How can the Perron-Frobenius theorem be used to show that for a connected graph, there is a simple eigenvector that is (i) real and (ii) smallest in magnitude and (iii) has an associated eigenvector that is positive?
The graph Laplacian is given as $L = D-A$, where $A$ is the non-negative adjacency matrix of the graph. The Perron-Frobenius theorem allows us to state that
$\rho(A) > 0$ and is a simple eigenvalue of $A$
$Ax = \rho(A)x$ with all elements of $x$ positive.
The matrix $D$ is diagonal with positive elements. It is well-known that for a connected graph, 0 is the smallest eigenvalue of $L$ and it is simple, and $x=\mathbb{1}$ (vector of all ones) is the associated positive eigenvector.
I am confused mainly because $L$ is no longer a non-negative (or non-positive) matrix. Any ideas?