Substructuring, FETI-DP and BDDC

If $V=\{0\}$, the identity $LT=I_U$ forces $U=\{0\}$ and $G$ has no eigenvalues, so the assertion holds with the stated zero-space norm convention. Assume henceforth $V\mathrm{\ne }\{0\}$. Put $D=T^{T}AT$, which is SPD because $T$ is injective (if $Tu=0$, then $u=LTu=0$), and $C=L{A}^{-1}{L}^T$, which is SPD because $L$ is onto (it has right inverse $T$). Hence $G=CD$ is similar to the symmetric positive definite matrix $D^{1\mathrm{/}2}C{D}^{1\mathrm{/}2}$ and has positive real eigenvalues. For an eigenvector $u\mathrm{\ne }0$ with eigenvalue $\lambda$, let $w=A^{-1}{L}^{T}Du$, so $Lw=Gu=\lambda u$ and $\mathrm{\parallel }w{\mathrm{\parallel }}_A^2=⟨Du,Gu⟩=\lambda ⟨Du,u⟩$.

By [A1], $⟨Du,u⟩=⟨Du,LTu⟩=⟨Aw,Tu⟩$. Cauchy–Schwarz in the $A$ inner product gives $⟨Du,u{⟩}^2\le \mathrm{\parallel }w{\mathrm{\parallel }}_A^2\mathrm{\parallel }Tu{\mathrm{\parallel }}_A^2=\lambda ⟨Du,u{⟩}^2$. Dividing by the positive square gives $\lambda \ge 1$.

Also $\mathrm{\parallel }Gu{\mathrm{\parallel }}_D^2=\mathrm{\parallel }TLw{\mathrm{\parallel }}_A^2\le \mathrm{\parallel }TL{\mathrm{\parallel }}_A^2\mathrm{\parallel }w{\mathrm{\parallel }}_A^2$. For the eigenvector, the left side is ${\lambda }^2⟨Du,u⟩$ and step 1.1 makes the right side $\mathrm{\parallel }TL{\mathrm{\parallel }}_A^2\lambda ⟨Du,u⟩$. Divide by the positive factor to obtain $\lambda \le \mathrm{\parallel }TL{\mathrm{\parallel }}_A^2$.

If $a_i(v,v)=0$, each nonnegative term ${\rho }_i{\int }_K\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}^2$ is zero. Since $\mathrm{\nabla }v$ is constant on each triangle, it vanishes on every triangle, so $v$ is constant there. Continuity across every shared edge makes neighboring constants equal; connectedness of the triangle adjacency graph makes $v$ one constant on all of ${\mathrm{\Omega }}_i$. Conversely, a constant has zero gradient, hence lies in the kernel. This proves the floating-block nullspace claim.

A vector in the interior block's kernel defines a local $P_1$ function with zero boundary trace and zero energy. Step 1.1 makes it constant and the zero boundary trace makes that constant zero. Thus the interior quadratic form is strictly positive on every nonzero vector, so its symmetric matrix $K_{II}^{(i)}$ is positive definite. If there are no interior vertices, the unique map on the zero vector space is invertible.

If the global energy vanishes, step 1.1 applied triangle by triangle gives one constant on every subdomain. Interface continuity makes adjacent subdomain constants equal; connectedness propagates one constant through $\mathrm{\Omega }$. The nonempty Dirichlet segment forces it to be zero. Hence only the zero global vector has zero quadratic energy, proving positive definiteness.

From [F3], block multiplication gives $K_i{H}_i{g}_i=(0,S_i{g}_i{)}^T$ and therefore $(H_i{g}_i{)}^T{K}_i(H_i{g}_i)=g_i^T{S}_i{g}_i$. Thus a zero local Schur energy is precisely zero energy of its harmonic lift. In a local dual subproblem the primal values are prescribed, so a zero-energy local vector, or the harmonic lift of a zero-energy dual trace, is constant by [F1]. A primal corner value is among its prescribed values; because the dual correction has zero primal values, that constant is zero. The local dual quadratic form therefore has trivial kernel and, being symmetric positive semidefinite, is positive definite.

For $g=(g_i)\in \tilde{W}$, [F3] and step 1.1 give $g^T\tilde{S}g={\sum }_i{g}_i^T{S}_i{g}_i={\sum }_i(H_i{g}_i{)}^T{K}_i(H_i{g}_i)$. Each summand is a nonnegative local energy. If the partially assembled energy is zero, every summand is zero, so [F1] makes each local harmonic extension constant on its subdomain. Neighboring local constants agree at their shared primal corner. The adjacency graph is connected, so all constants agree; the global Dirichlet boundary makes the common value zero. Thus the partially assembled trace is zero, and its assembled energy matrix is positive definite.

The BDDC local dual solves use the positive local restrictions from step 1.1 and its coarse solve uses the positive partially assembled energy from step 2.1. The FETI-DP operator uses the same invertible partially assembled energy and the jump quotient of [F2]. Hence the exact local, coarse and dual systems named in the Statement are well-defined.

Any local vector with trace $g_i$ has the form $v=H_i{g}_i+(z,0{)}^T$. By [F1], the interior block row of $K_i{H}_i{g}_i$ is $K_{II}(-K_{II}^{-1}{K}_{I\mathrm{\Gamma }}{g}_i)+K_{I\mathrm{\Gamma }}{g}_i=0$. Hence the energy cross term between $H_i{g}_i$ and $(z,0)$ vanishes, and $v^T{K}_{i}v=(H_i{g}_i{)}^T{K}_i(H_i{g}_i)+z^T{K}_{II}z$.

By [F2], the last term is nonnegative and is zero only for $z=0$. Thus $H_i{g}_i$ is the unique minimizer. Multiplying the block expression in [F1] gives $(H_i{g}_i{)}^T{K}_i(H_i{g}_i)=g_i^T(K_{\mathrm{\Gamma }\mathrm{\Gamma }}-K_{\mathrm{\Gamma }I}{K}_{II}^{-1}{K}_{I\mathrm{\Gamma }})g_i=g_i^T{S}_i{g}_i$.

Let $f_i=(f_I^{(i)},f_{\mathrm{\Gamma }}^{(i)}{)}^T$ be local load contributions whose assembly is the chosen global Galerkin load vector, as specified above. The global interior rows are local to each subdomain and give $u_I^{(i)}=(K_{II}^{(i)}{)}^{-1}(f_I^{(i)}-K_{I\mathrm{\Gamma }}^{(i)}{u}_{\mathrm{\Gamma }}^{(i)})\mathrm{.}$ The local interface residual contribution is $K_{\mathrm{\Gamma }I}^{(i)}{u}_I^{(i)}+K_{\mathrm{\Gamma }\mathrm{\Gamma }}^{(i)}{u}_{\mathrm{\Gamma }}^{(i)}-f_{\mathrm{\Gamma }}^{(i)}$. Substituting the interior formula makes it $S_i{u}_{\mathrm{\Gamma }}^{(i)}-(f_{\mathrm{\Gamma }}^{(i)}-K_{\mathrm{\Gamma }I}^{(i)}(K_{II}^{(i)}{)}^{-1}{f}_I^{(i)})$, precisely the local Schur contribution in [F3]. The global interface rows are the assembled sum of these contributions, not separate local interface equations.

Sum the reduced equations from step 1.1 over subdomains and identify copies of each shared nodal value as in [F1]. Because both the stiffness and the local load vectors assemble by summing their shared interface components, this gives the assembled interface Schur system with the correspondingly reduced global right-hand side. Any solution of the original global system yields this trace. Conversely, a solution of the assembled interface system and the uniquely defined interior values from step 1.1 satisfy every local interior equation and the assembled sum of interface equations, hence the assembled global system.

The global stiffness is positive definite by [F4], so the original solution is unique. Once the interface trace is solved, each local interior vector is exactly the reconstruction formula in step 1.1, which is unique because $K_{II}^{(i)}$ is invertible. Its trace-dependent homogeneous part is $H_i{u}_{\mathrm{\Gamma }}^{(i)}$ from [F2]. Therefore static condensation is exact in both directions.

On the upper half-plane, Fourier transformation in the tangential variable gives, for a smooth lift $V$ with trace $g$, $\mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }\widehat{g}(\xi ){\mathrm{\mid }}^2\le {\int }_0^{\mathrm{\infty }}(\mathrm{\mid }{\mathrm{\partial }}_t\widehat{V}(\xi ,t){\mathrm{\mid }}^2+\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{\mid }\widehat{V}(\xi ,t){\mathrm{\mid }}^2)dt\mathrm{.}$ Indeed, integrate $-{\mathrm{\partial }}_t\mathrm{\mid }\widehat{V}{\mathrm{\mid }}^2$ and apply $2ab\le a^2+b^2$. Integration in $\xi$ proves the trace seminorm bound. Conversely, $\widehat{V}(\xi ,t)=e^{-t\mathrm{\mid }\xi \mathrm{\mid }}\widehat{g}(\xi )$ has gradient energy $\int \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }\widehat{g}(\xi ){\mathrm{\mid }}^{2}d\xi$. In a bounded chart, multiply the half-plane extension by a fixed cutoff in the inward coordinate; the added gradient term is controlled by the localized $L^2$ boundary norm, hence by the full H1/2 norm after constants are removed. A finite collection of Lipschitz graph charts covers the polygon boundary, including its corners. Flattening each chart is bi-Lipschitz, so it changes the $H^1$ gradient and boundary Gagliardo seminorm by bounded factors. Applying a partition of unity and the half-plane trace inequality to arbitrary chart-localized lifts gives $\mathrm{\mid }g{\mathrm{\mid }}_{1\mathrm{/}2,\mathrm{\partial }{\mathrm{\Omega }}_i}\le C\mathrm{\parallel }V{\mathrm{\parallel }}_{H^1({\mathrm{\Omega }}_i)}$. Subtract the domain mean of $V$ and use scaled Poincar'e to replace the full $H^1$ norm by $\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2}$. In the other direction, apply a partition of unity to boundary data, use the half-plane extensions, and sum to obtain a right inverse of the trace. First subtract a boundary constant; the fractional Poincaré bound on the whole boundary controls the localized full norms: for the boundary mean $c$, $\mathrm{\parallel }g-c{\mathrm{\parallel }}_{L^2(\mathrm{\partial }{\mathrm{\Omega }}_i)}^2\le H_i^2\mathrm{\mid }g{\mathrm{\mid }}_{1\mathrm{/}2,\mathrm{\partial }{\mathrm{\Omega }}_i}^2\mathrm{/}(2\mathrm{\mid }\mathrm{\partial }{\mathrm{\Omega }}_i\mathrm{\mid })$, by integrating $\mathrm{\mid }s-t\mathrm{\mid }\le H_i$. Add that constant back as a constant lift, which has zero gradient. The chart count, scaled radii, Lipschitz constants and mean-zero Poincaré bounds are precisely the fixed shape parameters in [F1]; a pointwise interior-cone condition alone would not justify this uniformity.

Construct an interpolant $I_h:H^1({\mathrm{\Omega }}_i)\to V_h$ as follows. For each interior mesh vertex choose one incident triangle and set its coefficient to the average of $V$ on that triangle. For each boundary vertex choose one incident boundary edge and set its coefficient to the integral of the trace of $V$ against the $L^2$-dual function to the endpoint nodal basis on that edge. These dual functionals reproduce every affine edge trace, so the trace of $I_{h}V$ is exactly $g$ whenever the trace of $V$ is $g$. All coefficients reproduce constants. For a triangle $K$, the chosen elements and edges for its vertices lie in a connected patch ${\omega }_K$ of diameter at most $Ch$ containing a uniformly bounded number of triangles; these patches have bounded overlap. Let $c_K$ be the mean of $V$ on ${\omega }_K$. Scaled Poincaré gives $\mathrm{\parallel }V-c_K{\mathrm{\parallel }}_{L^2({\omega }_K)}\le Ch\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2({\omega }_K)}$, and the scaled trace inequality gives $\mathrm{\parallel }V-c_K{\mathrm{\parallel }}_{L^2(e)}\le C{h}^{1\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2({\omega }_K)}$ on a selected boundary edge $e$. Thus each interior average coefficient of $V-c_K$ is bounded by $C\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2({\omega }_K)}$. A boundary dual function has $L^2(e)$ norm at most $C{h}^{-1\mathrm{/}2}$, so its coefficient has the same bound. Since $I_h{c}_K=c_K$ and the affine basis functions have uniformly bounded $L^2(K)$ gradient norms in two dimensions, $\mathrm{\parallel }\mathrm{\nabla }{I}_{h}V{\mathrm{\parallel }}_{L^2(K)}\le C\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2({\omega }_K)}$. Summing and using bounded overlap proves $\mathrm{\parallel }\mathrm{\nabla }{I}_{h}V{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}\le C\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}$.

For any finite-element lift $v_h$ of $g$, step 1.1 gives $\mathrm{\mid }g{\mathrm{\mid }}_{1\mathrm{/}2,\mathrm{\partial }{\mathrm{\Omega }}_i}^2\le C\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}^2$. For the floating matrix, put $H^{\mathrm{f}\mathrm{l}}g=(-K_{II}^{-1}{K}_{I\mathrm{\Gamma }}g,g{)}^T$. Symmetry of the local Poisson form and its zero interior row give $a_i(H^{\mathrm{f}\mathrm{l}}g+(z,0),H^{\mathrm{f}\mathrm{l}}g+(z,0))=g^T{S}_i^{\mathrm{f}\mathrm{l}}g+z^T{K}_{II}z$. The positive interior block from [F2] therefore makes $g^T{S}_i^{\mathrm{f}\mathrm{l}}g$ the unique fixed-trace minimum, by the same identity as [F3]. Taking that minimum gives the lower Schur bound. For the upper bound, step 1.1 supplies a continuous lift $V$ with gradient energy at most $C\mathrm{\mid }g{\mathrm{\mid }}_{1\mathrm{/}2,\mathrm{\partial }{\mathrm{\Omega }}_i}^2$, and step 1.2 gives a boundary-preserving finite-element lift $I_{h}V$ with the same bound up to a uniform factor. The just-proved floating minimality bounds its energy by this lift. Multiplication by ${\rho }_i$ proves both inequalities. If $g$ is admissible, the fixed-trace feasible lifts already vanish on the outer Dirichlet portion; [F2]–[F3] identify the same minimum with $g^T{S}_{i}g$.

Differentiating the finite-dimensional quadratic in a direction $z\in \tilde{W}$ gives $J_f^{\mathrm{\prime }}(w)[z]=⟨\tilde{S}w,z⟩-⟨f,E_{0}z⟩$. Thus its critical point solves $\tilde{S}w=E_0^{T}f$, whose unique solution is $w_{\ast }={\tilde{S}}^{-1}{E}_0^{T}f$.

For any $z\in \tilde{W}$, expansion about the critical point yields $J_f(w_{\ast }+z)=J_f(w_{\ast })+\frac{1}{2}⟨\tilde{S}z,z⟩$. By [F2] the added term is positive unless $z=0$, so $w_{\ast }$ is the unique global minimizer. Applying $E_0$ gives the displayed BDDC correction by [F1].

If $E$ has no interior boundary node, then ${\theta }_E=0$. If $\mathrm{\mid }E\mathrm{\mid }\mathrm{/}h\le 4$, scaling $E$ to unit length gives a uniform bound on both terms of $\mathrm{\parallel }{\theta }_E{\mathrm{\parallel }}_{00,E}^2$. Otherwise the first and last mesh intervals have length comparable to $h$, ${\theta }_E$ is linear from $0$ to $1$ and from $1$ to $0$ there, and it equals $1$ between them. The weighted integral on either endpoint interval is bounded by ${\int }_0^h(s\mathrm{/}h{)}^{2}ds\mathrm{/}s\le C$; the plateau contributes at most $C+\log (\mathrm{\mid }E\mathrm{\mid }\mathrm{/}h)\le C+\log (H\mathrm{/}h)$. In the double integral, pairs in the plateau contribute zero. Pairs within one endpoint interval have integrand at most $C{h}^{-2}$ and hence contribute at most a constant. For one point $s$ in an endpoint interval and one outside it, integrating $(1-s\mathrm{/}h{)}^2\mathrm{/}(t-s{)}^2$ in the outside variable and then in $s$ gives a constant; the two endpoint intervals are treated in the same way. Therefore $\mathrm{\parallel }{\theta }_E{\mathrm{\parallel }}_{00,E}^2\le C(1+\log (H\mathrm{/}h))$.

Put $v_h=u_h-{\bar{u}}_{{\mathrm{\Omega }}_i}$, $H=H_i$, $h=h_i$ and $G=\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}$. Choose a triangle $K$ containing a point where $\mathrm{\mid }{v}_h\mathrm{\mid }$ is maximal, and let $x_0$ be its centroid. Affine scaling gives $\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }{v}_h(x_0)\mathrm{\mid }+CG$. The uniform interior-cone condition in [F1] supplies at $x_0$ a cone of radius $R=cH$ and angular interval $J$ of fixed positive length, contained in ${\mathrm{\Omega }}_i$. Quasiuniformity and affine scaling give the global inverse bound $\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{h}^{-1}G$. If $H\mathrm{/}h$ is bounded, the inverse bound $\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{h}^{-1}\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{L^2}$ and mean-zero scaled Poincaré prove the claim directly. Otherwise choose $\epsilon =c^{\mathrm{\prime }}h<R\mathrm{/}2$. Along each cone ray, the integral of $\mathrm{\mid }\mathrm{\nabla }{v}_h\mathrm{\mid }$ from $0$ to $\epsilon$ is at most $CG$, while weighted Cauchy–Schwarz gives ${\mid {\int }_{\epsilon }^r{\mathrm{\partial }}_s{v}_h(x_0+s{e}_{\theta })ds\mid }^2\le \log (R\mathrm{/}\epsilon ){\int }_{\epsilon }^R\mathrm{\mid }\mathrm{\nabla }{v}_h(x_0+s{e}_{\theta }){\mathrm{\mid }}^{2}sds\mathrm{.}$ Express $v_h(x_0)$ by its value at $x_0+r{e}_{\theta }$ minus these two radial integrals, square, and average with measure $rdrd\theta$ over $R\mathrm{/}2<r<R$, $\theta \in J$. The annulus has area comparable to $H^2$; the ray-energy term is independent of $r$ and integrates in $\theta$ to at most $G^2$. Hence $\mathrm{\mid }{v}_h(x_0){\mathrm{\mid }}^2\le C{H}^{-2}\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{L^2}^2+C(1+\log (H\mathrm{/}h))G^2\mathrm{.}$ Mean-zero scaled Poincaré bounds the first term by $C{G}^2$, and the centroid comparison proves the asserted $L^{\mathrm{\infty }}$ estimate. This is the radial argument underlying Widlund's estimate (5), with the centroid-to-maximum comparison included explicitly.

Put $z=u_h-{\bar{u}}_E$. In the first Gagliardo term use $\theta (s)z(s)-\theta (t)z(t)=\theta (s)(z(s)-z(t))+z(t)(\theta (s)-\theta (t))\mathrm{.}$ The weighted term is bounded by $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}^2{\int }_E{\theta }_E^2\mathrm{/}\mathrm{dist}(s,\mathrm{\partial }E)ds$. Together these give $\mathrm{\parallel }{\theta }_{E}z{\mathrm{\parallel }}_{00,E}^2\le C(\mathrm{\mid }z{\mathrm{\mid }}_{H^{1\mathrm{/}2}(E)}^2+(1+\log (H\mathrm{/}h))\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}^2)\mathrm{.}$ On the two endpoint mesh intervals, $I_h({\theta }_{E}z)-{\theta }_{E}z$ is a quadratic interpolation error vanishing at both endpoints; scaling the interval to unit length bounds its $00,E$ norm by $C\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}^2$. The same estimate therefore holds for $I_h({\theta }_{E}z)$. The floating local Schur energy of the trace of $u_h$ is at most ${\rho }_i\mathrm{\parallel }\mathrm{\nabla }{u}_h{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}^2$ because it is the minimum over finite-element lifts; the lower trace-Schur bound in [F2] therefore gives $\mathrm{\mid }z{\mathrm{\mid }}_{H^{1\mathrm{/}2}(E)}^2\le C\mathrm{\parallel }\mathrm{\nabla }{u}_h{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}^2$. Also $\mathrm{\mid }{\bar{u}}_E-{\bar{u}}_{{\mathrm{\Omega }}_i}\mathrm{\mid }\le \mathrm{\parallel }{u}_h-{\bar{u}}_{{\mathrm{\Omega }}_i}{\mathrm{\parallel }}_{L^{\mathrm{\infty }}({\mathrm{\Omega }}_i)}$, so $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2\mathrm{\parallel }{u}_h-{\bar{u}}_{{\mathrm{\Omega }}_i}{\mathrm{\parallel }}_{L^{\mathrm{\infty }}({\mathrm{\Omega }}_i)}$. Combining this with steps 1.1 and 1.2 proves the claimed squared-logarithmic multiplier estimate.

The Lagrangian is $L(w,\lambda )=\frac{1}{2}⟨\tilde{S}w,w⟩-⟨E_0^{T}f,w⟩+⟨\lambda ,Bw⟩$. Its derivative in arbitrary $z\in \tilde{W}$ is $⟨\tilde{S}w+B^T\lambda -E_0^{T}f,z⟩$, and its multiplier derivative is $Bw$. Setting both derivatives to zero gives the displayed saddle equations.

By the assumed positive definiteness of $\tilde{S}$, the first equation has the unique solution $w={\tilde{S}}^{-1}(E_0^{T}f-B^T\lambda )$. Substitution into $Bw=0$ gives the displayed dual system.

If $\lambda \in \mathrm{\Lambda }$ and $B^T\lambda =0$, write $\lambda =Bv$ using [F2]. Then $⟨\lambda ,\lambda ⟩=⟨Bv,\lambda ⟩=⟨v,B^T\lambda ⟩=0$, hence $\lambda =0$. Thus $B^T$ is injective on $\mathrm{\Lambda }$, and $⟨F\lambda ,\lambda ⟩=⟨{\tilde{S}}^{-1}{B}^T\lambda ,B^T\lambda ⟩>0$ for every nonzero $\lambda$. The exact solution satisfies $Bw=0$, so by [F2] it lies in $\widehat{W}$ and solves the assembled variational problem.

The products $G_1$ and $G_2$ act on $U_1$ and $U_2^{\mathrm{\prime }}$, respectively. Set $X=L_1{A}^{-1}{L}_2^T:U_2^{\mathrm{\prime }}\to U_1$ and $Y=T_2^{T}A{T}_1:U_1\to U_2^{\mathrm{\prime }}$. By the given hypotheses, $L_1{T}_1=I$, $L_2{T}_2=I$, and $T_1{L}_1+T_2{L}_2=I$. Expanding the products gives $G_1-I=-XY$ and $G_2-I=-YX$. In detail, $G_1-I=L_1{A}^{-1}((T_1{L}_1{)}^T-I)A{T}_1=-L_1{A}^{-1}{L}_2^T{T}_2^{T}A{T}_1$; also $YX=T_2^{T}A(I-T_2{L}_2)A^{-1}{L}_2^T=I-G_2$.

If $u\in U_1$ satisfies $G_{1}u=\lambda u$ with $\lambda \mathrm{\ne }1$, then $XYu=(1-\lambda )u$, so $Yu\mathrm{\ne }0$. The identity $G_2-I=-YX$ gives $(G_2-I)Yu=-YXYu=(\lambda -1)Yu$, hence $G_2(Yu)=\lambda Yu$. For $v\in U_2^{\mathrm{\prime }}$, the reverse map is $v\mapsto -Xv\mathrm{/}(\lambda -1)$; the same identities show it is the inverse on the $\lambda$-eigenspaces. Thus these eigenspaces are isomorphic.

Choose bases of the finite-dimensional spaces and their dual bases. Both operators are diagonalizable: $T_1^{T}A{T}_1$ and $L_1{A}^{-1}{L}_1^T$ are positive definite, so, writing $D_1=T_1^{T}A{T}_1$, $G_1$ is similar to the symmetric positive definite product $D_1^{1\mathrm{/}2}(L_1{A}^{-1}{L}_1^T)D_1^{1\mathrm{/}2}$; likewise $G_2$ is similar to $D_2^{1\mathrm{/}2}(L_2{A}^{-1}{L}_2^T)D_2^{1\mathrm{/}2}$, where $D_2=T_2^{T}A{T}_2$. The left-inverse identities make $T_i$ injective and $L_i$ surjective, proving each factor is positive definite. Therefore equal eigenspace dimensions from step 2.1 give equal algebraic multiplicities for every $\lambda \mathrm{\ne }1$.