In the background of this question is a matrix $A$, all of whose elements are positive. The Perron-Frobenius theorem implies 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 be said about the dominant eigenvector?

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?