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Reduction of affine variational inequalities

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Abstract

We consider an affine variational inequality posed over a polyhedral convex set in n-dimensional Euclidean space. It is often the case that this underlying set has dimension less than n, or has a nontrivial lineality space, or both. We show that when the variational inequality satisfies a well known regularity condition, we can reduce the problem to the solution of an affine variational inequality in a space of smaller dimension, followed by some simple linear-algebraic calculations. The smaller problem inherits the regularity condition from the original one, and therefore it has a unique solution. The dimension of the space in which the smaller problem is posed equals the rank of the original set: that is, its dimension less the dimension of the lineality space.

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Correspondence to Stephen M. Robinson.

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This material is based on work supported by the Air Force Research Laboratory under awards FA9550-10- 1-0101 and FA9550-15-1-0212. Any opinions, findings, and conclusions or recommendations expressed in this publication are those of the author and do not necessarily reflect the views of the sponsoring agency.

Appendix

Appendix

This appendix contains the proofs of two lemmas used in the analysis of Section 2.4.

Lemma 2.2

Let S be a nonempty convex subset of \(\mathbb {R}^n\) having dimension r, and let \(V\in \mathbb {R}^{n\times r}\) with orthonormal columns such that \({{\mathrm{im}}}V={{\mathrm{par}}}S\). Let \(a_0\in {{{\mathrm{aff}}}}{S}\) and define two maps by

$$\begin{aligned} \phi :{{{\mathrm{aff}}}}{S}\rightarrow \mathbb {R}^k,\quad \phi (x):=V^*(x-a_0);\qquad \theta :\mathbb {R}^k\rightarrow {{{\mathrm{aff}}}}{S},\quad \theta (y):=a_0+Vy. \end{aligned}$$

Then \(\phi \) and \(\theta \) are mutually inverse affine isometries between \({{{\mathrm{aff}}}}{S}\) and \(\mathbb {R}^k\). If \(T:=\phi (S)\) then the nonempty faces F of T are in one-to-one correspondence with the nonempty faces G of S through the equations

$$\begin{aligned} \phi (F) = G,\qquad \theta (G) = F. \end{aligned}$$

Proof

Evidently \(\phi \) takes \({{{\mathrm{aff}}}}{S}\) into \(\mathbb {R}^k\). As \(a_0+{{\mathrm{par}}}S={{{\mathrm{aff}}}}{S}\), \(\theta \) carries \(\mathbb {R}^r\) into \({{{\mathrm{aff}}}}{S}\). For each \(y\in \mathbb {R}^r\),

$$\begin{aligned} (\phi \circ \theta )(y) = V^*[(a_0+Vy)-a_0] = V^*Vy = y, \end{aligned}$$

because \(V^*V=I_r\). Therefore

$$\begin{aligned} \phi \circ \theta =I_r. \end{aligned}$$
(4.1)

For each \(x\in {{{\mathrm{aff}}}}{S}\), \((\theta \circ \phi )(x) = a_0+V[V^*(x-a_0)]\). We have \(x-a_0\in {{\mathrm{par}}}S\) and

$$\begin{aligned} VV^*|_{{{\mathrm{par}}}S}={{\mathrm{id}}}_{{{\mathrm{par}}}S}, \end{aligned}$$
(4.2)

so \((\theta \circ \phi )(x) = x\), and therefore

$$\begin{aligned} (\theta \circ \phi )={{\mathrm{id}}}_{{{{\mathrm{aff}}}}{S}}. \end{aligned}$$
(4.3)

This shows that \(\phi \) and \(\theta \) are mutually inverse affine isomorphisms between \({{{\mathrm{aff}}}}{S}\) and \(\mathbb {R}^r\).

If y and \(y'\) belong to \(\mathbb {R}^r\) then

$$\begin{aligned} \Vert \theta (y)-\theta (y')\Vert ^2=\Vert V(y-y')\Vert ^2=\Vert y-y'\Vert ^2, \end{aligned}$$

so \(\theta \) is an isometry. If x and \(x'\) belong to \({{{\mathrm{aff}}}}{S}\) then

$$\begin{aligned} \Vert \phi (x)-\phi (x')\Vert ^2=\Vert V^*(x-x')\Vert ^2=\Vert x-x'\Vert ^2, \end{aligned}$$

where we used \(x-x'\in {{\mathrm{par}}}S\) and (4.2).

Now let \(T=\phi (S)\), a convex subset of \(\mathbb {R}^r\). Let \({\mathscr {F}}(S)\) and \({\mathscr {F}}(T)\) be the collections of nonempty faces of S and T respectively. If \(F\in {\mathscr {F}}(S)\) then let \(G:=\phi (F)\). Suppose t and \(t'\) belong to T and \(\lambda \in (0,1)\) with \((1-\lambda )t+\lambda t'=:g\in G\). Let \(s:=\theta (t)\) and \(s':=\theta (t')\). Then

$$\begin{aligned} (1-\lambda )s+\lambda s'= & {} (1-\lambda )\theta (t)+\lambda \theta (t') = \theta [(1-\lambda )t+\lambda t']\\= & {} \theta (g)\in \theta (G)=(\theta \circ \phi )(F)=F, \end{aligned}$$

and as F is a face of S, both s and \(s'\) must be in F. This implies that

$$\begin{aligned} t = (\phi \circ \theta )(t)=\phi (s)\in \phi (F)=G, \end{aligned}$$

and similarly \(t'\in G\), showing that \(G\in {\mathscr {F}}(T)\). Therefore \(\phi :{\mathscr {F}}(S)\rightarrow {\mathscr {F}}(T)\). A parallel argument yields \(\theta :{\mathscr {F}}(T)\rightarrow {\mathscr {F}}(S)\), and then (4.1) and (4.3) show that the correspondence is one-to-one. \(\square \)

Lemma 2.3

Let A be an \(n\times n\) matrix with \(n>0\) and let D be a nonempty polyhedral convex subset of \(\mathbb {R}^n\) containing the origin and having parallel subspace P and lineality space K. Let \(H=P^\perp \) and \(R=P\cap K^\perp \), with \(\dim R=r\), and suppose that no more than one of H, K, and R is zero. Let \(D':=\varPi _{K^\perp }(D)\) and \(E:=\phi _R(D')\in \mathbb {R}^r\), where \(\phi _R\) and \(\theta _R\) are the operators obtained from Lemma 2.2 by taking \(a_0\) to be the origin and \(V\in \mathbb {R}^{n\times r}\) to have orthonormal columns constituting a basis for R.

If \(A\in {\mathscr {C}}_D\) then \(A^K\) is nonsingular and \((A^P/A^K)\in {\mathscr {C}}_E\).

Proof

We give the proof under the assumption that none of R, K, and H is the zero subspace. At the end, we indicate the changes needed if this assumption does not hold.

The subspace parallel to D is \(P=R+K\). As \(R\subset K^\perp \),

$$\begin{aligned} {{\mathrm{par}}}D' = \varPi _{K^\perp }({{\mathrm{par}}}D) = \varPi _{K^\perp }(R+K) = R, \end{aligned}$$

and therefore \(\dim D'=r\) as required by Lemma 2.2. We have a one-to-one correspondence between faces F of D and faces \(F'\) of \(D'\) given by \(F'=\varPi _{K^\perp }(F)\) and \(F=F'+K\), and by combining this with the one-to-one correspondence between faces \(F'\) of \(D'\) and faces G of E given by Lemma 2.2 we see that

$$\begin{aligned} G=\rho (F):=(\phi _R\circ \varPi _{K^\perp })(F), \qquad F=\sigma (G):=\theta _R(G)+K \end{aligned}$$
(4.4)

is a one-to-one correspondence between the faces of D and those of E.

Let F be a nonempty face of D having parallel subspace M of dimension m. Each nonempty face of D has K as its lineality space, so \(K\subset M\); also, \(M\subset R+K=P\). Therefore M has the form \(Y+K\), where Y is a subspace of R having dimension y.

We know from (4.4) that F is in one-to-one correspondence with a face G of E. To complete the proof we need to:

  1. 1.

    Determine an orthonormal basis for the subspace Z of \(\mathbb {R}^r\) that is parallel to E, and in the process establish the nonsingularity of \(A^K\);

  2. 2.

    Compute the compression \((A^P/A^K)^Z\) of the operator \((A^P/A^K)\in \mathbb {R}^{r\times r}\) into Z;

  3. 3.

    Show that the sign of the determinant of \((A^P/A^K)^Z\) is nonzero and independent of the subspace F from which we started the calculation.

For the first task, we start with the subspace \(M=Y+K\) parallel to F. Let \(V_Y\in \mathbb {R}^{n\times y}\) have orthonormal columns forming a basis of Y, so that \(Y=V_Y(\mathbb {R}^y)\). If we apply \(\rho \) to M the first step is to apply \(\varPi _{K^\perp }\), which annihilates K and fixes Y. The second step is to apply \(\phi _R\) and obtain

$$\begin{aligned} \phi _R(Y) = (V^*V_Y)(\mathbb {R}^y)={{\mathrm{im}}}V^*V_Y\subset \mathbb {R}^r. \end{aligned}$$

We can reverse this process by applying \(\sigma \) to \((V^*V_Y)(\mathbb {R}^y)\). The first step is multiplication by V, which changes the set to \(VV^*V_Y(\mathbb {R}^y)\). As \(VV^*=\varPi _R\) and \(Y\subset R\), this new set is in fact \({{\mathrm{im}}}V_Y=Y\). The second step adds K to this, so we recover \(Y+K=M\). Therefore the parallel subspace of G in \(\mathbb {R}^r\) is \(Z={{\mathrm{im}}}V^*V_Y\). If we write U for \(V^*V_Y\) then we see that

$$\begin{aligned} U^*U = V_Y^*(VV^*)V_Y, \end{aligned}$$

and as \(VV^*=\varPi _R\) and \(Y\subset R\) this is actually \(V_Y^*V_Y=I_y\). Therefore the columns of U form an orthonormal basis of Z.

The set E contains no line, and it is therefore the convex hull of its extreme points and extreme rays; in particular, it must have an extreme point e. As \(\{e\}\) is a face of E, \(\sigma (\{e\})\) is a face of F whose parallel subspace is K. The compression of A into the parallel subspace of that face is \(A_{KK}\). As the face is nonempty, the hypothesis requires \({{\mathrm{sgn}}}\det A_{KK}=:\tau \in \{-1,+1\}\), where \(\tau \) is the common nonzero determinantal sign of the compressions of A into the nonempty faces of D. This shows that \(A^K=A_{KK}\) is nonsingular.

The next step is to compress \((A^P/A^K)\) into Z. Let \(Q_R=V\) and use the mutual orthogonality of R, K, and H noted in (2.5) to find matrices \(Q_K\) and \(Q_H\) such that \({{\mathrm{im}}}Q_K=K\), \({{\mathrm{im}}}Q_H=H\), and

$$\begin{aligned} Q:=\begin{bmatrix} Q_R&Q_K&Q_H\end{bmatrix} \end{aligned}$$

is an orthogonal matrix. If we adopt the columns of Q as a basis for \(\mathbb {R}^n\) then in that coordinate system the representation of A is

$$\begin{aligned} Q^*AQ = \begin{bmatrix} A_{RR}&\quad A_{RK}&\quad A_{RH}\\ A_{KR}&\quad A_{KK}&\quad A_{KH}\\ A_{HR}&\quad A_{HK}&\quad A_{HH} \end{bmatrix}, \end{aligned}$$
(4.5)

where \(A_{\alpha \beta }=Q_\alpha ^* A Q_\beta \) for \(\alpha \) and \(\beta \) in \(\{R, K, H\}\). Bearing in mind that \(P=R+K\), we see that

$$\begin{aligned} \Big (A^P/A^K \Big ) = \left( \begin{bmatrix} A_{RR}&A_{RK} \\ A_{KR}&A_{KK} \end{bmatrix} \Big /A_{KK} \right) =A_{RR}-A_{RK}A_{KK}^{-1}A_{KR} \in \mathbb {R}^{r\times r}. \end{aligned}$$
(4.6)

To compress this into Z we compute \(U^*(A^P/A^K)U\). Recalling that \(U=V^*V_Y\), that \(Q_R=V\), and that \(VV^*=\varPi _R\) so that \(VV^*V_Y=V_Y\), we obtain

$$\begin{aligned} \Big (A^P/A^K \Big )^Z = U^* \Big (A^P/A^K \Big )U = V_Y^*AV_Y - \Big (V_Y^*AQ_K \Big )A_{KK}^{-1} \Big (Q_K^*AV_Y \Big ), \end{aligned}$$

which, if we expand our subscript notation from \(\{R, K, H\}\) to \(\{R, K, H, Y\}\) to include multiplication by \(V_Y\), becomes

$$\begin{aligned} \Big (A^P/A^K \Big )^Z = A_{YY} - A_{YK}A_{KK}^{-1}A_{KY} =\left( \begin{bmatrix} A_{YY}&\quad A_{YK} \\ A_{KY}&\quad A_{KK} \end{bmatrix} \Big /A_{KK} \right) . \end{aligned}$$

For the third task, recall that the parallel subspace M of the face F was \(Y+K\). We can take the orthonormal columns of the matrix \(J=\begin{bmatrix} V_Y Q_K\end{bmatrix}\) as a basis of M and then compute the compression \(A^M\) of A into M as

$$\begin{aligned} A^M = J^*AJ = \begin{bmatrix} A_{YY}&\quad A_{YK}\\ A_{KY}&\quad A_{KK} \end{bmatrix}. \end{aligned}$$
(4.7)

This shows that

$$\begin{aligned} \Big (A^P/A^K \Big )^Z = \Big (A^M/A_{KK} \Big ). \end{aligned}$$
(4.8)

Now we apply Schur’s determinantal theorem to the right side of (4.8) to obtain

$$\begin{aligned} \det A^M = \left[ \det \Big (A^M/A_{KK} \Big ) \right] \det A_{KK}, \end{aligned}$$
(4.9)

and recall that as M is parallel to the face F of D, the hypothesis requires that \({{\mathrm{sgn}}}\det A^M=\tau \). Then we combine (4.8) and (4.9) to obtain

$$\begin{aligned} {{\mathrm{sgn}}}\det (A^P/A^K)^Z = {{\mathrm{sgn}}}(\det A^M)/{{\mathrm{sgn}}}(\det A_{KK})=\tau /\tau = +1. \end{aligned}$$
(4.10)

Therefore the determinantal sign of the compression of \((A^P/A^K)\) into the subspace parallel to each nonempty face of E is \(+1\), and it follows that \((A^P/A^K)\in {\mathscr {C}}_E\).

Finally, we consider the case in which one of R, K, and H is the zero subspace. If that subspace is H, then \(P=\mathbb {R}^n\), so that \(A^P=A\). The analysis works as above except that elements involving H do not appear.

If K is zero then the faces F of D and G of E each have parallel subspace Y, so there is nothing to prove.

If R is zero then each of W and Y is zero so that \(H=K^\perp \). The set \(D\cap K^\perp \) is the origin, \(D=K\), and \(E=\{0\}\). Each of D and E has only one nonempty face (itself), so the only thing requiring verification is that neither the compression of A into the parallel subspace K of C nor the compression of \((A^P/A^K)\) into \(\{0\}\) has determinant zero. The second of these compressions has determinant \(+1\) by convention. As \(D=K\) is a nonempty face of itself, the hypothesis requires \(\det A^K\) to be nonzero, which disposes of the first verification. \(\square \)

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Robinson, S.M. Reduction of affine variational inequalities. Comput Optim Appl 65, 493–509 (2016). https://doi.org/10.1007/s10589-015-9796-7

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