Is Spectral Radius Equal to Operator Norm for Positive Valued Matrix?

Question

For any real-valued square matrix with all positive entries, by Perron-Frobenius theory, we have that the matrix has a dominant eigenvalue that is real, positive, and of multiplicity 1. Thus, the spectral radius is equal to the largest eigenvalue.

Q: Is the operator norm for such a matrix always equal to the spectral radius, or can it be strictly larger?

Answer 1

The operator norm of a positive matrix can be larger than the spectral radius. E.g. when $A=uv^T$ for some two non-parallel positive vectors $u$ and $v$, we have $$ \|A\|_2=\|u\|_2\|v\|_2>v^Tu=\rho(A). $$ We do know a few things about the relationship between $\|A\|_2$ and $\rho(A)$:

1. For any complex square matrix $A$, we have $\rho(A)\le\|A\|_2$. In fact, $\rho(A)$ is the infimum of all submultiplicative matrix norms of $A$. This is usually proved alongside Gelfand's formula.
2. A complex square matrix $A$ is said to be radial when $\rho(A)=\|A\|_2$. There is a complete characterisation of radial matrices.
3. If $A$ is column/row stochastic, then $\rho(A)=\|A\|_2$ if and only if $A$ is doubly stochastic. (See Characterize stochastic matrices such that max singular value is less or equal one.) The analogous holds for scalar multiples of stochastic matrices.
4. In general, for any complex square matrix $A$, all its eigenvalues of maximum moduli are semi-simple if and only if $\rho(A)=\|A\|$ for some submultiplicative matrix norm $\|\cdot\|$. (See here for a proof.) In your case, while it is not necessarily true that $A$ is radial, since $\rho(A)$ is a dominant simple eigenvalue, we do know that $\rho(A)=\|A\|$ for some matrix norm.