The nature of eigenvectors of a given eigenvalue 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?
 A: 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).
A: 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.
