Worked solution: Marcus–Spielman–Srivastava proof via interlacing families (2013)

Kadison and Singer's 1959 question is about states: a pure state on the diagonal algebra $D \cong \ell^\infty$ inside $B(\ell^2)$ is determined by an ultrafilter on $\mathbb{N}$, and the question was whether its extension by Hahn–Banach to a pure state on all of $B(\ell^2)$ is forced to be unique. The hypothesis $\sum_i v_i v_i^* = I_d$ in Weaver's $KS_2$ is the finite-dimensional shadow of a tight frame decomposition of the identity, and $\|v_i\|^2 \le \delta$ is the smallness condition that replaces "the diagonal is diffuse": no single vector carries a macroscopic share of $I_d$.

Weaver's 2004 theorem (Discrete Mathematics 278) reduces the operator-algebra problem to a countable family of finite-dimensional paving statements, of which $KS_2$ is the sharp form: partition the index set into $S_1, S_2$ so that both operator norms $\|\sum_{i \in S_j} v_i v_i^*\|$ stay strictly below $1$. The constant matters quantitatively, not just qualitatively — a bound of $(1+\sqrt{2\delta})^2/2$ is what the eventual proof delivers, and it is $< 1$ precisely because $\|v_i\|^2 \le \delta$ lets $\delta \to 0$ when one needs it (by first subdividing the frame).

The rest of the proof never touches $\ell^2$ or $B(\ell^2)$ again. Every step operates on finitely many vectors $v_i$ and rank-one matrices $v_i v_i^*$: this is why a polynomial-method argument can settle a problem that began life inside $C^*$-algebra theory.