This is a sort of reverse eigenvector problem. Usually, given a matrix, we want to describe its eigenvalues. Here -- given a vector, we'd like to determine matrices (satisfying some conditions) for which the vector is invariant.
More precisely :
Given a positive vector
v, when does there exist a non-negative square matrix
A whose columns sum to
0's on the main diagonal, so that
Av = v ? Always ? (if so, an algorithm would be nice). If
v is a constant vector, then there are clearly available
A's to choose from. But if
v is not constant, I think there might be no such matrices. Certainly if
v is not positive, then (1,0,0,0,0,0) will never be invariant for such a matrix. I'm not sure under what conditions a positive vector will have solutions, though.
For example, provided
A is also irreducible (a fine assumption in my case), then the Perron-Frobenius eigenvalue will have only one eigenvector (up to a scalar). Is it possible to find a positive vector
v so that no irreducible non-negative matrix has
v as it's Perron-Frobenius eigenvector?