# Spectral Radius vs Row/Col sum

Let $A=(a_{ij})$ be an irreducible $n\times n$ matrix of nonnegative real numbers with spectral radius $\rho(A)$, namely the largest modulus of the eigenvalues of $A$. By the Perron-Frobenius theorem [1], $\rho(A)>0$ is the largest eigenvalue of $A$, such that the associated eigenvector $v$ has strictly positive entries and that any nonnegative eigenvectors is a multiple of $v$.

## Row/Column-sum bound on spectral radius
If $A$ is nonnegative, then
$$\min_{i} \sum_{j=1}^n A_{ij} \leq \rho(A) \leq \max_{i} \sum_{j=1}^n A_{ij}$$ and $$\min_{j} \sum_{i=1}^n A_{ij} \leq \rho(A) \leq \max_{j} \sum_{i=1}^n A_{ij}.$$

### Proof
It suffices to consider the case $A>0$ (and hence irreducible), as $\rho(A)$ is a polynomial (and hence continuous function) in the entries of $A$. Also we only need to prove the first formula as eigenvalues are invariant under transpose.

Let $v>0$ be the eigenvector associated with $\rho(A)$. Then for any $i=1,...,n$ we have $\sum_{j} A_{ij}v_j = \rho(A)v_i$. Taking $i$ to be the largest and smallest component of $v$ finishes the proof.

# References
[1] [The Perron-Frobenius theorem](https://www.math.miami.edu/~armstrong/685fa12/sternberg_perron_frobenius.pdf)