TITLE: The nature of eigenvectors of a given eigenvalue
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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).