Nested-basis H² multiplication is exact for its represented matrix

At a column leaf $\sigma$, the algorithm sets $z_{\sigma }=V_{\sigma }^{\ast }x{\mathrm{\mid }}_{\sigma }$. Suppose this identity holds for both children ${\sigma }_1,{\sigma }_2$ of an internal cluster $\sigma$. By the nestedness in [F1] and the disjoint union of child indices, $V_{\sigma }^{\ast }x{\mathrm{\mid }}_{\sigma }={\sum }_{j=1}^2(E_{{\sigma }_j}^V{)}^{\ast }{V}_{{\sigma }_j}^{\ast }x{\mathrm{\mid }}_{{\sigma }_j}={\sum }_{j=1}^2(E_{{\sigma }_j}^V{)}^{\ast }{z}_{{\sigma }_j}$, exactly the upward update in [F4]. Induction gives $z_{\sigma }=V_{\sigma }^{\ast }x{\mathrm{\mid }}_{\sigma }$ for every column cluster.

Fix a far block $(\tau ,\sigma )$. Its coupling contributes $S_{\tau \sigma }{z}_{\sigma }=S_{\tau \sigma }{V}_{\sigma }^{\ast }x{\mathrm{\mid }}_{\sigma }$ to $w_{\tau }$. Each downward transfer from a parent to a child applies the corresponding $E^U$; repeated nestedness [F1] therefore turns $U_{\tau }{S}_{\tau \sigma }{V}_{\sigma }^{\ast }x{\mathrm{\mid }}_{\sigma }$ into exactly its restriction on every descendant row leaf. Summing the finite far-block contributions gives their full matrix action.

By [F2] and [F4], every near block contributes its exact dense product $D_{\tau \sigma }x{\mathrm{\mid }}_{\sigma }$. The far and near block leaves partition $I\times J$ by [F3], so adding their actions gives every entry of $\tilde{A}x$ exactly once in the matrix sum. Thus the returned vector is $\tilde{A}x$.