Vrftjkry

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?

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 $|A|=sup_{xin H-{0}}frac{|Ax|}{|x|}$, we will have this result: $|A|=sup_{lambdainsigma(A)}lambda=sup_{|x|=1}left<x,Axright>$, where $sigma(A)$ is the spectrum of $A$. This is a special case of a more general result: $|A|=sup_{lambdainsigma(A)}|lambda|$ 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 $sup_{lambdainsigma(A)}|lambda|=sup_{lambdainsigma(A)}lambda$.

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.