In the background of this question is a matrix $A$, all of whose elements are positive. The Perron-Frobenius theorem tells us that the eigenvalue with largest absolute value is real, and that there is an associated *dominant eigenvector*, all of whose elements are positive.


Suppose we don't actually observe $A$, but are told what its first row sum is. We're also told the first row sum of $A^{2}$, $A^{3}$, $A^{4}$, ... . In other words, writing $e$ for the vector of ones, we're told the first element of $Ae$, $A^{2}e$, $A^{3}e$, and so on. If, for example, $A$ is a stochastic matrix then $Ae = e$ so that the information given is simply a list of ones: $(Ae)\_{1}=1$, $(A^{2}e)\_{1}=1$, etc.


This information is enough to work out the dominant eigenvalue of $A$ via the [power method](http://en.wikipedia.org/wiki/Power_method): simply compute $\lim\_{n \to \infty} \left( (A^{n} e)\_{1} \right)^{1/n}$.


My question is:



> 
> Can anything at all be said about the dominant eigenvector?
> 
> 
> 



Calculating the Perron-Frobenius eigenvector of a positive matrix from limited information
The information you have does not determine the dominant eigenvector.


Let $G$ be the graph with vertex set $\{0,1,\ldots,7\}$ and adjacency matrix
$$
\left(\begin{array}{rrrrrrrr}
0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
\end{array}\right)
$$
Construct a second graph $H\_2$ by joining a new vertex to the vertex 2, and a third
graph $H5$ by joining a new vertex to vertex 5. Then
$$
 (A(H\_2)^k e)\_2 = (A(H\_5)^k e)\_5
$$
for all $k$. (For $k=0,\ldots,8$ the actual numbers are 
$$
1,\\ 4,\\ 8,\\ 25,\\ 57,\\ 163,\\ 392,\\ 1073,\\ 2656)
$$
The Perron vectors are
$$
(1, 1, 1.579071, 1.460275, 1.019079, 1.168003, 0.5330099, 0.206667, 0.612263)
$$
for $H\_2$ and, for $H\_5$,
$$
(1, 1, 1.579071, 2.0725388, 1.631342, 2.134811, 0.974205, 0.377735, 0.827744)
$$


If you want positive matrices, take the sixth powers of $A(H\_2)$ and $A(H\_5)$. The relevant
property of $G$ is that the graphs $G\setminus2$ and $G\setminus5$ are cospectral, and their
complements are cospectral too.


All computations carried out in sage.


In a sense the problem is that you are getting a bit of information about each eigenspace,
whereas you want detailed information about a particular eigenspace.


