Let $\mathbf{A}$ be any $n\times n$ symmetric positive matrix ($A\_{ij}>0$). It is easy to show that the solution to the following optimization problem 


\begin{align}
\max\_{\mathbf{x}}~~\mathbf{x^TAx}\,\,;s.t.~~\mathbf{x}\geq \mathbf{0},~~\|\mathbf{x}\|\_2=1
\end{align}


is given by the so-called Perron vector of $\mathbf{A}$, which will the eigenvector corresponding to the largest eigenvalue (known as the Perron root). It will also turn out that the Perron vector has all its entries as positive. I need to see if Perron vector is still a solution if I replace the $2$-norm constraint with the $1$-norm constraint. I need to know if the Perron vector will be a solution to 


\begin{align}
\max\_{\mathbf{x}}~~\mathbf{x^TAx}\,\,;s.t.~~\mathbf{x}\geq \mathbf{0},~~\|\mathbf{x}\|\_1=1
\end{align}


If it is not, how badly does it miss out? Is there any research on this? This comes from an engineering problem I am working on.



Does the Perron vector maximize $x^TAx$ in the simplex?
No. The Perron vector is in general very far from optimizing the quantity you're looking at. 
Here is an example: 
Let $A$ be the $n\times n$ tridiagonal matrix with $\frac 13$ on the diagonal and the off-diagonals as well as in the $(1,n)$ and $(n,1)$ entries. (I think of this as a Markov transition matrix on a ring of size $n$ with probability $\frac 13$ of staying still or moving left or right). 


The Perron vector of norm 1 is the vector $x=(\frac 1n,\ldots,\frac 1n)^T$. This is an eigenvector with eigenvalue 1, so that $x^TAx=1/n$. On the other hand, if $y$ is the coordinate vector $(1,0,\ldots,0)^T$, then $y^TAy=\frac 13$ so the Perron vector is nowhere close to optimizing $x^TAx$ over the $\ell^1$ unit ball.


