Let $A$ be an irreducible non-negative matrix. Is it true that the eigenvectors of $A$ can span the $R^n$ ?


Or are all the eigenvalues of $A$ distinct?



The dimension of the eigenvector space of non-negative irreducible matrices
You don't need to search for complicated counterexamples; just consider the matrix with all elements equal to 1.


[EDIT: removed a second counterexample after a comment pointed out it was reducible. If you want an example with all distinct eigenvalues, you can take the cyclic shift matrix.]


