= Gershgorin bounds for symmetric matrices
{c}
{title2=$\min_i(a_{ii}-r_i)\leq\lambda\leq\max_i(a_{ii}+r_i)$}
For a real <symmetric matrix>, or more generally a <Hermitian matrix>, the <finite-dimensional spectral theorem> makes all <eigenvalues> real. Intersecting its <Gershgorin discs> with the real axis gives
$$
\lambda\in\bigcup_{i=1}^n[a_{ii}-r_i,a_{ii}+r_i],\qquad \min_i(a_{ii}-r_i)\leq\lambda\leq\max_i(a_{ii}+r_i).
$$
The union preserves gaps that the single outer <interval> loses. If every $a_{ii}-r_i>0$, all <eigenvalues> are positive, so the <matrix> is a <positive-definite matrix>. This is the positive-diagonal <strictly diagonally dominant matrix> criterion. For the three-by-three example, $1.7\leq\lambda\leq9.4$ gives a <positive-definite matrix> certificate and bounds its <spectral condition number of a positive-definite matrix> by $9.4/1.7\approx5.529412$; the computed value is about $4.544364$.
Strictness matters. The <symmetric matrix> $\left(\begin{smallmatrix}1&-1\\-1&1\end{smallmatrix}\right)$ has positive diagonal and is a <diagonally dominant matrix>, but its <eigenvalues> are $0$ and $2$: it is singular. For a general complex <matrix>, the condition $\operatorname{Re}a_{ii}>r_i$ only puts the <eigenvalues> in the open <right half-plane>. The <Hermitian matrix> assumption is what converts that statement into positive definiteness.
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