Singular-value transform is independent of SVD choices

For each distinct positive singular value $\sigma$, define $E_{\sigma }=\ker (A^{\ast }A-{\sigma }^{2}I)$ in $H_R$ and $F_{\sigma }=\ker (A{A}^{\ast }-{\sigma }^{2}I)$ in $H_L$. The adjoint rule [F4] and adjoint algebra [F5] make $A^{\ast }A$ and $A{A}^{\ast }$ self-adjoint, and [F2]–[F3] identify these eigenspaces with the right and left singular subspaces. If $v\in E_{\sigma }$, then $Av\in F_{\sigma }$ because $A{A}^{\ast }Av=A(A^{\ast }Av)={\sigma }^{2}Av$. Also [F4] gives $\mathrm{\parallel }Av{\mathrm{\parallel }}^2=⟨v,A^{\ast }Av⟩={\sigma }^2\mathrm{\parallel }v{\mathrm{\parallel }}^2$. Thus $J_{\sigma }:=A\mathrm{/}\sigma$ is an isometry from $E_{\sigma }$ into $F_{\sigma }$. For $y\in F_{\sigma }$, [F5] gives $A^{\ast }y\in E_{\sigma }$, and $J_{\sigma }(A^{\ast }y\mathrm{/}\sigma )=y$; hence $J_{\sigma }$ maps $E_{\sigma }$ onto $F_{\sigma }$.

Group the positive singular terms in the odd transform by their common value $\sigma$. In any orthonormal basis $(v_j)$ of $E_{\sigma }$, the paired left vectors are $u_j=J_{\sigma }{v}_j$, so ${\sum }_{j:{\sigma }_j=\sigma }\mathrm{\mid }{u}_j⟩⟨v_j\mathrm{\mid }=J_{\sigma }{\mathrm{\Pi }}_{E_{\sigma }}$, where ${\mathrm{\Pi }}_{E_{\sigma }}$ is the unique projection from [F6]. The grouped term is therefore $p(\sigma )J_{\sigma }{\mathrm{\Pi }}_{E_{\sigma }}$, independent of the basis.

For the even right transform, grouping by positive singular value gives $p(\sigma ){\mathrm{\Pi }}_{E_{\sigma }}$, and the zero-singular part is $p(0){\mathrm{\Pi }}_{\ker A}$; both are canonical projections. For the even left transform the corresponding terms are $p(\sigma ){\mathrm{\Pi }}_{F_{\sigma }}$ and $p(0){\mathrm{\Pi }}_{\ker A^{\ast }}$. The zero right and left singular vectors may be chosen independently, but the even definitions use only these unique subspace projections, while the odd definition omits zero singular pairs. Thus all three transforms are independent of the SVD choices.