Let $\mathcal{T}$ be a finite nonempty set of $3$-element subsets of a totally ordered set $V$. We view $\mathcal{T}$ as the set of triangles in the support graph. Let $δ_{1,\mathcal{T}}$ be the signed edge-triangle incidence matrix, and $λ(\mathcal{T})$ the spectral gap of $δ_{1,\mathcal{T}}^Tδ_{1,\mathcal{T}}.$ Our main results show that large $λ(\mathcal{T})$ forces strong overlap and a large minimum degree in the support graph. In particular, every support edge lies in at least $\lceil λ(\mathcal{T})\rceil-2$ triangles in $\mathcal{T}$ and hence the graph has minimum degree at least $\lceil λ(\mathcal{T})\rceil-1$. We further prove that $\binom{n}{3}$ is the exact threshold for attaining level $n:$ if $|\mathcal{T}|< \binom{n}{3}$, then $λ(\mathcal{T}) \leq n-1,$ while if $|\mathcal{T}|=\binom{n}{3}$ and $λ(\mathcal{T}) > n-1,$ then $\mathcal{T}$ is exactly the full set of triangles on an $n$-vertex clique. Moreover, this clique peak is isolated in a strong interval-scale sense: letting $φ(t)=\max_{|\mathcal{T}|=t} λ(\mathcal{T})$, immediately above $\binom{n}{3}$ there is a forbidden interval on which $φ(t) \leq n-1$, and the first passage above the level $n-1$ is delayed by $Θ(n^2)$ additional triangles. Since $\binom{n+1}{3} - \binom{n}{3}=Θ(n^2),$ this implies that after the peak at $\binom{n}{3}$ one must traverse a nonzero proportion of the full gap until the next clique threshold before substantial recovery can occur. In particular, $φ$ is not monotone. However, $φ(t)=Θ(t^{\frac{1}{3}}).$ Finally, if $Λ(t):=\max_{1 \leq s \leq t}φ(s),$ then $Λ(t)=\max\{n \in \mathbb{N}:\binom{n}{3} \leq t\}.$ Thus complete triple systems are the unique minimal spectral extremizers, but their peaks are isolated on the natural scale between consecutive clique thresholds.