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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?$

Thank you in advance.

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Hint: Consider $\|A-I\|_{\infty}$, where $\|\cdot\|_{\infty}$ denotes the operator norm induced by the max-norm.

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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.

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