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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 1, with 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?

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    $\begingroup$ You can use the implicit function theorem to show that the set of all vectors for which solutions exist is open, provided n is larger than 2. $\endgroup$ – user75463 Jan 15 '14 at 1:56
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Here is a preliminary partial result. Let $s=\sum v_j$ be the sum of the entries in $v$. Then for every $i$, since $A$ has zeros on the diagonal, $$v_i=\sum_{j} a_{ij} v_j=\sum_{i\neq j} a_{ij} v_j\leq \sum_{i\neq j} v_j =s-v_i.$$ From this we conclude that $v_i\leq s/2$, so that no individual entry makes up more than half the sum. So there are lots of "impossible" vectors.

In particular, in the $2\times 2$ case the only possibility is $v=(s/2,s/2)$.

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  • $\begingroup$ shouldn't the first sum be over j not i? $\endgroup$ – gabe Jan 15 '14 at 18:06
  • $\begingroup$ Yep, thanks!!!! $\endgroup$ – user940 Jan 15 '14 at 18:25

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