Is spectral radius = operator norm for a positive valued matrix? 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?

 A: A few comments about the general situation:
The spectral radius is not a norm on the space of all $n \times n$ matrices, for $n>1$. This is because there exist nonzero matrices with zero spectral radius, e.g. 
$$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.$$
However, given a matrix $A$ with spectral radius $\rho(A)$, for every $\varepsilon > 0$ there exists an operator norm $\| \cdot \|$ such that $\| A \| \leq \rho(A) + \varepsilon$.
At the same time, if a matrix is self-adjoint with respect to an inner product, then its operator norm with respect to this inner product and its spectral radius correspond.
In view of the preceding comments, to find a positive matrix whose Euclidean operator norm is larger than its spectral radius, you should look for something which is very asymmetric, like
$$\begin{bmatrix} 1 & 1 \\ \varepsilon & 1 \end{bmatrix}$$
where $0<\varepsilon \ll 1$. This has trace $2$ and determinant $1-\varepsilon$, so the eigenvalues are $1+\sqrt{\varepsilon}$ and $1-\sqrt{\varepsilon}$. On the other hand the Euclidean operator norm is at least $\sqrt{2}$ independent of $\varepsilon$, as you can see by multiplying this matrix with the unit vector $e_2$. So this gives an example whenever $\sqrt{\varepsilon}<\sqrt{2}-1$; concretely, you can take $\varepsilon=0.1$.
A: 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 $\rho(A)$:


*

*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.

*A complex square matrix $A$ is said to be radial when $\rho(A)=\|A\|_2$. There is a complete characterisation of radial matrices.

*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.

*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.

