infinite dimensional positive matrix Suppose $A$ is a infinite dimensional positive complex matrix,what is the operator norm of $A$?
In the finite dimensional case,we can use the spectral theorm,$\|A\|=sup|\lambda_i|$,where $\lambda_i$ is the eigenvalue of $A$.If $A$ is infinite dimensional,can we have a similar conclusion?
 A: I think you need to further specify your problem.
Generally speaking, in a Hilbert space $H$ (since you are asking about positive operators I will put it in a Hilbert space, but some notations also carry over to Banach spaces), if we use the operator norm $\|A\|=\sup_{x\in H-\{0\}}\frac{\|Ax\|}{\|x\|}$, we will have this result: $\|A\|=\sup_{\lambda\in\sigma(A)}\lambda=\sup_{\|x\|=1}\left<x,Ax\right>$, where $\sigma(A)$ is the spectrum of $A$. This is a special case of a more general result: $\|A\|=\sup_{\lambda\in\sigma(A)}|\lambda|$ given $A$ is normal. Then take $A$ to be self-adjoint we know its spectrum is a compact subset of $\mathbb{R}$, and if $A$ is furthermore positive we know $\sup_{\lambda\in\sigma(A)}|\lambda|=\sup_{\lambda\in\sigma(A)}\lambda$.
We have to be careful since in infinite dimensional space the spectrum of an operator has really subtle differences from that of a finite dimensional one, thus the "eigenvalues" (actually we call it the point spectrum $P_\sigma(A)$ of $A$) need to be replaced by the spectrum $\sigma(A)$. If the operator is compact, we can furthermore know the nonzero spectrum $\sigma(A)\backslash\{0\}$ of $A$ is identical to its point spectrum. Every finite rank operator is compact, so this will give you the same result in a finite dimensional Hilbert space.
