Hermitian dilation for a rectangular linear system

The block matrix of $H_A$ is $\left(\begin{array}{cc}0& A\\ A^{\ast }& 0\end{array}\right)$. Its adjoint, by [F2], is the same block matrix, so $H_A^{\ast }=H_A$.

Multiplying the SVD in [F1] gives $A{v}_j={\sigma }_j{u}_j$ and $A^{\ast }{u}_j={\sigma }_j{v}_j$. Therefore $H_A(u_j,v_j)={\sigma }_j(u_j,v_j)$ and $H_A(u_j,-v_j)=-{\sigma }_j(u_j,-v_j)$. After division by $\sqrt{2}$ these are unit eigenvectors. Orthonormality of the singular pairs makes all such vectors mutually orthogonal.

In the SVD bases, write $A=U\mathrm{\Sigma }{V}^{\ast }$ with positive diagonal entries ${\sigma }_j$. The diagonal matrix ${\mathrm{\Sigma }}^{+}$ that replaces each positive entry by $1\mathrm{/}{\sigma }_j$ and leaves zero entries zero satisfies $\mathrm{\Sigma }{\mathrm{\Sigma }}^{+}\mathrm{\Sigma }=\mathrm{\Sigma }$, ${\mathrm{\Sigma }}^{+}\mathrm{\Sigma }{\mathrm{\Sigma }}^{+}={\mathrm{\Sigma }}^{+}$, and both products $\mathrm{\Sigma }{\mathrm{\Sigma }}^{+}$ and ${\mathrm{\Sigma }}^{+}\mathrm{\Sigma }$ are diagonal projections. Unitary transport shows $V{\mathrm{\Sigma }}^{+}{U}^{\ast }$ satisfies the four equations in [F3], so uniqueness gives $A^{+}=V{\mathrm{\Sigma }}^{+}{U}^{\ast }$ and $A^{+}{u}_j=v_j\mathrm{/}{\sigma }_j$.

The equation $H_A(x,y)=0$ is equivalent to $Ay=0$ and $A^{\ast }x=0$, so $\ker H_A=\ker (A^{\ast })\oplus \ker (A)$. Together with step 1.2, the nonzero eigenvalue magnitudes are exactly the positive singular values, each appearing with both signs. Thus when there is positive rank their maximum-to-minimum ratio is unchanged.

The vectors from step 1.2 together with orthonormal bases of $\ker (A^{\ast })$ and $\ker (A)$ form an orthonormal eigenbasis of the direct sum. Define $G$ on this basis by $G{w}_{j,+}=w_{j,+}\mathrm{/}{\sigma }_j$, $G{w}_{j,-}=-w_{j,-}\mathrm{/}{\sigma }_j$, and $G=0$ on the kernel. Then $H_{A}G$ and $G{H}_A$ are self-adjoint projections, while diagonal multiplication gives $H_{A}G{H}_A=H_A$ and $G{H}_{A}G=G$. Thus $G$ satisfies [F3], and uniqueness gives $G=H_A^{+}$. Since $(u_j,0)=(w_{j,+}+w_{j,-})\mathrm{/}\sqrt{2}$, this definition gives $H_A^{+}(u_j,0)=(0,v_j\mathrm{/}{\sigma }_j)$. Complete the positive left singular vectors to the left singular basis from [F1]. The remaining basis vectors have zero singular coordinates, so the diagonal SVD gives $A^{\ast }{u}_j=0$ for each; hence they span the complementary part $\ker (A^{\ast })$. Thus every $b\in {\mathbb{C}}^m$ has a decomposition $b={\sum }_{j:{\sigma }_j>0}{\beta }_j{u}_j+b_0$ with $b_0\in \ker (A^{\ast })$. Since $A^{\ast }{b}_0=0$, its positive left-singular coordinates vanish, so ${\mathrm{\Sigma }}^{+}{U}^{\ast }{b}_0=0$ and the formula in step 1.3 gives $A^{+}{b}_0=0$. By step 2.1 $(b_0,0)\in \ker (H_A)$, so $H_A^{+}(b_0,0)=G(b_0,0)=0$. Applying the positive-pair calculation to the other summands gives $H_A^{+}(b,0)=(0,A^{+}b)$ for arbitrary $b$, including $b\mathrm{\notin }\mathrm{ran}(A)$.