ be an eigenvalue of $A$, then $A u_\mu = \mu u_\mu$ and $v_{\mu} A = \mu v_{\mu}$. Now $$\lambda v_\lambda u_\mu = v_{\lambda} A u_\mu = \mu v_\lambda u_\mu$$ hence $(\lambda - \mu) v_\lambda u_\mu = 0$, hence $v_\lambda u_\mu = 0$, because $\mu \not= \lambda$. Similarly $u_\lambda v_\mu = 0$.

Because $u_\lambda$ and $v_\lambda$ are positive component-wise, the only way the dot product can be zero if $v_\mu$ and $u_\mu$ have negative components, since they are not identically zero as eigenvectors.

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I don't understand what you're saying. Are you assuming $A$ is symmetric? –  Qiaochu Yuan Sep 15 '11 at 18:47
    
Eigenvectors corresponding to different eigenvalues are not orthogonal, in general. However, the rest of your argument does not rely on this fact. It's a bit confusing. –  Calle Sep 15 '11 at 19:35
    
No, I am not assuming $A$ is symmetric. What I am saying is identical to Robert's answer, except that I write $v_\lambda A u_\mu$, while he writes $v_\lambda^T A u_\mu$. –  Sasha Sep 15 '11 at 21:39
    
I see, you mean that you're showing that "eigenvectors corresponding to different eigenvalues should be orthogonal". Well, still, you haven't show that - you have shown that the left eigenvectors are orthogonal to the right eigenvectors when they correspond to different eigenvalues. –  Calle Sep 16 '11 at 0:13
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@Calle Good point, thanks! I will correct the statement –  Sasha Sep 16 '11 at 0:18

If $w^T$ is a left eigenvector and $v$ a right eigenvector of the matrix $A$ for eigenvalues $\lambda$ and $\mu$ respectively, then $\lambda w^T v = w^T A v = \mu w^T v$. So if $\lambda \ne \mu$ we must have $w^T v = 0$. But if all $w_i > 0$ and $v_i \ge 0$, the only way that can happen is if $v = 0$ (which isn't allowed for an eigenvector).

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