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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?

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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.
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  • $\begingroup$ Thanks for the answer as well as for the related references! $\endgroup$ – Hedonist Sep 6 '15 at 20:12
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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 $\varepsilon=0.1$.

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