2
$\begingroup$

The adjacency matrix $A(G)$ of an infinite undirected graph $G$ is considered as a bounded self-adjoint linear operator $A$ on the Hilbert Space $l^2(G)$ (last section of https://en.wikipedia.org/wiki/Spectral_radius). Then the spectral radius of the $A(G)$ would be the spectral radius of the $A$ (I presume).

The spectral radius of the matrix $A(G)$ by Rayleigh quotient formula $R(A(G),x)=\frac{x^*A(G)x}{x^*x}=\frac{<x,A(G)x>}{<x,x>}$ is $R(A(G),v_{max})=\lambda_{max}$, that is, $\rho(A(G))=\sup_{x\neq0}\frac{<x,A(G)x>}{<x,x>}$.

And the spectral radius of the linear operator by Gelfand's formula is $\rho(A)=\lim_{k\to\infty}\|A^k\|^{\frac1k}$ which converts into $\rho(A)=\|A\|$ for self-adjoint operators because they are normal. The operator norm for the p-norm vectors of $l^2 (G)$ is $\|A\|=\|A\|_2=\sup_{x\neq0}\frac{\|Ax\|_2}{\|x\|_2}$(https://en.wikipedia.org/wiki/Matrix_norm). Which means $\rho(A)=\sup_{x\neq0}\frac{<Ax,Ax>^\frac12}{<x,x>^\frac12}$.

I am confused: why $\rho(A(G))$ and $\rho(A)$ are not equal? I am surely making a mistake or missing something important. Please help.

$\endgroup$
0
$\begingroup$

Your two expressions for the spectral radius are not identical, but their values on (finite) symmetric matrices are equal.

One comment. For an infinite graph of bounded degree, it's perfectly reasonable to view its adjacency matrix as an operator on $\ell^2$, but it's not clear to me that this is even an established convention, let alone a rule.

$\endgroup$
  • $\begingroup$ Shouldn't the both spectral radius be equal? After all, adjacency matrix and the operator are two different ways of seeing the same thing. $\endgroup$ – Crystal Jul 27 '15 at 22:59
0
$\begingroup$

As, $R(A(G),x)=\frac{<x,A(G)x>}{<x,x>}$,

$R(A^2(G),x)=\frac{<x,A^2(G)x>}{<x,x>}=\frac{<A(G)x,A(G)x>}{<x,x>}=\frac{\|A(G)x\|_2^2}{\|x\|_2^2}=\|A(G)\|_2^2$ for $x=v_{max}$.

Also, $\rho(A^2)=\rho(A)^2=\|A\|^2_2$.

So, $R(A^2(G),x)=\rho(A^2)$ for $x=v_{max}$.

If $\mu$ is the spectral radius of $A^2$, then $\sqrt{\mu}$ is the spectral radius of $A$.(similar argument holds for A(G)).

Which means that the both spectral radii are equal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.