= Solution
Suppose a <feature map> $\phi:\mathbb R^d\to\mathbb R^p$ represented the <Gaussian kernel>. Choose $n>p$, a unit vector $e$, and points $x_i=iae$ for $i=0,\ldots,n-1$. Their <kernel matrix> is
$$
K_{ij}=\exp\!\left(-\frac{a^2(i-j)^2}{2\sigma^2}\right)=r^{(i-j)^2},
\qquad r=e^{-a^2/(2\sigma^2)}.
$$
For sufficiently large $a$, hence sufficiently small $r$, every row satisfies
$$
\sum_{j\ne i}|K_{ij}|\leq2\sum_{m\geq1}r^{m^2}<1=K_{ii}.
$$
Thus $K$ is a symmetric <strictly diagonally dominant matrix> with positive diagonal and is therefore a <positive-definite matrix>, so $\operatorname{rank}K=n$.
On the other hand, if $\Phi$ is the $n\times p$ matrix whose $i$th row is $\phi(x_i)^T$, then $K=\Phi\Phi^T$ and $\operatorname{rank}K\leq p<n$, a contradiction. Hence every feature-space realization of the Gaussian kernel requires an <infinite-dimensional vector space>.
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