Let $A$ be any matrix with all entries positive (which means Perron-Frobenius theorem can be applied). Then its numerical range is defined as the set of complex numbers $$W(A)=\{x^HAx\lvert ~x^Hx=1\}$$ where $x^H$ is the conjugate transpose of complex vector $x$. Now, by Perron-Frobenius theorem, it follows that $A$ has a eigenvalue which will be strictly greater than all other eigenvalues in the absolute value. Also, it will have a corresponding eigenvector which has all its entries positive. Given this, does the following hold $$\max_{x^Hx=1}|x^HAx|\,=\,\max_{u^Tu=1,u> 0}u^TAu$$ where $u>0$ implies all entries of $u$ are positive and $u^T$ is the transpose of $u$.

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    $\begingroup$ That's true: see Theorem 3.1 in Maroulas, Psarrakos,Tsatsomeros , Perron-Frobenius type results on the numerical range, Linear Algebra Apps, 2002. $\endgroup$ – user75485 Sep 23 '15 at 10:33
  • $\begingroup$ @Josep also, does it mean that the solution of the RHS is given by the leading eigenvector? $\endgroup$ – dineshdileep Oct 1 '15 at 7:40
  • $\begingroup$ The solution is the leading egenvector of $(A+A^*)/2.$ $\endgroup$ – user75485 Oct 6 '15 at 7:41

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