The Gram pseudoinverse reconstructs a spanning Hermitian frame

By [F2], $G$ is real symmetric positive semidefinite, $G=S^{\ast }S$, and $\ker G=\ker S$. The coordinate formula for the adjoint is $(S^{\ast }X{)}_a=⟨F_a,X{⟩}_{\mathrm{H}\mathrm{S}}=\mathrm{Tr}(X{F}_a)$ by [F1] and [F2].

Regard the real symmetric positive semidefinite matrix $G$ as a complex Hermitian matrix and apply [F3]. Its eigenvalues are real. If $z=x+iy$ is in the complex eigenspace for a real eigenvalue $\lambda$, then realness of $G$ gives $Gx=\lambda x$ and $Gy=\lambda y$; thus that eigenspace is the complexification of its real eigenspace. Real eigenspaces for distinct eigenvalues are orthogonal, since $Gu=\lambda u$, $Gv=\mu v$, and symmetry give $(\lambda -\mu )u^{\mathsf{T}}v=0$. The real eigenspaces form a direct sum of dimension $m$: their complexifications are the orthogonal complex eigenspaces, which span ${\mathbb{C}}^m$ by [F3]. Each eigenvalue is nonnegative because, for a nonzero real eigenvector $x$ with $Gx=\lambda x$, $\lambda \mathrm{\parallel }x{\mathrm{\parallel }}_2^2=x^{\mathsf{T}}Gx\ge 0$. In each real eigenspace, take a finite basis $v_1,\dots ,v_k$ and set $u_1=v_1\mathrm{/}\mathrm{\parallel }{v}_1{\mathrm{\parallel }}_2$ and $u_j=(v_j-{\sum }_{i<j}(v_j^{\mathsf{T}}{u}_i)u_i)\mathrm{/}\mathrm{\parallel }{v}_j-{\sum }_{i<j}(v_j^{\mathsf{T}}{u}_i)u_i{\mathrm{\parallel }}_2$ for $2\le j\le k$; basis independence makes every denominator nonzero, and these vectors form a real orthonormal basis. These bases give a real spectral resolution $G={\sum }_{\lambda \ge 0}\lambda P_{\lambda }$ with real orthogonal projections whose sum is $I$. Put $H_0={\sum }_{\lambda >0}{\lambda }^{-1}{P}_{\lambda }$. Orthogonality gives $G{H}_{0}G=G$, $H_{0}G{H}_0=H_0$, and $G{H}_0=H_{0}G={\sum }_{\lambda >0}{P}_{\lambda }$, which are self-adjoint. Thus $H_0$ satisfies the four Penrose equations [F4], and uniqueness in [F5] gives $H_0=G^{+}$. Consequently $Q=G^{+}G={\sum }_{\lambda >0}{P}_{\lambda }$ is an orthogonal projection and $\ker Q=\ker G$, so $\mathrm{Ran}(I-Q)=\ker G=\ker S$ by [F2].

The synthesis map of the canonical dual is $D=S{G}^{+}$ by [F1]. For any $X\in {\mathsf{H}}_d$, the spanning hypothesis [A1] supplies $c\in {\mathbb{R}}^m$ with $X=Sc$. Then $D(S^{\ast }X)=S{G}^{+}{S}^{\ast }Sc=S{G}^{+}Gc=SQc=Sc=X,$ because $c-Qc\in \mathrm{Ran}(I-Q)=\ker S$ by step 2.1. The coordinates of $S^{\ast }X$ are $\mathrm{Tr}(X{F}_a)$ by step 1.1, which proves the stated sum.