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: 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?
Best Answer
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.