= Solution
The <Gershgorin circle theorem> states that every <eigenvalue> $\lambda$ of a complex matrix $A=(a_{ij})$ lies in at least one disk
$$
\boxed{
|\lambda-a_{ii}|
\leq\sum_{j\ne i}|a_{ij}|}.
$$
To prove it, choose a nonzero <eigenvector> $x$ for $\lambda$ and an index $i$ such that $|x_i|=\max_j|x_j|>0$. The $i$th component of $Ax=\lambda x$ gives
$$
(\lambda-a_{ii})x_i
=\sum_{j\ne i}a_{ij}x_j.
$$
Taking absolute values and using $|x_j|\leq|x_i|$ yields
$$
|\lambda-a_{ii}|\,|x_i|
\leq\sum_{j\ne i}|a_{ij}|\,|x_j|
\leq |x_i|\sum_{j\ne i}|a_{ij}|.
$$
Division by $|x_i|$ proves the theorem.
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