Ideal inverse amplitude and HHL postselection probability

In the eigenbasis from [F1], if $H{u}_j={\lambda }_j{u}_j$, self-adjointness and [F5] give ${\lambda }_j\mathrm{\parallel }{u}_j{\mathrm{\parallel }}^2=⟨H{u}_j,u_j⟩=⟨u_j,H^{\ast }{u}_j⟩=⟨u_j,H{u}_j⟩=\overline{{\lambda }_j}\mathrm{\parallel }{u}_j{\mathrm{\parallel }}^2,$ so every ${\lambda }_j$ is real. Define $G{u}_j={\lambda }_j^{-1}{u}_j$ for ${\lambda }_j\mathrm{\ne }0$ and $G{u}_j=0$ for ${\lambda }_j=0$. Diagonal multiplication gives $HGH=H$, $GHG=G$, and $HG=GH$ equal to the orthogonal projection onto the nonzero-eigenvalue span; both products are self-adjoint. Thus $G$ satisfies [F2], and uniqueness gives $H^{+}=G$.

Expand $\mathrm{\mid }b⟩={\sum }_j{\beta }_j\mathrm{\mid }{u}_j⟩$ in [F1], with ${\sum }_j\mathrm{\mid }{\beta }_j{\mathrm{\mid }}^2=1$. Put $J=\{j:{\beta }_j\mathrm{\ne }0\}$; then ${\sum }_{j\in J}\mathrm{\mid }{\beta }_j{\mathrm{\mid }}^2=1$ and every $j\in J$ has $1\mathrm{/}\kappa \le \mathrm{\mid }{\lambda }_j\mathrm{\mid }\le 1$, so [F3] multiplies it by $(2\kappa {\lambda }_j{)}^{-1}$. Step 1.1 therefore gives the well vector $v={\sum }_{j\in J}{\beta }_j(2\kappa {\lambda }_j{)}^{-1}\mathrm{\mid }{u}_j⟩=(2\kappa {)}^{-1}{H}^{+}\mathrm{\mid }b⟩$. Its normalization is the state in [F4].

Orthonormality gives $p={\sum }_{j\in J}\mathrm{\mid }{\beta }_j{\mathrm{\mid }}^2\mathrm{/}(4{\kappa }^2\mathrm{\mid }{\lambda }_j{\mathrm{\mid }}^2)$. For $j\in J$, $1\le \mathrm{\mid }{\lambda }_j{\mathrm{\mid }}^{-1}\le \kappa$, so each weight factor lies in $[1\mathrm{/}(4{\kappa }^2),1\mathrm{/}4]$. Their convex average lies in the same interval, proving $1\mathrm{/}(4{\kappa }^2)\le p\le 1\mathrm{/}4$.