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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 givesThe 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, $1.7\le \lambda \le 9.4$ gives a positive-definite matrix certificate and bounds its spectral condition number of a positive-definite matrix by $9.4/1.7\approx 5.529412$; the computed value is about $4.544364$.

Strictness matters. The symmetric matrix $\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$ 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 $\mathrm{Re}{a}_{ii}>r_i$ only puts the eigenvalues in the open right half-plane. The Hermitian matrix assumption is what converts that statement into positive definiteness.

Suppose a feature map $\varphi :{\mathbb{R}}^d\to {\mathbb{R}}^p$ represented the Gaussian kernel. Choose $n>p$, a unit vector $e$, and points $x_i=iae$ for $i=0,\dots ,n-1$. Their kernel matrix isFor sufficiently large $a$, hence sufficiently small $r$, every row satisfiesThus $K$ is a symmetric strictly diagonally dominant matrix with positive diagonal and is therefore a positive-definite matrix, so $\mathrm{rank}K=n$.

On the other hand, if $\Phi$ is the $n\times p$ matrix whose $i$th row is $\varphi (x_i{)}^T$, then $K=\Phi {\Phi }^T$ and $\mathrm{rank}K\le p<n$, a contradiction. Hence every feature-space realization of the Gaussian kernel requires an infinite-dimensional vector space.