= Solution
{c}
For a complex $n\times n$ <matrix> $B=(b_{ij})$, put
$$
R_i=\sum_{j\ne i}|b_{ij}|,\qquad
D_i=\{z\in\mathbb C:|z-b_{ii}|\leq R_i\}.
$$
The <Gershgorin circle theorem> states that
$$
\boxed{\operatorname{spec}(B)\subseteq\bigcup_{i=1}^nD_i.}
$$
To prove it, let $Bv=\lambda v$ with $v\ne0$, and choose $i$ for which $|v_i|=\max_j|v_j|>0$. The $i$th coordinate of the <eigenvalue equation> gives
$$
(\lambda-b_{ii})v_i=\sum_{j\ne i}b_{ij}v_j.
$$
The <triangle inequality>, followed by division by $|v_i|$, gives
$$
|\lambda-b_{ii}|\leq\sum_{j\ne i}|b_{ij}|\frac{|v_j|}{|v_i|}
\leq\sum_{j\ne i}|b_{ij}|=R_i.
$$
Thus $\lambda\in D_i$, proving the assertion for every <eigenvalue>.
The second theorem, <Gershgorin eigenvalue counting>, says that if a union of $k$ indexed <Gershgorin discs> is disjoint from the remaining $n-k$ <Gershgorin discs>, that union contains exactly $k$ <eigenvalues>, counted with <algebraic multiplicity>. In particular,
$$
\boxed{\text{An isolated Gershgorin disc contains exactly one simple eigenvalue.}}
$$
Discs that touch are not disjoint; overlapping <Gershgorin discs> cannot individually be assigned one <eigenvalue> each.
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