# Prove that for every eigen value $\lambda$ of A , $0 \leq \lambda \leq 2.$

Let A be be $n \times n$ square matrix whose all diagonal entries are 1. Suppose that sum of the abosolute values of each row is less than equal to 2. With this setting , I am looking for the following question:

Question: Prove that for every eigen value $\lambda$ of A , $0 \leq \lambda \leq 2.$

It is easy to see that $|\lambda| \leq 2.$ But how can I show that $\lambda \geq 0?$

Hint: Consider $\|A-I\|_{\infty}$, where $\|\cdot\|_{\infty}$ denotes the operator norm induced by the max-norm.
The Gershgorin circle theorem says that every eigenvalue of a matrix $(a_{ij})_{i,j}$ lies within distance $\sum_{j \ne i} |a_{ij}|$ from $a_{ii}$ for some index $i$. This gives you both inequalities.