Normalization and nonzero success weight of the supported pseudoinverse target

By [F1], in ordered orthonormal singular bases write $A=U\mathrm{\Sigma }{V}^{\ast }$ with positive diagonal entries ${\sigma }_1,\dots ,{\sigma }_r$. Define $B=V{\mathrm{\Sigma }}^{+}{U}^{\ast }$ by replacing each positive ${\sigma }_j$ with $1\mathrm{/}{\sigma }_j$ and leaving zeros zero. Direct multiplication gives $ABA=A$, $BAB=B$, $AB=U\mathrm{diag}(I_r,0)U^{\ast }$ and $BA=V\mathrm{diag}(I_r,0)V^{\ast }$, both self-adjoint. Thus [F3] makes $B$ a Moore–Penrose inverse, and [F4] gives $B=A^{+}={\sum }_j{\sigma }_j^{-1}\mathrm{\mid }{v}_j⟩⟨u_j\mathrm{\mid }$.

Since $b$ is normalized and its coefficients in the orthonormal singular vectors are ${\beta }_j$, ${\sum }_j\mathrm{\mid }{\beta }_j{\mathrm{\mid }}^2=1$. Each $v_j$ is a unit vector and $A{v}_j={\sigma }_j{u}_j$, so [F1] and [F5] give ${\sigma }_j=\mathrm{\parallel }A{v}_j\mathrm{\parallel }\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \alpha$. Step 1.1 therefore yields $\mathrm{\parallel }{A}^{+}b{\mathrm{\parallel }}^2={\sum }_j\mathrm{\mid }{\beta }_j{\mathrm{\mid }}^2\mathrm{/}{\sigma }_j^2\ge {\alpha }^{-2}>0$. Thus its normalized state is defined.

The singular values of $\bar{A}=A\mathrm{/}\alpha$ are ${\sigma }_j\mathrm{/}\alpha$ with the same left and right vectors. Repeating the construction in step 1.1 gives $(A\mathrm{/}\alpha {)}^{+}=\alpha A^{+}$ because $\alpha >0$. Hence $(\delta \mathrm{/}2){\bar{A}}^{+}b=(\gamma \mathrm{/}2)A^{+}b$, and multiplication by the positive scalar $\gamma \mathrm{/}2$ preserves the normalized vector.

By [F2], coefficients outside the supported positive singular subspace vanish. By [F1] and [F5], each unit right singular vector satisfies ${\sigma }_j=\mathrm{\parallel }A{v}_j\mathrm{\parallel }\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \alpha$. Since $\delta =\gamma \mathrm{/}\alpha$, step 2.2 gives $\mathrm{\parallel }(\delta \mathrm{/}2){\bar{A}}^{+}b{\mathrm{\parallel }}^2={\sum }_{j=1}^r\mathrm{\mid }{\beta }_j{\mathrm{\mid }}^2(\gamma \mathrm{/}(2{\sigma }_j){)}^2\ge {\sum }_{j=1}^r\mathrm{\mid }{\beta }_j{\mathrm{\mid }}^2{\gamma }^2\mathrm{/}(4{\alpha }^2)={\delta }^2\mathrm{/}4$. This is the ideal QSVT flag weight; any implementation error requires a separate error budget.