Skip to main content
MathOverflow

Return to Question

Aperiodicity
Source Link
A. Batsis
A. Batsis
  • 21
  • 3

Let $A$ be a non-negative (entrywise) matrix andsuch that $u$ a non-negative vector$A(1,1)>0$. Set $u=(1,0,0,...,0)^T$. Is it always true that there exists a non-negative eigenvector $v$ of $A$ such that $\lim_{n\rightarrow\infty}\frac{A^nu}{||A^nu||_1}=\frac{v}{||v||_1}$?

Let $A$ be a non-negative (entrywise) matrix and $u$ a non-negative vector. Is it always true that there exists a non-negative eigenvector $v$ of $A$ such that $\lim_{n\rightarrow\infty}\frac{A^nu}{||A^nu||_1}=\frac{v}{||v||_1}$?

Let $A$ be a non-negative (entrywise) matrix such that $A(1,1)>0$. Set $u=(1,0,0,...,0)^T$. Is it always true that there exists a non-negative eigenvector $v$ of $A$ such that $\lim_{n\rightarrow\infty}\frac{A^nu}{||A^nu||_1}=\frac{v}{||v||_1}$?

Source Link
A. Batsis
A. Batsis
  • 21
  • 3

Matrix iteration for non-negative matrices. Does it converge to some eigenvector?

Let $A$ be a non-negative (entrywise) matrix and $u$ a non-negative vector. Is it always true that there exists a non-negative eigenvector $v$ of $A$ such that $\lim_{n\rightarrow\infty}\frac{A^nu}{||A^nu||_1}=\frac{v}{||v||_1}$?