Regular spectral characterization and notation convention

The row sums of $A$ are $d^{\ast }$, so $A\mathbf{1}=d^{\ast }\mathbf{1}$. Since $A$ is nonnegative, if $Av=\lambda v$ and $\mathrm{\mid }{v}_x\mathrm{\mid }$ is maximal among the coordinates of a nonzero eigenvector, then $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\mid }{v}_x\mathrm{\mid }\le {\sum }_y{A}_{xy}\mathrm{\mid }{v}_y\mathrm{\mid }\le d^{\ast }\mathrm{\mid }{v}_x\mathrm{\mid }$. Thus every eigenvalue has absolute value at most $d^{\ast }$, and the eigenvalue belonging to $\mathbf{1}$ can be placed first in the nonincreasing list. Any additional copies of $d^{\ast }$ remain in the list after that first copy.

If $d^{\ast }>0$ and $n\ge 2$, [F3] gives $\mathrm{\parallel }A-(d^{\ast }\mathrm{/}n)J\mathrm{\parallel }={\max }_{i\mathrm{\ne }1}\mathrm{\mid }{\lambda }_i(A)\mathrm{\mid }$: the constant eigenvector accounts for one occurrence of $d^{\ast }$, and all remaining eigenvalues, including repeated copies of $d^{\ast }$, occur on its orthogonal complement. If $d^{\ast }>0$ and $n=1$, [F3] gives norm zero, agreeing with the stated empty-index convention.

If $d^{\ast }=0$, nonnegativity and zero row sums force every entry of $A$ to be zero. Both the centered norm and every adjacency eigenvalue are therefore zero, including the stipulated empty maximum when $n=1$. In all cases the centered norm equals the stated maximum; substituting this equality into [F2] proves the equivalence.