Maximally monotone linear subspace extensions of ... - Springer Link
Math. Program., Ser. B (2013) 139:327–352 DOI 10.1007/s10107-013-0674-8 FULL LENGTH PAPER
Maximally monotone linear subspace extensions of monotone subspaces: explicit constructions and characterizations Xianfu Wang · Liangjin Yao
Received: 7 March 2011 / Accepted: 11 September 2011 / Published online: 20 April 2013 © Springer-Verlag Berlin Heidelberg and Mathematical Optimization Society 2013
Abstract Monotone linear relations play important roles in variational inequality problems and quadratic optimizations. In this paper, we give explicit maximally monotone linear subspace extensions of a monotone linear relation in finite dimensional spaces. Examples are provided to illustrate our extensions. Our results generalize a recent result by Crouzeix and Ocaña-Anaya. Keywords Adjoint of linear relation · Linear relation · Monotone operator · Maximally monotone extensions · Minty parametrization Mathematics Subject Classification (2000)
47H05 · 47B65 · 47A06 · 49N15
1 Introduction Throughout this paper, we assume that Rn (n ∈ N = {1, 2, 3, . . .}) is an Euclidean space with the inner product ·, ·, and induced Euclidean norm ·. Let G : Rn ⇒ Rn be a set-valued operator from Rn to Rn , i.e., for every x ∈ Rn , Gx ⊆ Rn , and let
Dedicated to Jonathan Borwein on the occasion of his 60th birthday. X. Wang was partially supported by the Natural Sciences and Engineering Research Council of Canada. X. Wang Mathematics, Irving K. Barber School, University of British Columbia, Kelowna, BC V1V 1V7, Canada e-mail: [email protected] L. Yao (B) CARMA, University of Newcastle, Newcastle, New South Wales 2308, Australia e-mail: [email protected]
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graG = (x, x ∗ ) ∈ Rn × Rn | x ∗ ∈ Gx be the graph of G. Recall that G is monotone if ∀(x, x ∗ ) ∈ graG ∀(y, y ∗ ) ∈ graG x − y, x ∗ − y ∗ ≥ 0,
(1.1)
and maximally monotone if G is monotone and G has no proper monotone extension (in the sense of graph inclusion). We say that G is a linear relation if graG is a linear subspace. While linear relations have been extensively studies [2,3,6,9,18–20], monotone operators are ubiquitous in convex optimization and variational analysis [1,6,7,16]. Let R p×n denote the set of all p ×n matrices, for n, p ∈ N; and rank(M) denote the rank of M ∈ R p×n . The central object of this paper is to consider the linear relation G : Rn ⇒ Rn : graG = {(x, x ∗ ) ∈ Rn × Rn | Ax + Bx ∗ = 0} where
(1.2)
A, B ∈ R , rank(A B) = p.
(1.3) (1.4)
p×n
Our main concern is to find explicit maximally monotone linear subspace extensions of G. Recently, finding constructive maximal monotone extensions instead of using Zorn’s lemma has been a very active topic [5,4,10–12]. In [10], Crouzeix and OcañaAnaya gave an algorithm to find maximally monotone linear subspace extensions of G, but it is not clear what the maximally monotone extensions are analytically. In this paper, we provide some maximally monotone extensions of G with closed analytical forms. Along the way, we also give a new proof to Crouzeix and Ocaña-Anaya’s characterizations on monotonicity and maximal monotonicity of G. Our key tool is the Brezis-Browder characterization of maximally monotone linear relations. The paper is organized as follows. In the remainder of this introductory section, we describe some central notions fundamental to our analysis. In Sect. 2, we collect some auxiliary results for future reference and for the reader’s convenience. Section 3 provides explicit self-dual maximally monotone extensions by using subspaces on which AB T + B AT is negative semidefinite, and obtain a complete characterization of all maximally monotone extensions. Section 4 deals with Minty’s parameterizations of monotone operator G. In Sect. 5, we get some explicit maximally monotone extensions with the same domain or the same range by utilizing normal cone operators. In Sect. 6, we illustrate our maximally monotone extensions by considering three examples. Our notations are standard. We use domG = x ∈ Rn | Gx = ∅ for the domain of G, ranG = G(Rn ) for the range of G and ker G = x ∈ Rn | 0 ∈ Gx for the kernel of G. Given a subset C of Rn , spanC is the span (the set of all finite linear combinations) of C. We set C ⊥ = {x ∗ ∈ Rn | (∀c ∈ C) x ∗ , c = 0}. Then the adjoint of G, denoted by G ∗ , is defined by graG ∗ = {(x, x ∗ ) ∈ Rn × Rn | (x ∗ , −x) ∈ (graG)⊥ }.
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Let Id : Rn → Rn denote the identity mapping, i.e., Idx = x for x ∈ Rn . We also set PX : Rn × Rn → Rn : (x, x ∗ ) → x, and PX ∗ : Rn × Rn → Rn : (x, x ∗ ) → x ∗ . If X , Y are subspaces of Rn , we let X + Y = x + y | x ∈ X, y ∈ Y . Counting multiplicities, let λ1 , λ2 , . . . , λk be all positive eigenvalues of (AB T + B AT ) ∈ R p× p and (1.5) λk+1 , λk+2 , . . . , λ p be nonpositive eigenvalues of (AB T + B AT ). (1.6) Moreover, let vi ∈ R p be an eigenvector of eigenvalue λi of (AB T + B AT ) satisfying vi = 1, and vi , v j = 0 for 1 ≤ i = j ≤ p. It will be convenient to put ⎛
λ1 ⎜0 ⎜ ⎜ Idλ = diag(λ1 , . . . , λ p ) = ⎜ ⎜0 ⎜ .. ⎝. 0
⎞ 0 0 ··· 0 λ2 0 · · · 0 ⎟ ⎟ .. ⎟ p× p . ⎟ 0 λ3 . ⎟ , V = v1 v2 . . . v p ∈ R ⎟ .. .0 ⎠ 0 0 0 0 0 λp (1.7)
2 Auxiliary results on linear relations In this section, we collect some facts and preliminary results which will be used in sequel. We first provide a result about subspaces on which a linear operator from Rn → Rn , i.e, an n × n matrix, is monotone. For M ∈ Rn×n , define three subspaces of Rn , namely, the positive eigenspace, null eigenspace and negative eigenspace associated with M + M T by ⎧ ⎫ ⎨ w1 , . . . , ws : wi is an eigenvector of positive eigenvalue ⎬ V+ (M) = span αi of M + M T , wi , w j = 0, ∀ i = j ⎩ ⎭ wi = 1, i, j = 1, . . . , s. ⎧ ⎫ ⎨ ws+1 , . . . , wl : wi is an eigenvector of 0 eigenvalue of ⎬ V0 (M) = span M + M T , wi , w j = 0, ∀ i = j ⎩ ⎭ wi = 1, i, j = s + 1, . . . , l. ⎧ ⎫ ⎨ wl+1 , . . . , wn : wi is an eigenvector of negative ⎬ V− (M) = span eigenvalue αi of M + M T , wi , w j = 0 ⎩ ⎭ ∀ i = j, wi = 1, i, j = l + 1, . . . , n. which is possible since a symmetric matrix always has a complete orthonormal set of eigenvectors, [15, pp. 547–549].
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Proposition 2.1 Let M be an n × n matrix. Then (i) M is strictly monotone on V+ (M). Moreover, M + M T : V+ (M) → V+ (M) is a bijection. (ii) M is monotone on V+ (M) + V0 (M). (iii) −M is strictly monotone on V− (M). Moreover, −(M + M T ) : V− (M) → V− (M) is a bijection. (iv) −M is monotone on V− (M) + V0 (M). (v) For every x ∈ V0 (M), (M + M T )x = 0 and x, M x = 0. In particular, the orthogonal decomposition holds: Rn = V+ (M)⊕V0 (M)⊕V− (M). s li wi for some (l1 , . . . , ls ) ∈ Rs . Since Proof (i): Let x ∈ V+ (M). Then x = i=1 {w1 , . . . , ws } is a set of orthonormal vectors, they are linearly independent so that x = 0 ⇔ (l1 , . . . , ls ) = 0. Note that αi > 0 when i = 1, . . . , s and wi , w j = 0 for i = j. We have
T
2x, M x = x, (M + M )x = =
s
s
T
li wi , (M + M )
i=1
li wi ,
i=1
s
li αi wi =
i=1
s
li wi
i=1 s
αi li2 > 0
i=1
if x = 0. s For every x ∈ V+ (M) with x = i=1 li wi , we have (M + M T )x =
s
li (M + M T )wi =
i=1
s
αi li wi ∈ V+ (M).
i=1
As αi > 0 for i = 1, . . . , s and {w1 , . . . , ws } is an orthonormal basis of V+ (M), we conclude that M + M T : V+ (M) → V+ (M) is a bijection. (ii): Let x ∈ V+ (M) + V0 (M). Then x = li=1 li wi for some (l1 , . . . , ll ) ∈ Rl . Note that αi ≥ 0 when i = 1, . . . , l and wi , w j = 0 for i = j. We have 2x, M x = x, (M + M T )x = =
l
li wi , (M + M T )
i=1
l i=1
li wi ,
l i=1
li αi wi =
li wi
i=1 l
αi li2 ≥ 0.
i=1
The proofs for (iii), (iv) are similar as (i), (ii). (v): Obvious.
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Corollary 2.2 The following hold: (i) graT = {(B T u, AT u) | u ∈ V+ (B AT )} is strictly monotone. (ii) graT = {(B T u, AT u) | u ∈ V+ (B AT ) + V0 (B AT )} is monotone. (iii) graT = {(B T u, −AT u) | u ∈ V− (B AT )} is strictly monotone. (iv) graT = {(B T u, −AT u) | u ∈ V− (B AT ) + V0 (B AT ))} is monotone. Proof As B T u, AT u = u, B AT u ∀u ∈ Rn , the result follows from Proposition 2.1 by letting M = B AT . Lemma 2.3 For every subspace S ⊆ R p , the following hold. dim{(B T u, AT u) | u ∈ S} = dim S. T
T
dim{(B u, −A u) | u ∈ S} = dim S. Proof See [15, p. 208, Exercise 4.4.9].
(2.1) (2.2)
The following fact is straightforward from the definition of V . Fact 2.4 We have (AB T + B AT )V = V Idλ . Two key criteria concerning maximally monotone linear relations come as follows: Fact 2.5 (See [2, Corollary 6.7], [3, Proposition 2.10].) Let T : Rn ⇒ Rn be a monotone linear relation. The following are equivalent: (i) T is maximally monotone. (ii) dim graT = n. (iii) domT = (T 0)⊥ .
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Fact 2.6 (Brezis-Browder) (See [8, Theorem 2], [14, Theorem 10], [20] or [17].) Let T : Rn ⇒ Rn be a monotone linear relation. Then the following statements are equivalent. (i) T is maximally monotone. (ii) T ∗ is maximal monotone. (iii) T ∗ is monotone. Some basic properties of G are: Lemma 2.7 (i) graG = ker(A B). (ii) G0 = ker B, G −1 (0) = ker A. (iii) domG = PX (ker(A B)) and ranG = PX ∗ (ker(A B)). (iv) ran(G + Id) = PX ∗ (ker(A − B B)) = PX (ker(A B − A)), and domG = PX (ker(A − B B)), ranG = PX ∗ (ker(A (B − A)). (v) dim gra G = 2n − p. Proof (i), (ii), (iii) follow from definition of G. Since Ax + Bx ∗ = 0 ⇔ (A − B)x + B(x + x ∗ ) = 0 ⇔ A(x + x ∗ ) + (B − A)x ∗ = 0, (iv) holds. (v): We have 2n = dim ker(A B) + dim ran
AT BT
= dim graG + p.
Hence dim graG = 2n − p.
The following result summarizes the monotonicities of G ∗ and G. Lemma 2.8 The following hold. (i) graG ∗ = {(B T u, −AT u) | u ∈ R p }. (ii) G ∗ is monotone ⇔ the matrix AT B + B T A ∈ R p× p is negative-semidefinite. (iii) Assume G is monotone. Then n ≤ p. Moreover, G is maximally monotone if and only dim graG = n = p. Proof (i): By Lemma 2.7(i), we have (x, x ∗ ) ∈ graG ∗ ⇔ (x ∗ , −x) ∈ graG ⊥ = ran
AT BT
= {(AT u, B T u) | u ∈ R p }.
Thus graG ∗ = {(B T u, −AT u) | u ∈ R p }. (ii): Since graG ∗ is a linear subspace, by (i), G ∗ is monotone ⇔ B T u, −AT u ≥ 0, ∀u ∈ R p ⇔ u, −B AT u ≥ 0, ∀u ∈ R p
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⇔ u, B AT u ≤ 0, ∀u ∈ R p ⇔ u, (AT B + B T A)u ≤ 0, ∀u ∈ R p ⇔ (AT B + B T A) is negative semidefinite. (iii): By Fact 2.5 and Lemma 2.7(v), 2n − p = dim graG ≤ n ⇒ n ≤ p. By Fact 2.5 and Lemma 2.7(v) again, G is maximally monotone ⇔ 2n − p = dim graG = n ⇔ dim graG = p = n. 2.1 One linear relation: two equivalent formulations The linear relation G given by (1.2)–(4): graG = {(x, x ∗ ) ∈ Rn × Rn | Ax + Bx ∗ = 0}
(2.3)
is an intersection of p linear hyperplanes. It can be equivalently described as a span of q = 2n − p points in Rn × Rn . Indeed, for (2.3) we can use Gaussian elimination to reduce (A B) to row echelon form. Then back substitution to solve basic variables in terms of the free variables, see [15, p. 61]. Row-echelon form gives
x x∗
= h 1 y1 + · · · + h 2n− p y2n− p =
C D
y
where y ∈ R2n− p and
C D
= (h 1 , . . . , h 2n− p )
with C, D being n × (2n − p) matrices. Therefore, graG =
Cy Dy
| y∈R
2n− p
C = ran D
(2.4)
which is a span of 2n − p points in Rn × Rn . The two formulations (2.3) and (2.4) coincide when p = q = n, Id = −B = C and D = A in which Id ∈ Rn×n . 3 Explicit maximally monotone extensions of monotone linear relations In this section, we give explicit maximally monotone linear subspace extensions of G by using V+ (AB T ) or Vg . A characterization of all maximally monotone extensions of G is also given. We also provide a new proof to Crouzeix and Ocaña-Anaya’s characterizations on monotonicity and maximal monotonicity of G. We shall use notations given in (1.2)–(1.7), in particular, G is in the form of (2.3).
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and G be defined by Lemma 3.1 Let M ∈ R p× p , and linear relations G = {(x, x ∗ ) | M T Ax + M T Bx ∗ = 0} gra G = {(B T u, −AT u) | u ∈ ranM}. gra G ∗ = G. Then (G) Proof Let (y, y ∗ ) ∈ Rn × Rn . Then we have ∗ (y, y ∗ ) ∈ gra(G) ⊥ = (ker M T A M T B ⊥ = ran ⇔ (y ∗ , −y) ∈ (graG)
AT M BT M
⇔ (y, y ∗ ) ∈ graG. ∗ = G. Hence (G)
and G by Lemma 3.2 Define linear relations G = {(x, x ∗ ) | Vg Ax + Vg Bx ∗ = 0} gra G = {(B T u, −AT u) | u ∈ V− (B AT ) + V0 (B AT )}, gra G where Vg is ( p − k) × p matrix defined by ⎛
T vk+1
⎞
⎜ T ⎟ ⎜v ⎟ k+2 ⎟ . Vg = ⎜ ⎜ .. ⎟ ⎝. ⎠ v Tp Then is monotone. (i) G ∗ = G. (ii) (G) T B T u | u ∈ V+ (B A ) . (iii) gra G = graG + AT Proof (i): Apply Corollary 2.2(iv). (ii): Notations are as in (1.7). Define the p × p matrix N by N=
00 0 Id
in which Id ∈ R( p−k)×( p−k) . Then we have T 00 0 . N T V T = (v1 . . . vk VgT ) = 0 Id Vg
123
(3.1)
Maximally monotone linear subspace extensions
335
Then we have Vg Ax + Vg Bx ∗ = 0 ⇔
0 Vg Ax + Vg Bx ∗
=0
⇔ N T V T Ax + N T V T Bx ∗ = 0, ∀(x, x ∗ ) ∈ Rn × Rn . Hence = {(x, x ∗ ) | N T V T Ax + N T V T Bx ∗ = 0}. graG Thus by Lemma 3.1 with M = V N , ∗ = {(B T u, −AT u) | u ∈ ranV N = ran 0 VgT gra(G) = V− (B AT ) + V0 (B AT )} = graG. ∗ = (G) ∗∗ = G. Hence (G) (iii): Let J be defined by
BT AT
graJ = graG +
u | u ∈ V+ (B AT ) .
Then we have (graJ )⊥ = (graG)⊥ ∩
BT AT
⊥ u | u ∈ V+ (B AT )
.
By Lemma 2.7(i), ⊥
graG =
AT BT
w|w∈R
p
Then
AT BT
w∈
BT AT
⊥ T
u | u ∈ V+ (B A )
if and only if (AT w, B T w), (B T u, AT u) = 0 ∀ u ∈ V+ (B AT ), that is, AT w, B T u+B T w, AT u = w, (AB T + B AT )u = 0 ∀u ∈ V+ (AB T ). (3.2)
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Because AB T + B AT : V+ (AB T ) → V+ (AB T ) is onto by Proposition 2.1(i), we obtain that (3.2) holds if and only if w ∈ V− (AB T ) + V0 (AB T ). Hence (graJ )⊥ = {(AT w, B T w) | w ∈ V− (B AT ) + V0 (B AT )}, Then by (i), from which graJ ∗ = graG. = gra(G) ∗ = graJ ∗∗ = graJ. graG We are ready to apply Brezis-Browder Theorem, namely Fact 2.6, to improve Crouzeix and Ocaña-Anaya’s characterizations of monotonicity and maximal monotonicity of G and provide a different proof. G be defined in Lemma 3.2. The following are equivalent: Theorem 3.3 Let G, (i) (ii) (iii) (iv) (v)
G is monotone; is monotone; G is maximally monotone; G is maximally monotone; G dim V+ (B AT ) = p − n, equivalently, AB T + B AT has exactly p − n positive eigenvalues (counting multiplicity).
Proof (i)⇔(ii): Lemma 3.2(iii) andCorollary 2.2(i). = G ∗ and G is always a monotone linear relation (ii)⇔(iii)⇔(iv): Note that G by Corollary 2.2(iv). It suffices to combine Lemma 3.2 and Fact 2.6. is monotone by Lemma 3.2(iii) and (i)⇒(v): Assume that G is monotone. Then G is maximally Corollary 2.2(i). By Lemma 3.2(ii), Corollary 2.2(iv) and Fact 2.6, G = p−k = n by Fact 2.5 and Lemma 2.3, thus k = p−n. monotone, so that dim(graG) Note that for each eigenvalue of a symmetric matrix, its geometric multiplicity is the same as its algebraic multiplicity [15, p. 512]. = p − k = n by Lemma 2.3, (v)⇒(i): Assume that k = p − n. Then dim(graG) is maximally monotone by Fact 2.5(i)(ii). By Lemma 3.2(ii) and Fact 2.6, so that G is monotone, which implies that G is monotone. G Corollary 3.4 Assume that G is monotone. Then = graG + gra G
BT AT
u | u ∈ V+ (B A T
= {(x, x ∗ ) | Vg Ax + Vg Bx ∗ = 0} is a maximally monotone extension of G, where ⎛
v Tp−n+1
⎞
⎜ T ⎟ ⎜ v p−n+2 ⎟ ⎟. Vg = ⎜ ⎜ .. ⎟ ⎝. ⎠ v Tp
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Proof Combine Theorem 3.3 and Lemma 3.2(iii) directly.
Note that Corollary 3.4 gives both types of maximally monotone extensions of G, namely, type (2.3) and type (2.4). A remark is in order to compare our extension with the one by Crouzeix and Ocaña-Anaya. Remark 3.5 (i). Crouzeix and Ocaña-Anaya [10] defines the union of monotone extension of G as T B u|u∈K , S = graG + AT where K = {u ∈ Rn | u, (AB T + B AT )u ≥ 0}. Although this is the set monotonically related to G, it is not monotone in general as long as (AB T + B AT ) has both positive eigenvalues and negative eigenvalues. Indeed, let (α1 , u 1 ) and (α2 , u 2 ) be eigen-pairs of (AB T + B AT ) with α1 > 0 and α2 < 0. We have u 1 , (AB T + B AT )u 1 = α1 u 1 2 > 0, u 2 , (AB T + B AT )u 2 = α2 u 2 2 < 0. Choose > 0 sufficiently small so that u 1 + u 2 , (AB T + B AT )(u 1 + u 2 ) > 0. Then
However,
BT AT
BT AT
u1,
BT AT
(u 1 + u 2 ) −
(u 1 + u 2 ) ∈ S.
BT AT
u1 =
BT AT
u2
has B T u 2 , AT u 2 = 2 u 2 , B AT u 2 = 2
u 2 , (AB T + B AT )u 2 < 0. 2
Therefore S is not monotone. By using V+ (B AT ) ⊆ K , we have obtained a maximally monotone extension of G. (ii). Crouzeix and Ocaña-Anaya [10] find a maximally monotone linear subspace extension of G algorithmically by using u˜ k ∈ graG k \graG k and construct graG k+1 = graG k + Ru˜ k where T Bk u˜ k = u k , u k , (Ak BkT + Bk ATk )u k ≥ 0. ATk This recursion is done until dim graG k = n. In particular, each u k may be chosen as an eigenvector associated with a positive eigenvalue of Ak BkT + Bk ATk , which is
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possible since p > n when G k is not maximally monotone. Their construction uses both formulations, namely, (2.3) and (2.4). No computation indications are given on the passage from one formulation to the other one. The following result extends the characterization of maximally monotone linear relations given by Crouzeix and Ocaña-Anaya [10]. G be defined in Lemma 3.2. The following are equivalent: Theorem 3.6 Let G, (i) (ii) (iii) (iv)
G is maximally monotone; p = n and G is monotone; p = n and AB T + B AT is negative semidefinite. is maximally monotone. p = n and G
Proof (i)⇒(ii): Apply Lemma 2.8(iii). (ii)⇒(iii): Apply directly Theorem 3.3(i)(v). (iii)⇒(i): Assume that p = n and (AB T + B AT ) is negative semidefinite. Then = G. It follows that dim(graG) = p − k = n by Lemma 2.3, so that G k = 0 and G ∗ is maximally monotone by Corollary 2.2(iv) and Fact 2.5(i)(ii). Since G = G by = G is maximally monotone. Lemma 3.2(ii), Fact 2.6 gives that G (iii)⇒(iv): Assume that p = n and (AB T + B AT ) is negative semidefinite. We have = p − k = n − 0 = n. Hence (iv) holds by Corollary 2.2(iv) k = 0 and dim(graG) and Fact 2.5(i)(ii). is maximally monotone and p = n. We have (iv)⇒(iii): Assume that G = p−k =n−k =n dim(graG) so that k = 0. Hence (AB T + B AT ) is negative semidefinite.
Corollary 3.4 supplies only one maximally monotone linear subspace extension of G. Can we find all of them? Surprisingly, we may give a characterization of all the maximally monotone linear subspace extensions of G when it is given in the form of (2.3). is a maximally monotone extension of G Theorem 3.7 Let G be monotone. Then G p× p with rank of n such that N T Idλ N is negative if and only if there exists N ∈ R semidefinite and = {(x, x ∗ ) | N T V T Ax + N T V T Bx ∗ = 0}. gra G
(3.3)
Proof “⇒”: By Lemma 2.8(i), we have graG ∗ = {(B T u, −AT u) | u ∈ R p }.
(3.4)
and thus gra(G) ∗ is a subspace of graG ∗ . Since graG ⊆ graG Thus by (3.4), there exists a subspace F of R p such that ∗ = {(B T u, −AT u) | u ∈ F}. gra(G)
123
(3.5)
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339
By Facts 2.6, 2.5 and Lemma 2.3, we have dim F = n.
(3.6)
Thus, there exists N ∈ R p× p with rank n such that ranV N = F and ∗ = {(B T V N y, −AT V N y) | y ∈ R p }. gra(G)
(3.7)
is maximally monotone, (G) ∗ is maximally monotone by Fact 2.6, so As G N T V T (B AT + AB T )V N is negative semidefinite. Using Fact 2.4, we have N T Idλ N = N T V T V Idλ N = N T V T (AB T + B AT )V N
(3.8)
which is negative semidefinite. (3.3) follows from (3.7) by Lemma 3.1 using M = V N . “⇐”: By Lemma 3.1, we have ∗ = {(B T V N u, −AT V N u) | u ∈ R p }. gra(G)
(3.9)
∗ is monotone because N T V T (AB T + B AT )V N = N T Idλ N is negObserve that (G) ative semidefinite by Fact 2.4 and the assumption. As rank(V N ) = n, it follows from ∗ is maximally monotone ∗ = n. Therefore (G) (3.9) and Lemma 2.3 that dim gra(G) ∗ = (G) ∗∗ is maximally by Fact 2.5. Applying Fact 2.6 for T = (G) yields that G monotone. From the above proof, we see that to find a maximally monotone extension of G one essentially need to find subspace F ⊆ R p such that dim F = n and AB T + B AT is negative semidefinite on F. If F = ranM and M ∈ R p× p with rankM = n, one can let N = V T M. The maximally monotone linear subspace extension of G is = {(x, x ∗ ) | M T Ax + M T Bx ∗ = 0}. G In Corollary 3.4, one can choose M = 0 · · · 0$ v p−n+1 . . . v p . ! 0"# n
is a maximally monotone extension of G Corollary 3.8 Let G be monotone. Then G p× p if and only if there exists M ∈ R with rank of n such that M T (AB T + B AT )M is negative semidefinite and = {(x, x ∗ ) | M T Ax + M T Bx ∗ = 0}. gra G
(3.10)
Note that G may have different representations in terms of A, B. The maximally given in Theorem 3.7 and Corollary 3.4 relies on A, B matrimonotone extension of G ces and N . This might leads different maximally monotone extensions, see Sect. 6.
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Remark 3.9 A referee points out that there is a shorter way to see Theorem 3.7. Consider the maximally monotone linear subspace extension of G of type: + = {(x, x ∗ ) ∈ Rn × Rn | Ax Bx ∗ = 0} ⊇ graG graG where A, B ∈ Rn×n . With the nonsingular p × p matrix V given as in (1.7), an equivalent formulation of G is graG = {(x, x ∗ ) ∈ Rn × Rn | V T Ax + V T Bx ∗ = 0}. is maximally monotone, the n × 2n matrix has rank( A, As G B) = n and the matrix T ∈ Rn×n BA A BT + ⊇ graG, we have is negative semidefinite. Since graG ran
T A BT
T T (V A) ⊥ ⊥ = (graG) ⊆ (graG) = ran . (V T B)T
Therefore, there exists a p × n matrix N with rank N = n such that
T A BT
=
(V T A)T (V T B)T
N=
(V T A)T N (V T B)T N
= N T V T A, from which A B = N T V T B. Then the n × n matrix T = N T V T A(N T V T B)T + N T V T B(N T V T A)T BA A BT + T
T
T
T
= N V (AB + B A )V N T
= N Idλ N .
(3.11) (3.12) (3.13)
Therefore, all maximally monotone linear subspace extensions of G can be obtained by using = {(x, x ∗ ) ∈ Rn × Rn | N T V T Ax + N T V T Bx ∗ = 0} graG in which the p × n matrix N satisfying rankN = n and N T Idλ N being negative semidefinite. 4 Minty parameterizations Although G is set-valued in general, when G is monotone it has a beautiful Minty parametrization in terms of A, B, which is what we are going to show in this section.
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341
Lemma 4.1 The linear relation G is monotone if and only if y2 − y ∗ 2 ≥ 0, whenever ∗
(A + B)y + (B − A)y = 0.
(4.1) (4.2)
Consequently, if G is monotone then the p × n matrix B − A must have full column rank, namely n. Proof Define the 2n × 2n matrix P=
0 Id Id 0
where Id ∈ Rn×n . It is easy to see that G is monotone if and only if %
∗
(x, x ), P
x x∗
& ≥ 0,
whenever Ax + Bx ∗ = 0. Define the orthogonal matrix 1 Q=√ 2
Id −Id Id Id
and put
x x∗
=Q
y y∗
.
Then G is monotone if and only if y2 − y ∗ 2 ≥ 0, whenever ∗
(A + B)y + (B − A)y = 0.
(4.3) (4.4)
If (B − A) does not have full column rank, then there exists y ∗ = 0 such that (B − A)y ∗ = 0. Then (0, y ∗ ) satisfies (4.4) but (4.3) fails. Therefore, B − A has to be full column rank. Theorem 4.2 (Minty parametrization) Assume that G is a monotone operator. Then (x, x ∗ ) ∈ graG if and only if 1 [Id + (B − A)† (B + A)]y 2 1 x ∗ = [Id − (B − A)† (B + A)]y 2 x=
(4.5) (4.6)
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for y = x + x ∗ ∈ ran(Id + G). Here the Moore-Penrose inverse (B − A)† = [(B − A)T (B − A)]−1 (B − A)T . In particular, when G is maximally monotone, we have graG = {((B − A)−1 By, −(B − A)−1 Ay) | y ∈ Rn }. Proof As (B − A) is full column rank, (B − A)T (B − A) is invertible. It follows from (4.2) that (B − A)T (A + B)y + (B − A)T (B − A)y ∗ = 0 so that y ∗ = −((B − A)T (B − A))−1 (B − A)T (A + B)y = −(B − A)† (A + B)y. Then 1 x = √ (y − y ∗ ) = 2 1 x ∗ = √ (y + y ∗ ) = 2 ∗ x+x √ 2
where y =
1 √ [Id + (B − A)† (B + A)]y 2 1 √ [Id − (B − A)† (B + A)]y 2
with (x, x ∗ ) ∈ graG. Since ran(Id + G) is a subspace, we have 1 [Id + (B − A)† (B + A)] y˜ 2 1 x ∗ = [Id − (B − A)† (B + A)] y˜ 2 x=
with y˜ = x + x ∗ ∈ ran(Id + G). If G is maximally monotone, then p = n by Theorem 3.6 and hence B − A is invertible, thus (B − A)† = (B − A)−1 . Moreover, ran(G + Id) = Rn . Then (4.5) and (4.6) transpire to 1 (B − A)−1 [B − A + (B + A)]y = (B − A)−1 By 2 1 x ∗ = (B − A)−1 [(B − A) − (B + A)]y = −(B − A)−1 Ay 2 x=
for y ∈ Rn .
(4.7) (4.8)
Remark 4.3 See Lemma 2.7 for ran(G + Id). Note that as G is a monotone linear relation, the mapping z → ((G + Id)−1 , Id − (G + Id)−1 )(z) is bijective and linear from ran(G + Id) to graG, therefore dim(ran(G + Id)) = dim(graG).
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343
defined in Corollary 3.4, the Corollary 4.4 Let G be a monotone operator. Then G maximally monotone extension of G, has its Minty parametrization given by = {((Vg B − Vg A)−1 Vg By, −(Vg B − Vg A)−1 Vg Ay) | y ∈ Rn } gra G where Vg is given as in Corollary 3.4. Proof Since rank(Vg ) = n and rank(A B) = p, by Lemma 2.3(2.1), rank(Vg A Vg B) = n. Apply Corollary 3.4 and Theorem 4.2 directly.
Corollary 4.5 When G is maximally monotone, domG = (B − A)−1 (ranB), ranG = (B − A)−1 (ran A). Recall that T : Rn → Rn is firmly nonexpansive if T x − T y2 ≤ T x − T y, x − y ∀ x, y ∈ domT. In terms of matrices Corollary 4.6 Suppose that p = n, AB T + B AT is negative semidefinite. Then (B − A)−1 B and −(B − A)−1 A are firmly nonexpansive. Proof By Theorem 3.6, G is maximally monotone. Theorem 4.2 gives that (B − A)−1 B = (Id + G)−1 , −(B − A)−1 A = (Id + G −1 )−1 . Being resolvents of monotone operators G, G −1 , they are firmly nonexpansive, see [1,13] or [4, Fact 2.5]. 5 Maximally monotone extensions with the same domain or the same range How do we find maximally monotone linear subspace extensions of G if it is given in the form of (2.4)? The purpose of this section is to find maximally monotone linear subspace extensions of G which keep either domG or ranG unchanged. For a closed convex set S ⊆ Rn , let N S denote its normal cone mapping. Proposition 5.1 Assume that T : Rn ⇒ Rn is a monotone linear relation. Then (i) T1 = T + NdomT , i.e., x → T1 x =
T x + (domT )⊥ ∅
if x ∈ domT otherwise
is maximally monotone. In particular, domT1 = domT . (ii) T2 = (T −1 + NranT )−1 is a maximally monotone extension of T and ranT2 = ranT .
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Proof (i): Since 0 ∈ T 0 ⊆ (domT )⊥ by [2, Proposition 2.2(i)], we have T1 0 = T 0 + (domT )⊥ = (domT )⊥ so that domT1 = domT = (T1 0)⊥ . Hence T1 is maximally monotone by Fact 2.5. (ii): Apply (i) to T −1 to see that T −1 + NranT is a maximally monotone extension of T −1 with dom(T −1 + NranT ) = ranT . Therefore, T2 is a maximally monotone extension of T with ranT2 = ranT . Define linear relations E i : Rn ⇒ Rn (i = 1, 2) by
Cy Dy
0 (ranC)⊥
, Cy (ranD)⊥ | y ∈ R2n− p . graE 2 = + Dy 0
graE 1 =
+
| y∈R
2n− p
(5.1) (5.2)
Theorem 5.2 (i) E 1 is a maximally monotone extension of G with domE 1 = domG. Moreover, 0 0 C C = ran + . (5.3) graE 1 = ran + D D (ranC)⊥ ker C T (ii) E 2 is a maximally monotone extension of G with ranE 2 = ranG. Moreover, C C (ranD)⊥ ker D T . (5.4) = ran graE 2 = ran + + D D 0 0 Proof (i): Note that domG = ranC. The maximal monotonicity follows from Proposition 5.1. (5.3) follows from (5.1) and that (ranC)⊥ = ker C T [15, page 405]. (ii): Apply (i) to G −1 , i.e., Dy (5.5) graG −1 = | y ∈ R2n− p Cy and followed by taking the set-valued inverse. Apparently, both extensions E 1 , E 2 rely on graG, domG, ranG, not on the A, B. In this sense, E 1 , E 2 are intrinsic maximally monotone linear subspace extensions. Remark 5.3 Theorem 5.2 is much easier to use than Corollary 3.8 when G is written in the form of (2.4). Indeed, it is not hard to check that gra(E 1∗ ) = {(B T u, −AT u) | B T u ∈ domG, u ∈ R p }.
(5.6)
gra(E 2∗ )
(5.7)
T
T
T
= {B u, −A u) | A u ∈ ranG, u ∈ R }. p
According to Fact 2.6, E i∗ is maximally monotone and dim E i∗ = n. This implies that dim{u ∈ R p | B T u ∈ domG} = n,
123
dim{u ∈ R p | AT u ∈ ranG} = n.
Maximally monotone linear subspace extensions
345
Let Mi ∈ R p× p with rankMi = n and {u ∈ R p | B T u ∈ domG} = ranM1 ,
(5.8)
T
{u ∈ R | A u ∈ ranG} = ranM2 . p
(5.9)
Corollary 3.8 shows that graE i = {(x, x ∗ ) | MiT Ax + MiT Bx ∗ = 0}. However, finding Mi from (5.8) and (5.9) may not be easy as it seems. Remark 5.4 Unfortunately, we do not know how to determine all maximally monotone linear subspace extensions of G if it is given in the form of (2.4). 6 Examples In the final section, we illustrate our maximally monotone extensions by considering rely three examples. In particular, they show that maximally monotone extensions G on the representation of G in terms of A, B and choices of N we shall use. However, the maximally monotone extensions E i are intrinsic, only depending on graG. Example 6.1 Consider Id 0 ∗ n n ∗ graG = (x, x ) ∈ R × R | x+ x =0 0 C where C ∈ Rn×n is symmetric and positive definite, and Id ∈ Rn×n . Clearly, graG =
0 . 0
We have α defined by (i) For every α ∈ [−1, 1] , G ' α = graG
n {(0, ( R )}, ) 1+α −1 (x, 1−α C x) | x ∈ Rn ,
if α = 1; otherwise
is a maximally monotone linear extension of G. 1 and E 2 = G −1 . (ii) E 1 = G α , we need eigenvectors of Proof (i): To find G A=
0 Id 0 C (Id 0) = (0 C T ) + . C 0 C0
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Counting multiplicity, the positive definite matrix C has eigen-pairs (λi , wi ) (i = 1, . . . , n) such that λi > 0, wi = 1 and wi , w j = 0 for i = j. As such, the matrix A has 2n eigen-pairs, namely wi λi , wi and wi −λi , −wi with i = 1, . . . , n. Put W = (w1 . . . wn ) ∈ Rn×n and write W W V = . W −W Then W T C W = D = diag(λ1 , λ2 , . . . , λn ). In Theorem 3.7, take Nα =
0 αId 0 Id
∈ R2n×2n
where Id ∈ Rn×n . We have rank Nα = n, 00 00 = NαT Idλ Nα = 0 (α 2 − 1)D 0 (α 2 − 1)W T C W being negative semidefinite, and V Nα =
0 (1 + α)W 0 (α − 1)W
.
α Then by Theorem 3.7, we have an maximally monotone linear extension G given by 0 ∗ n n =0 graG α = (x, x ) ∈ R × R | (1 + α)W T x + (α − 1)W T C x ∗ = (x, x ∗ ) ∈ Rn × Rn | (1 + α)x + (α − 1)C x ∗ = 0 ' {(0, Rn )}, if α = 1; ( ) = 1+α −1 n (x, 1−α C x) | x ∈ R , otherwise. Hence we get the result as desire.
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347
(ii): It is immediate from Theorem 5.2 and (i). Example 6.2 Consider ⎧ ⎨
⎛ ⎞ 1 −1 0 x graG = (x, x ∗ ) ∈ R2 × R2 | ⎝ 0 0 ⎠ 1 + ⎝ 0 x2 ⎩ 0 0 −1 ⎛
⎫ ⎞ 0 ∗ ⎬ x 1 ⎠ 1∗ = 0 . x2 ⎭ 1
Then i : R2 ⇒ R2 for i = 1, 2 given by (i) the linear operators G 1 = G
10 √ 0 −1+√ 2
2 = , G
2− 2
1 0
2 5 √
2 10
are two maximally monotone extensions of G. (ii) E 1 (x1 , 0) = (x1 , R) ∀x1 ∈ R. (iii) E 2 (x1 , y) = (x1 , 0) ∀x1 , y ∈ R. Proof We have ⎫ ⎧⎛ ⎞ x1 ⎪ ⎪ ⎪ ⎪ ⎬ ⎨⎜ ⎟ 0 ⎟ | x1 ∈ R graG = ⎜ ⎝ ⎠ ⎪ ⎪ ⎪ ⎪ x1 ⎭ ⎩ 0 is monotone. Since dim G = 1, G is not maximally monotone by Fact 2.5. The matrix ⎛ ⎞ −2 0 0 T T AB + B A = ⎝ 0 0 −1 ⎠ 0 −1 −2 has a positive eigenvalue −1 +
√ 2 with an eigenvector
⎛
⎞ 0 u = ⎝1 √ ⎠ 1− 2
so that
T
B AT
⎞ 0 √ ⎜2 − 2 ⎟ ⎟. u=⎜ ⎠ ⎝0 √ −1 + 2 ⎛
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Then by Corollary 3.4, ⎧⎛ ⎞ ⎫ ⎧⎛ ⎫ ⎞ 0 √ x1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎟ ⎬ ⎨⎜ ⎬ ⎟ 2 − 0 2 ⎟ x2 | x2 ∈ R 1 = ⎜ ⎟ | x1 ∈ R + ⎜ graG ⎝ ⎝ ⎠ ⎠ x1 ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ √ ⎩ ⎩ ⎭ ⎪ ⎭ 0 −1 + 2 ⎧⎛ ⎫ ⎞ x1 √ ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎬ ⎟ 2)x (2 − 2 ⎟ ⎜ | x = ⎝ . , x ∈ R 1 2 ⎠ x1 ⎪ ⎪ ⎪ ⎪ √ ⎩ ⎭ (−1 + 2)x2 Therefore, 1 = G
10 √ 0 −1+√ 2
2− 2
is a maximally monotone extension of G. Now we have √ ⎞ ⎛ ⎛ ⎞ 0 0 1 −1 + 2 0√ 0 1 1 Idλ = ⎝ 0 −1 − 2 0 ⎠ , V = ⎝ − −1+√2 − −1−√2 0 ⎠ . 0 0 −2 1 1 0
(6.1)
Take ⎛
⎞ 0 −1 1 N = ⎝ 0 2 −1 ⎠ . 0 1 1
(6.2)
We have rank N = 2 and ⎞ 0 0 √ 0√ 3 2 1 + 2⎠, N T Idλ N = ⎝ 0 −7 − √ 0 1+ 2 −4 ⎛
(6.3)
being negative semidefinite. By Theorem 3.7, with V, N given in (6.1) and (6.2), we use the NullSpace command in Maple to solve (V N )T Ax + (V N )T Bx ∗ = 0, and get ⎧ ⎛ ⎞ ⎛ √ ⎞⎫ −2 2 ⎪ 1 ⎪ ⎪ ⎬ ⎨ ⎜ ⎟ ⎜ √ ⎟⎪ 0 2 = span ⎜ ⎟ ⎜ 5 2 ⎟ . graG ⎝1⎠⎝0 ⎠⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0 1
123
Maximally monotone linear subspace extensions
2 = Thus G
√ −1 1 1 −2 √ 2 = 05 2 0
2 5 √
349
is another maximally monotone extension of
2 10
G. On the other hand, ⎧⎛ ⎞ ⎫ ⎛ ⎞ ⎧⎛ ⎞ ⎫ x1 0 x1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎟ ⎬ ⎜ ⎟ ⎨⎜ ⎟ ⎬ 0⎟ 0 ⎟ 0 ⎟ ⎜ ⎜ ⎜ graE 1 = ⎝ ⎠ | x1 ∈ R + ⎝ ⎠ = ⎝ ⎠ | x1 ∈ R 0 x1 x1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ 0 R R gives E 1 (x1 , 0) = (x1 , R) ∀x1 ∈ R. And ⎧⎛ ⎞ ⎫ x1 ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎟ ⎬ R⎟ ⎜ graE 2 = ⎝ ⎠ | x1 ∈ R . x1 ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0 gives E 2 (x1 , y) = (x1 , 0) ∀x1 , y ∈ R. In [5], the authors use autoconjugates to find maximally monotone extensions of monotone operators. In general, it is not clear whether the maximally monotone extensions of a linear relation is still a linear relation. As both monotone operators in Examples 6.2 and 6.1 are subset of {(x, x) | x ∈ Rn }, [5, Example 5.10] shows that the maximally monotone extension obtained by autoconjugates must be Id, which is different from the ones given here. Example 6.3 Consider graG = {(x, x ∗ ) ∈ R2 × R2 | Ax + Bx ∗ = 0} where ⎛
1 A = ⎝2 3
⎞ ⎛ 1 1 0⎠, B = ⎝1 1 0
⎞ ⎛ ⎞ 5 1115 7 ⎠ , thus (A B) = ⎝ 2 0 1 7 ⎠ . 2 3102
(6.4)
i : Rn ⇒ Rn for i = 1, 2 given by Then linear operators G ⎛ 1 = G
√ √ −117+17 √ 201 −107+7√ 201 2(−1+ 201) ⎝ 2(−1+ 201) √ √ − −23+3√ 201 − −21+√201 2(−1+ 201) 2(−1+ 201)
⎞
2 = ⎠, G
√ 201 201 13 6√ 4 − 6√ 201 201 29 9 − 20 + 30 − 20 + 30 33 4
−
√
are two maximally monotone linear extensions of G.
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Moreover, ⎧⎛ ⎫ ⎞ ⎛ ⎞ 0 ⎪ ⎪ ⎪ −1 ⎪ ⎨ ⎬ ⎜1 ⎟ ⎜0⎟ ⎜ ⎟ ⎜ ⎟ , x + | x , x ∈ R graE 1 = ⎝ x 1 2 1 2 ⎝1⎠ ⎪ −5 ⎠ ⎪ ⎪ ⎪ ⎩ ⎭ 1 1 ⎧⎛ ⎫ ⎞ ⎛ ⎞ −1 1 ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎬ ⎜5⎟ 1 ⎟ ⎜ ⎟ ⎜ ⎟ graE 2 = ⎝ . x + | x , x ∈ R x 1 2 1 2 ⎝0⎠ −5 ⎠ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 1 0 Proof We have rank(A B) = 3 and ⎛ Idλ = ⎝
13 +
√ ⎞ ⎛ 20 √ 0 201 0 0 1+ 201 ⎠, V = ⎝ 1 0 −6 0√ −1 0 0 13 − 201 1 1
20 √ 1− 201
1 1
⎞ ⎠,
(6.5)
and Vg =
0 20 √ 1− 201
−1 1 1 1
.
(6.6)
Clearly, here p = 3, n = 2 and AB T + B AT has exactly p − n = 3 − 2 = 1 positive eigenvalue. By Theorem 3.3(i)(v), G is monotone. Since AB T + B AT is not negative semidefinite, by Theorem 3.6(i)(iii), G is not maximally monotone. With Vg given in (6.6) and A, B in (6.4), use the NullSpace command in maple to 1 defined by solve Vg Ax + Vg Bx ∗ = 0 and obtain G ⎧⎛ ⎞ ⎛ ⎞⎫ √ √ −21+√201 −107+7√ 201 ⎪ ⎪ − − ⎪ ⎪ ⎪ 2(−1+ 2(−1+√201) ⎟⎪ ⎪ ⎪ √ 201) ⎟ ⎜ ⎬ ⎨⎜ ⎜ −23+3√ 201 ⎟ ⎜ −117+17 ⎟ √ 201 ⎟ ⎜ ⎜ ⎟ graG 1 = span ⎜ 2(−1+ 201) ⎟ , ⎜ 2(−1+ 201) ⎟ . ⎪ ⎪ ⎪ ⎝1 ⎠ ⎝0 ⎠⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0 1 1 is a maximally monotone linear subspace extension of G. Then By Corollary 3.4, G ⎛ 1 = ⎝ G
√ √ −21+√201 − −107+7√ 201 2(−1+ 2(−1+√201) √ 201) −117+17 −23+3√ 201 √ 201 2(−1+ 201) 2(−1+ 201)
−
⎛
⎞−1 ⎠
=
√ √ −117+17 √ 201 −107+7√ 201 2(−1+ 201) ⎝ 2(−1+ 201) √ √ − −23+3√ 201 − −21+√201 2(−1+ 201) 2(−1+ 201)
⎞ ⎠.
Let N be defined by ⎛
⎞ 0 0 15 N = ⎝0 1 0 ⎠. 001
123
(6.7)
Maximally monotone linear subspace extensions
351
Then rankN = 2 and ⎛
0 0 N T Idλ N = ⎝ 0 −6 0 0
0 0√
⎞ ⎠.
338−24 201 25
is negative semidefinite. With N in (6.7), A, B in (6.4) and V in (6.5), use the NullSpace command in maple to solve (V N )T Ax +(V N )T Bx ∗ = 0. By Theorem 3.7, we get a maximally monotone 2 , defined by linear extension of G, G 2 = G
√
√
201 201 9 − 20 +√ 30 − 13 4 + √ 6 201 201 33 29 20 − 30 4 − 6
−1 =
√ 201 201 13 − 6√ 4 6√ 201 201 29 9 − 20 + 30 − 20 + 30 33 4
−
√
.
To find E 1 and E 2 , using the LinearSolve command in Maple, we get graG = C ran , where D C=
−1 , 1
D=
−5 . 1
It follows from Theorem 5.2 that ⎧⎛ ⎫ ⎞ ⎛ ⎞ −1 0 ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎬ ⎟ ⎜0⎟ 1 ⎟ x1 + ⎜ ⎟ x2 | x1 , x2 ∈ R , graE 1 = ⎜ ⎝ −5 ⎠ ⎝1⎠ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 1 1 ⎧⎛ ⎫ ⎞ ⎛ ⎞ −1 1 ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎬ ⎜5⎟ 1 ⎟ ⎜ ⎜ ⎟ ⎟ graE 2 = ⎝ . + | x , x ∈ R x x 1 2 1 2 ⎝0⎠ −5 ⎠ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 1 0 Acknowledgments The authors thank two anonymous referees for their constructive comments which have improved the presentation of the paper. The authors also thank Dr. Heinz Bauschke for bring their attentions of [10] and many valuable discussions.
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4. Bauschke, H.H., Wang, X.: Firmly nonexpansive and Kirszbraun-Valentine extensions: a constructive approach via monotone operator theory. Nonlinear Analysis and Optimization I. Nonlinear Analysis, Contemp. Math., 513, pp. 55–64, American Mathematical Society, Providence, RI (2010) 5. Bauschke, H.H., Wang, X.: The kernel average for two convex functions and its applications to the extension and representation of monotone operators. Trans. Am. Math. Soc. 36, 5947–5965 (2009) 6. Borwein, J.M., Lewis, A.S. (2006) Convex Analysis and Nonlinear Optimization, Theory and Examples, 2nd edn. CMS Books in Mathematics/Ouvrages de Mathmatiques de la SMC, 3 Springer, New York 7. Borwein, J.M., Vanderwerff, J.D.: Convex Functions. Cambridge University Press, Cambridge (2010) 8. Brézis, H., Browder, F.E.: Linear maximal monotone operators and singular nonlinear integral equations of Hammerstein type. In: Nonlinear Analysis (collection of papers in honor of Erich H. Rothe), pp. 31–42, Academic Press (1978) 9. Cross, R.: Multivalued Linear Operators. Marcel Dekker, New York (1998) 10. Crouzeix, J.P., Ocaña, Anaya E.: Monotone and maximal monotone affine subspaces. Oper. Res. Lett. 38, 139–142 (2010) 11. Crouzeix, J.P., Ocaña, Anaya E.: Maximality is nothing but continuity. J. Convex Anal. 17, 521–534 (2010) 12. Crouzeix, J.P., Ocaña Anaya, E., Sandoval, W.S.: A construction of a maximal monotone extension of a monotone map. ESAIM Proc. 20, 93–104 (2007) 13. Eckstein, J., Bertsekas, D.P.: On the Douglas-Rachford splitting method and the proximal point algorithm for maximal monotone operators. Math. Program. 55, 293–318 (1992) 14. Haraux, A.: Nonlinear Evolution Equations—Global Behavior of Solutions. Springer, Berlin (1981) 15. Meyer, C.: Matrix Analysis and Applied Linear Algebra. Society for Industrial and Applied Mathematics (SIAM), Philadelphia (2000) 16. Rockafellar, R.T., Wets, R.J.-B.: Variational Analysis, corrected 3rd printing. Springer, Berlin (2009) 17. Simons S.: A Brezis-Browder theorem for SSDB spaces. September (2010). http://arxiv.org/abs/1004. 4251v3 18. Voisei, M.D., Zalinescu, C.: Linear monotone subspaces of locally convex spaces. Set-Valued Var. Anal. 18, 29–55 (2010) 19. Yao, L.: Decompositions and Representations of Monotone Operators with Linear Graphs. M. Sc. Thesis, The University of British Columbia Okanagan, December (2007) 20. Yao, L.: The Brézis-Browder Theorem revisited and properties of Fitzpatrick functions of order n. In: Fixed Point Theory for Inverse Problems in Science and Engineering (Banff 2009), vol. 49, pp. 391–402. Springer, Berlin (2011)
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Maximally monotone linear subspace extensions of monotone subspaces: explicit constructions and characterizations Xianfu Wang · Liangjin Yao
Received: 7 March 2011 / Accepted: 11 September 2011 / Published online: 20 April 2013 © Springer-Verlag Berlin Heidelberg and Mathematical Optimization Society 2013
Abstract Monotone linear relations play important roles in variational inequality problems and quadratic optimizations. In this paper, we give explicit maximally monotone linear subspace extensions of a monotone linear relation in finite dimensional spaces. Examples are provided to illustrate our extensions. Our results generalize a recent result by Crouzeix and Ocaña-Anaya. Keywords Adjoint of linear relation · Linear relation · Monotone operator · Maximally monotone extensions · Minty parametrization Mathematics Subject Classification (2000)
47H05 · 47B65 · 47A06 · 49N15
1 Introduction Throughout this paper, we assume that Rn (n ∈ N = {1, 2, 3, . . .}) is an Euclidean space with the inner product ·, ·, and induced Euclidean norm ·. Let G : Rn ⇒ Rn be a set-valued operator from Rn to Rn , i.e., for every x ∈ Rn , Gx ⊆ Rn , and let
Dedicated to Jonathan Borwein on the occasion of his 60th birthday. X. Wang was partially supported by the Natural Sciences and Engineering Research Council of Canada. X. Wang Mathematics, Irving K. Barber School, University of British Columbia, Kelowna, BC V1V 1V7, Canada e-mail: [email protected] L. Yao (B) CARMA, University of Newcastle, Newcastle, New South Wales 2308, Australia e-mail: [email protected]
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graG = (x, x ∗ ) ∈ Rn × Rn | x ∗ ∈ Gx be the graph of G. Recall that G is monotone if ∀(x, x ∗ ) ∈ graG ∀(y, y ∗ ) ∈ graG x − y, x ∗ − y ∗ ≥ 0,
(1.1)
and maximally monotone if G is monotone and G has no proper monotone extension (in the sense of graph inclusion). We say that G is a linear relation if graG is a linear subspace. While linear relations have been extensively studies [2,3,6,9,18–20], monotone operators are ubiquitous in convex optimization and variational analysis [1,6,7,16]. Let R p×n denote the set of all p ×n matrices, for n, p ∈ N; and rank(M) denote the rank of M ∈ R p×n . The central object of this paper is to consider the linear relation G : Rn ⇒ Rn : graG = {(x, x ∗ ) ∈ Rn × Rn | Ax + Bx ∗ = 0} where
(1.2)
A, B ∈ R , rank(A B) = p.
(1.3) (1.4)
p×n
Our main concern is to find explicit maximally monotone linear subspace extensions of G. Recently, finding constructive maximal monotone extensions instead of using Zorn’s lemma has been a very active topic [5,4,10–12]. In [10], Crouzeix and OcañaAnaya gave an algorithm to find maximally monotone linear subspace extensions of G, but it is not clear what the maximally monotone extensions are analytically. In this paper, we provide some maximally monotone extensions of G with closed analytical forms. Along the way, we also give a new proof to Crouzeix and Ocaña-Anaya’s characterizations on monotonicity and maximal monotonicity of G. Our key tool is the Brezis-Browder characterization of maximally monotone linear relations. The paper is organized as follows. In the remainder of this introductory section, we describe some central notions fundamental to our analysis. In Sect. 2, we collect some auxiliary results for future reference and for the reader’s convenience. Section 3 provides explicit self-dual maximally monotone extensions by using subspaces on which AB T + B AT is negative semidefinite, and obtain a complete characterization of all maximally monotone extensions. Section 4 deals with Minty’s parameterizations of monotone operator G. In Sect. 5, we get some explicit maximally monotone extensions with the same domain or the same range by utilizing normal cone operators. In Sect. 6, we illustrate our maximally monotone extensions by considering three examples. Our notations are standard. We use domG = x ∈ Rn | Gx = ∅ for the domain of G, ranG = G(Rn ) for the range of G and ker G = x ∈ Rn | 0 ∈ Gx for the kernel of G. Given a subset C of Rn , spanC is the span (the set of all finite linear combinations) of C. We set C ⊥ = {x ∗ ∈ Rn | (∀c ∈ C) x ∗ , c = 0}. Then the adjoint of G, denoted by G ∗ , is defined by graG ∗ = {(x, x ∗ ) ∈ Rn × Rn | (x ∗ , −x) ∈ (graG)⊥ }.
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Let Id : Rn → Rn denote the identity mapping, i.e., Idx = x for x ∈ Rn . We also set PX : Rn × Rn → Rn : (x, x ∗ ) → x, and PX ∗ : Rn × Rn → Rn : (x, x ∗ ) → x ∗ . If X , Y are subspaces of Rn , we let X + Y = x + y | x ∈ X, y ∈ Y . Counting multiplicities, let λ1 , λ2 , . . . , λk be all positive eigenvalues of (AB T + B AT ) ∈ R p× p and (1.5) λk+1 , λk+2 , . . . , λ p be nonpositive eigenvalues of (AB T + B AT ). (1.6) Moreover, let vi ∈ R p be an eigenvector of eigenvalue λi of (AB T + B AT ) satisfying vi = 1, and vi , v j = 0 for 1 ≤ i = j ≤ p. It will be convenient to put ⎛
λ1 ⎜0 ⎜ ⎜ Idλ = diag(λ1 , . . . , λ p ) = ⎜ ⎜0 ⎜ .. ⎝. 0
⎞ 0 0 ··· 0 λ2 0 · · · 0 ⎟ ⎟ .. ⎟ p× p . ⎟ 0 λ3 . ⎟ , V = v1 v2 . . . v p ∈ R ⎟ .. .0 ⎠ 0 0 0 0 0 λp (1.7)
2 Auxiliary results on linear relations In this section, we collect some facts and preliminary results which will be used in sequel. We first provide a result about subspaces on which a linear operator from Rn → Rn , i.e, an n × n matrix, is monotone. For M ∈ Rn×n , define three subspaces of Rn , namely, the positive eigenspace, null eigenspace and negative eigenspace associated with M + M T by ⎧ ⎫ ⎨ w1 , . . . , ws : wi is an eigenvector of positive eigenvalue ⎬ V+ (M) = span αi of M + M T , wi , w j = 0, ∀ i = j ⎩ ⎭ wi = 1, i, j = 1, . . . , s. ⎧ ⎫ ⎨ ws+1 , . . . , wl : wi is an eigenvector of 0 eigenvalue of ⎬ V0 (M) = span M + M T , wi , w j = 0, ∀ i = j ⎩ ⎭ wi = 1, i, j = s + 1, . . . , l. ⎧ ⎫ ⎨ wl+1 , . . . , wn : wi is an eigenvector of negative ⎬ V− (M) = span eigenvalue αi of M + M T , wi , w j = 0 ⎩ ⎭ ∀ i = j, wi = 1, i, j = l + 1, . . . , n. which is possible since a symmetric matrix always has a complete orthonormal set of eigenvectors, [15, pp. 547–549].
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Proposition 2.1 Let M be an n × n matrix. Then (i) M is strictly monotone on V+ (M). Moreover, M + M T : V+ (M) → V+ (M) is a bijection. (ii) M is monotone on V+ (M) + V0 (M). (iii) −M is strictly monotone on V− (M). Moreover, −(M + M T ) : V− (M) → V− (M) is a bijection. (iv) −M is monotone on V− (M) + V0 (M). (v) For every x ∈ V0 (M), (M + M T )x = 0 and x, M x = 0. In particular, the orthogonal decomposition holds: Rn = V+ (M)⊕V0 (M)⊕V− (M). s li wi for some (l1 , . . . , ls ) ∈ Rs . Since Proof (i): Let x ∈ V+ (M). Then x = i=1 {w1 , . . . , ws } is a set of orthonormal vectors, they are linearly independent so that x = 0 ⇔ (l1 , . . . , ls ) = 0. Note that αi > 0 when i = 1, . . . , s and wi , w j = 0 for i = j. We have
T
2x, M x = x, (M + M )x = =
s
s
T
li wi , (M + M )
i=1
li wi ,
i=1
s
li αi wi =
i=1
s
li wi
i=1 s
αi li2 > 0
i=1
if x = 0. s For every x ∈ V+ (M) with x = i=1 li wi , we have (M + M T )x =
s
li (M + M T )wi =
i=1
s
αi li wi ∈ V+ (M).
i=1
As αi > 0 for i = 1, . . . , s and {w1 , . . . , ws } is an orthonormal basis of V+ (M), we conclude that M + M T : V+ (M) → V+ (M) is a bijection. (ii): Let x ∈ V+ (M) + V0 (M). Then x = li=1 li wi for some (l1 , . . . , ll ) ∈ Rl . Note that αi ≥ 0 when i = 1, . . . , l and wi , w j = 0 for i = j. We have 2x, M x = x, (M + M T )x = =
l
li wi , (M + M T )
i=1
l i=1
li wi ,
l i=1
li αi wi =
li wi
i=1 l
αi li2 ≥ 0.
i=1
The proofs for (iii), (iv) are similar as (i), (ii). (v): Obvious.
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Corollary 2.2 The following hold: (i) graT = {(B T u, AT u) | u ∈ V+ (B AT )} is strictly monotone. (ii) graT = {(B T u, AT u) | u ∈ V+ (B AT ) + V0 (B AT )} is monotone. (iii) graT = {(B T u, −AT u) | u ∈ V− (B AT )} is strictly monotone. (iv) graT = {(B T u, −AT u) | u ∈ V− (B AT ) + V0 (B AT ))} is monotone. Proof As B T u, AT u = u, B AT u ∀u ∈ Rn , the result follows from Proposition 2.1 by letting M = B AT . Lemma 2.3 For every subspace S ⊆ R p , the following hold. dim{(B T u, AT u) | u ∈ S} = dim S. T
T
dim{(B u, −A u) | u ∈ S} = dim S. Proof See [15, p. 208, Exercise 4.4.9].
(2.1) (2.2)
The following fact is straightforward from the definition of V . Fact 2.4 We have (AB T + B AT )V = V Idλ . Two key criteria concerning maximally monotone linear relations come as follows: Fact 2.5 (See [2, Corollary 6.7], [3, Proposition 2.10].) Let T : Rn ⇒ Rn be a monotone linear relation. The following are equivalent: (i) T is maximally monotone. (ii) dim graT = n. (iii) domT = (T 0)⊥ .
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Fact 2.6 (Brezis-Browder) (See [8, Theorem 2], [14, Theorem 10], [20] or [17].) Let T : Rn ⇒ Rn be a monotone linear relation. Then the following statements are equivalent. (i) T is maximally monotone. (ii) T ∗ is maximal monotone. (iii) T ∗ is monotone. Some basic properties of G are: Lemma 2.7 (i) graG = ker(A B). (ii) G0 = ker B, G −1 (0) = ker A. (iii) domG = PX (ker(A B)) and ranG = PX ∗ (ker(A B)). (iv) ran(G + Id) = PX ∗ (ker(A − B B)) = PX (ker(A B − A)), and domG = PX (ker(A − B B)), ranG = PX ∗ (ker(A (B − A)). (v) dim gra G = 2n − p. Proof (i), (ii), (iii) follow from definition of G. Since Ax + Bx ∗ = 0 ⇔ (A − B)x + B(x + x ∗ ) = 0 ⇔ A(x + x ∗ ) + (B − A)x ∗ = 0, (iv) holds. (v): We have 2n = dim ker(A B) + dim ran
AT BT
= dim graG + p.
Hence dim graG = 2n − p.
The following result summarizes the monotonicities of G ∗ and G. Lemma 2.8 The following hold. (i) graG ∗ = {(B T u, −AT u) | u ∈ R p }. (ii) G ∗ is monotone ⇔ the matrix AT B + B T A ∈ R p× p is negative-semidefinite. (iii) Assume G is monotone. Then n ≤ p. Moreover, G is maximally monotone if and only dim graG = n = p. Proof (i): By Lemma 2.7(i), we have (x, x ∗ ) ∈ graG ∗ ⇔ (x ∗ , −x) ∈ graG ⊥ = ran
AT BT
= {(AT u, B T u) | u ∈ R p }.
Thus graG ∗ = {(B T u, −AT u) | u ∈ R p }. (ii): Since graG ∗ is a linear subspace, by (i), G ∗ is monotone ⇔ B T u, −AT u ≥ 0, ∀u ∈ R p ⇔ u, −B AT u ≥ 0, ∀u ∈ R p
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⇔ u, B AT u ≤ 0, ∀u ∈ R p ⇔ u, (AT B + B T A)u ≤ 0, ∀u ∈ R p ⇔ (AT B + B T A) is negative semidefinite. (iii): By Fact 2.5 and Lemma 2.7(v), 2n − p = dim graG ≤ n ⇒ n ≤ p. By Fact 2.5 and Lemma 2.7(v) again, G is maximally monotone ⇔ 2n − p = dim graG = n ⇔ dim graG = p = n. 2.1 One linear relation: two equivalent formulations The linear relation G given by (1.2)–(4): graG = {(x, x ∗ ) ∈ Rn × Rn | Ax + Bx ∗ = 0}
(2.3)
is an intersection of p linear hyperplanes. It can be equivalently described as a span of q = 2n − p points in Rn × Rn . Indeed, for (2.3) we can use Gaussian elimination to reduce (A B) to row echelon form. Then back substitution to solve basic variables in terms of the free variables, see [15, p. 61]. Row-echelon form gives
x x∗
= h 1 y1 + · · · + h 2n− p y2n− p =
C D
y
where y ∈ R2n− p and
C D
= (h 1 , . . . , h 2n− p )
with C, D being n × (2n − p) matrices. Therefore, graG =
Cy Dy
| y∈R
2n− p
C = ran D
(2.4)
which is a span of 2n − p points in Rn × Rn . The two formulations (2.3) and (2.4) coincide when p = q = n, Id = −B = C and D = A in which Id ∈ Rn×n . 3 Explicit maximally monotone extensions of monotone linear relations In this section, we give explicit maximally monotone linear subspace extensions of G by using V+ (AB T ) or Vg . A characterization of all maximally monotone extensions of G is also given. We also provide a new proof to Crouzeix and Ocaña-Anaya’s characterizations on monotonicity and maximal monotonicity of G. We shall use notations given in (1.2)–(1.7), in particular, G is in the form of (2.3).
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and G be defined by Lemma 3.1 Let M ∈ R p× p , and linear relations G = {(x, x ∗ ) | M T Ax + M T Bx ∗ = 0} gra G = {(B T u, −AT u) | u ∈ ranM}. gra G ∗ = G. Then (G) Proof Let (y, y ∗ ) ∈ Rn × Rn . Then we have ∗ (y, y ∗ ) ∈ gra(G) ⊥ = (ker M T A M T B ⊥ = ran ⇔ (y ∗ , −y) ∈ (graG)
AT M BT M
⇔ (y, y ∗ ) ∈ graG. ∗ = G. Hence (G)
and G by Lemma 3.2 Define linear relations G = {(x, x ∗ ) | Vg Ax + Vg Bx ∗ = 0} gra G = {(B T u, −AT u) | u ∈ V− (B AT ) + V0 (B AT )}, gra G where Vg is ( p − k) × p matrix defined by ⎛
T vk+1
⎞
⎜ T ⎟ ⎜v ⎟ k+2 ⎟ . Vg = ⎜ ⎜ .. ⎟ ⎝. ⎠ v Tp Then is monotone. (i) G ∗ = G. (ii) (G) T B T u | u ∈ V+ (B A ) . (iii) gra G = graG + AT Proof (i): Apply Corollary 2.2(iv). (ii): Notations are as in (1.7). Define the p × p matrix N by N=
00 0 Id
in which Id ∈ R( p−k)×( p−k) . Then we have T 00 0 . N T V T = (v1 . . . vk VgT ) = 0 Id Vg
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(3.1)
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Then we have Vg Ax + Vg Bx ∗ = 0 ⇔
0 Vg Ax + Vg Bx ∗
=0
⇔ N T V T Ax + N T V T Bx ∗ = 0, ∀(x, x ∗ ) ∈ Rn × Rn . Hence = {(x, x ∗ ) | N T V T Ax + N T V T Bx ∗ = 0}. graG Thus by Lemma 3.1 with M = V N , ∗ = {(B T u, −AT u) | u ∈ ranV N = ran 0 VgT gra(G) = V− (B AT ) + V0 (B AT )} = graG. ∗ = (G) ∗∗ = G. Hence (G) (iii): Let J be defined by
BT AT
graJ = graG +
u | u ∈ V+ (B AT ) .
Then we have (graJ )⊥ = (graG)⊥ ∩
BT AT
⊥ u | u ∈ V+ (B AT )
.
By Lemma 2.7(i), ⊥
graG =
AT BT
w|w∈R
p
Then
AT BT
w∈
BT AT
⊥ T
u | u ∈ V+ (B A )
if and only if (AT w, B T w), (B T u, AT u) = 0 ∀ u ∈ V+ (B AT ), that is, AT w, B T u+B T w, AT u = w, (AB T + B AT )u = 0 ∀u ∈ V+ (AB T ). (3.2)
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Because AB T + B AT : V+ (AB T ) → V+ (AB T ) is onto by Proposition 2.1(i), we obtain that (3.2) holds if and only if w ∈ V− (AB T ) + V0 (AB T ). Hence (graJ )⊥ = {(AT w, B T w) | w ∈ V− (B AT ) + V0 (B AT )}, Then by (i), from which graJ ∗ = graG. = gra(G) ∗ = graJ ∗∗ = graJ. graG We are ready to apply Brezis-Browder Theorem, namely Fact 2.6, to improve Crouzeix and Ocaña-Anaya’s characterizations of monotonicity and maximal monotonicity of G and provide a different proof. G be defined in Lemma 3.2. The following are equivalent: Theorem 3.3 Let G, (i) (ii) (iii) (iv) (v)
G is monotone; is monotone; G is maximally monotone; G is maximally monotone; G dim V+ (B AT ) = p − n, equivalently, AB T + B AT has exactly p − n positive eigenvalues (counting multiplicity).
Proof (i)⇔(ii): Lemma 3.2(iii) andCorollary 2.2(i). = G ∗ and G is always a monotone linear relation (ii)⇔(iii)⇔(iv): Note that G by Corollary 2.2(iv). It suffices to combine Lemma 3.2 and Fact 2.6. is monotone by Lemma 3.2(iii) and (i)⇒(v): Assume that G is monotone. Then G is maximally Corollary 2.2(i). By Lemma 3.2(ii), Corollary 2.2(iv) and Fact 2.6, G = p−k = n by Fact 2.5 and Lemma 2.3, thus k = p−n. monotone, so that dim(graG) Note that for each eigenvalue of a symmetric matrix, its geometric multiplicity is the same as its algebraic multiplicity [15, p. 512]. = p − k = n by Lemma 2.3, (v)⇒(i): Assume that k = p − n. Then dim(graG) is maximally monotone by Fact 2.5(i)(ii). By Lemma 3.2(ii) and Fact 2.6, so that G is monotone, which implies that G is monotone. G Corollary 3.4 Assume that G is monotone. Then = graG + gra G
BT AT
u | u ∈ V+ (B A T
= {(x, x ∗ ) | Vg Ax + Vg Bx ∗ = 0} is a maximally monotone extension of G, where ⎛
v Tp−n+1
⎞
⎜ T ⎟ ⎜ v p−n+2 ⎟ ⎟. Vg = ⎜ ⎜ .. ⎟ ⎝. ⎠ v Tp
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Proof Combine Theorem 3.3 and Lemma 3.2(iii) directly.
Note that Corollary 3.4 gives both types of maximally monotone extensions of G, namely, type (2.3) and type (2.4). A remark is in order to compare our extension with the one by Crouzeix and Ocaña-Anaya. Remark 3.5 (i). Crouzeix and Ocaña-Anaya [10] defines the union of monotone extension of G as T B u|u∈K , S = graG + AT where K = {u ∈ Rn | u, (AB T + B AT )u ≥ 0}. Although this is the set monotonically related to G, it is not monotone in general as long as (AB T + B AT ) has both positive eigenvalues and negative eigenvalues. Indeed, let (α1 , u 1 ) and (α2 , u 2 ) be eigen-pairs of (AB T + B AT ) with α1 > 0 and α2 < 0. We have u 1 , (AB T + B AT )u 1 = α1 u 1 2 > 0, u 2 , (AB T + B AT )u 2 = α2 u 2 2 < 0. Choose > 0 sufficiently small so that u 1 + u 2 , (AB T + B AT )(u 1 + u 2 ) > 0. Then
However,
BT AT
BT AT
u1,
BT AT
(u 1 + u 2 ) −
(u 1 + u 2 ) ∈ S.
BT AT
u1 =
BT AT
u2
has B T u 2 , AT u 2 = 2 u 2 , B AT u 2 = 2
u 2 , (AB T + B AT )u 2 < 0. 2
Therefore S is not monotone. By using V+ (B AT ) ⊆ K , we have obtained a maximally monotone extension of G. (ii). Crouzeix and Ocaña-Anaya [10] find a maximally monotone linear subspace extension of G algorithmically by using u˜ k ∈ graG k \graG k and construct graG k+1 = graG k + Ru˜ k where T Bk u˜ k = u k , u k , (Ak BkT + Bk ATk )u k ≥ 0. ATk This recursion is done until dim graG k = n. In particular, each u k may be chosen as an eigenvector associated with a positive eigenvalue of Ak BkT + Bk ATk , which is
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possible since p > n when G k is not maximally monotone. Their construction uses both formulations, namely, (2.3) and (2.4). No computation indications are given on the passage from one formulation to the other one. The following result extends the characterization of maximally monotone linear relations given by Crouzeix and Ocaña-Anaya [10]. G be defined in Lemma 3.2. The following are equivalent: Theorem 3.6 Let G, (i) (ii) (iii) (iv)
G is maximally monotone; p = n and G is monotone; p = n and AB T + B AT is negative semidefinite. is maximally monotone. p = n and G
Proof (i)⇒(ii): Apply Lemma 2.8(iii). (ii)⇒(iii): Apply directly Theorem 3.3(i)(v). (iii)⇒(i): Assume that p = n and (AB T + B AT ) is negative semidefinite. Then = G. It follows that dim(graG) = p − k = n by Lemma 2.3, so that G k = 0 and G ∗ is maximally monotone by Corollary 2.2(iv) and Fact 2.5(i)(ii). Since G = G by = G is maximally monotone. Lemma 3.2(ii), Fact 2.6 gives that G (iii)⇒(iv): Assume that p = n and (AB T + B AT ) is negative semidefinite. We have = p − k = n − 0 = n. Hence (iv) holds by Corollary 2.2(iv) k = 0 and dim(graG) and Fact 2.5(i)(ii). is maximally monotone and p = n. We have (iv)⇒(iii): Assume that G = p−k =n−k =n dim(graG) so that k = 0. Hence (AB T + B AT ) is negative semidefinite.
Corollary 3.4 supplies only one maximally monotone linear subspace extension of G. Can we find all of them? Surprisingly, we may give a characterization of all the maximally monotone linear subspace extensions of G when it is given in the form of (2.3). is a maximally monotone extension of G Theorem 3.7 Let G be monotone. Then G p× p with rank of n such that N T Idλ N is negative if and only if there exists N ∈ R semidefinite and = {(x, x ∗ ) | N T V T Ax + N T V T Bx ∗ = 0}. gra G
(3.3)
Proof “⇒”: By Lemma 2.8(i), we have graG ∗ = {(B T u, −AT u) | u ∈ R p }.
(3.4)
and thus gra(G) ∗ is a subspace of graG ∗ . Since graG ⊆ graG Thus by (3.4), there exists a subspace F of R p such that ∗ = {(B T u, −AT u) | u ∈ F}. gra(G)
123
(3.5)
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339
By Facts 2.6, 2.5 and Lemma 2.3, we have dim F = n.
(3.6)
Thus, there exists N ∈ R p× p with rank n such that ranV N = F and ∗ = {(B T V N y, −AT V N y) | y ∈ R p }. gra(G)
(3.7)
is maximally monotone, (G) ∗ is maximally monotone by Fact 2.6, so As G N T V T (B AT + AB T )V N is negative semidefinite. Using Fact 2.4, we have N T Idλ N = N T V T V Idλ N = N T V T (AB T + B AT )V N
(3.8)
which is negative semidefinite. (3.3) follows from (3.7) by Lemma 3.1 using M = V N . “⇐”: By Lemma 3.1, we have ∗ = {(B T V N u, −AT V N u) | u ∈ R p }. gra(G)
(3.9)
∗ is monotone because N T V T (AB T + B AT )V N = N T Idλ N is negObserve that (G) ative semidefinite by Fact 2.4 and the assumption. As rank(V N ) = n, it follows from ∗ is maximally monotone ∗ = n. Therefore (G) (3.9) and Lemma 2.3 that dim gra(G) ∗ = (G) ∗∗ is maximally by Fact 2.5. Applying Fact 2.6 for T = (G) yields that G monotone. From the above proof, we see that to find a maximally monotone extension of G one essentially need to find subspace F ⊆ R p such that dim F = n and AB T + B AT is negative semidefinite on F. If F = ranM and M ∈ R p× p with rankM = n, one can let N = V T M. The maximally monotone linear subspace extension of G is = {(x, x ∗ ) | M T Ax + M T Bx ∗ = 0}. G In Corollary 3.4, one can choose M = 0 · · · 0$ v p−n+1 . . . v p . ! 0"# n
is a maximally monotone extension of G Corollary 3.8 Let G be monotone. Then G p× p if and only if there exists M ∈ R with rank of n such that M T (AB T + B AT )M is negative semidefinite and = {(x, x ∗ ) | M T Ax + M T Bx ∗ = 0}. gra G
(3.10)
Note that G may have different representations in terms of A, B. The maximally given in Theorem 3.7 and Corollary 3.4 relies on A, B matrimonotone extension of G ces and N . This might leads different maximally monotone extensions, see Sect. 6.
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Remark 3.9 A referee points out that there is a shorter way to see Theorem 3.7. Consider the maximally monotone linear subspace extension of G of type: + = {(x, x ∗ ) ∈ Rn × Rn | Ax Bx ∗ = 0} ⊇ graG graG where A, B ∈ Rn×n . With the nonsingular p × p matrix V given as in (1.7), an equivalent formulation of G is graG = {(x, x ∗ ) ∈ Rn × Rn | V T Ax + V T Bx ∗ = 0}. is maximally monotone, the n × 2n matrix has rank( A, As G B) = n and the matrix T ∈ Rn×n BA A BT + ⊇ graG, we have is negative semidefinite. Since graG ran
T A BT
T T (V A) ⊥ ⊥ = (graG) ⊆ (graG) = ran . (V T B)T
Therefore, there exists a p × n matrix N with rank N = n such that
T A BT
=
(V T A)T (V T B)T
N=
(V T A)T N (V T B)T N
= N T V T A, from which A B = N T V T B. Then the n × n matrix T = N T V T A(N T V T B)T + N T V T B(N T V T A)T BA A BT + T
T
T
T
= N V (AB + B A )V N T
= N Idλ N .
(3.11) (3.12) (3.13)
Therefore, all maximally monotone linear subspace extensions of G can be obtained by using = {(x, x ∗ ) ∈ Rn × Rn | N T V T Ax + N T V T Bx ∗ = 0} graG in which the p × n matrix N satisfying rankN = n and N T Idλ N being negative semidefinite. 4 Minty parameterizations Although G is set-valued in general, when G is monotone it has a beautiful Minty parametrization in terms of A, B, which is what we are going to show in this section.
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341
Lemma 4.1 The linear relation G is monotone if and only if y2 − y ∗ 2 ≥ 0, whenever ∗
(A + B)y + (B − A)y = 0.
(4.1) (4.2)
Consequently, if G is monotone then the p × n matrix B − A must have full column rank, namely n. Proof Define the 2n × 2n matrix P=
0 Id Id 0
where Id ∈ Rn×n . It is easy to see that G is monotone if and only if %
∗
(x, x ), P
x x∗
& ≥ 0,
whenever Ax + Bx ∗ = 0. Define the orthogonal matrix 1 Q=√ 2
Id −Id Id Id
and put
x x∗
=Q
y y∗
.
Then G is monotone if and only if y2 − y ∗ 2 ≥ 0, whenever ∗
(A + B)y + (B − A)y = 0.
(4.3) (4.4)
If (B − A) does not have full column rank, then there exists y ∗ = 0 such that (B − A)y ∗ = 0. Then (0, y ∗ ) satisfies (4.4) but (4.3) fails. Therefore, B − A has to be full column rank. Theorem 4.2 (Minty parametrization) Assume that G is a monotone operator. Then (x, x ∗ ) ∈ graG if and only if 1 [Id + (B − A)† (B + A)]y 2 1 x ∗ = [Id − (B − A)† (B + A)]y 2 x=
(4.5) (4.6)
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for y = x + x ∗ ∈ ran(Id + G). Here the Moore-Penrose inverse (B − A)† = [(B − A)T (B − A)]−1 (B − A)T . In particular, when G is maximally monotone, we have graG = {((B − A)−1 By, −(B − A)−1 Ay) | y ∈ Rn }. Proof As (B − A) is full column rank, (B − A)T (B − A) is invertible. It follows from (4.2) that (B − A)T (A + B)y + (B − A)T (B − A)y ∗ = 0 so that y ∗ = −((B − A)T (B − A))−1 (B − A)T (A + B)y = −(B − A)† (A + B)y. Then 1 x = √ (y − y ∗ ) = 2 1 x ∗ = √ (y + y ∗ ) = 2 ∗ x+x √ 2
where y =
1 √ [Id + (B − A)† (B + A)]y 2 1 √ [Id − (B − A)† (B + A)]y 2
with (x, x ∗ ) ∈ graG. Since ran(Id + G) is a subspace, we have 1 [Id + (B − A)† (B + A)] y˜ 2 1 x ∗ = [Id − (B − A)† (B + A)] y˜ 2 x=
with y˜ = x + x ∗ ∈ ran(Id + G). If G is maximally monotone, then p = n by Theorem 3.6 and hence B − A is invertible, thus (B − A)† = (B − A)−1 . Moreover, ran(G + Id) = Rn . Then (4.5) and (4.6) transpire to 1 (B − A)−1 [B − A + (B + A)]y = (B − A)−1 By 2 1 x ∗ = (B − A)−1 [(B − A) − (B + A)]y = −(B − A)−1 Ay 2 x=
for y ∈ Rn .
(4.7) (4.8)
Remark 4.3 See Lemma 2.7 for ran(G + Id). Note that as G is a monotone linear relation, the mapping z → ((G + Id)−1 , Id − (G + Id)−1 )(z) is bijective and linear from ran(G + Id) to graG, therefore dim(ran(G + Id)) = dim(graG).
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343
defined in Corollary 3.4, the Corollary 4.4 Let G be a monotone operator. Then G maximally monotone extension of G, has its Minty parametrization given by = {((Vg B − Vg A)−1 Vg By, −(Vg B − Vg A)−1 Vg Ay) | y ∈ Rn } gra G where Vg is given as in Corollary 3.4. Proof Since rank(Vg ) = n and rank(A B) = p, by Lemma 2.3(2.1), rank(Vg A Vg B) = n. Apply Corollary 3.4 and Theorem 4.2 directly.
Corollary 4.5 When G is maximally monotone, domG = (B − A)−1 (ranB), ranG = (B − A)−1 (ran A). Recall that T : Rn → Rn is firmly nonexpansive if T x − T y2 ≤ T x − T y, x − y ∀ x, y ∈ domT. In terms of matrices Corollary 4.6 Suppose that p = n, AB T + B AT is negative semidefinite. Then (B − A)−1 B and −(B − A)−1 A are firmly nonexpansive. Proof By Theorem 3.6, G is maximally monotone. Theorem 4.2 gives that (B − A)−1 B = (Id + G)−1 , −(B − A)−1 A = (Id + G −1 )−1 . Being resolvents of monotone operators G, G −1 , they are firmly nonexpansive, see [1,13] or [4, Fact 2.5]. 5 Maximally monotone extensions with the same domain or the same range How do we find maximally monotone linear subspace extensions of G if it is given in the form of (2.4)? The purpose of this section is to find maximally monotone linear subspace extensions of G which keep either domG or ranG unchanged. For a closed convex set S ⊆ Rn , let N S denote its normal cone mapping. Proposition 5.1 Assume that T : Rn ⇒ Rn is a monotone linear relation. Then (i) T1 = T + NdomT , i.e., x → T1 x =
T x + (domT )⊥ ∅
if x ∈ domT otherwise
is maximally monotone. In particular, domT1 = domT . (ii) T2 = (T −1 + NranT )−1 is a maximally monotone extension of T and ranT2 = ranT .
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Proof (i): Since 0 ∈ T 0 ⊆ (domT )⊥ by [2, Proposition 2.2(i)], we have T1 0 = T 0 + (domT )⊥ = (domT )⊥ so that domT1 = domT = (T1 0)⊥ . Hence T1 is maximally monotone by Fact 2.5. (ii): Apply (i) to T −1 to see that T −1 + NranT is a maximally monotone extension of T −1 with dom(T −1 + NranT ) = ranT . Therefore, T2 is a maximally monotone extension of T with ranT2 = ranT . Define linear relations E i : Rn ⇒ Rn (i = 1, 2) by
Cy Dy
0 (ranC)⊥
, Cy (ranD)⊥ | y ∈ R2n− p . graE 2 = + Dy 0
graE 1 =
+
| y∈R
2n− p
(5.1) (5.2)
Theorem 5.2 (i) E 1 is a maximally monotone extension of G with domE 1 = domG. Moreover, 0 0 C C = ran + . (5.3) graE 1 = ran + D D (ranC)⊥ ker C T (ii) E 2 is a maximally monotone extension of G with ranE 2 = ranG. Moreover, C C (ranD)⊥ ker D T . (5.4) = ran graE 2 = ran + + D D 0 0 Proof (i): Note that domG = ranC. The maximal monotonicity follows from Proposition 5.1. (5.3) follows from (5.1) and that (ranC)⊥ = ker C T [15, page 405]. (ii): Apply (i) to G −1 , i.e., Dy (5.5) graG −1 = | y ∈ R2n− p Cy and followed by taking the set-valued inverse. Apparently, both extensions E 1 , E 2 rely on graG, domG, ranG, not on the A, B. In this sense, E 1 , E 2 are intrinsic maximally monotone linear subspace extensions. Remark 5.3 Theorem 5.2 is much easier to use than Corollary 3.8 when G is written in the form of (2.4). Indeed, it is not hard to check that gra(E 1∗ ) = {(B T u, −AT u) | B T u ∈ domG, u ∈ R p }.
(5.6)
gra(E 2∗ )
(5.7)
T
T
T
= {B u, −A u) | A u ∈ ranG, u ∈ R }. p
According to Fact 2.6, E i∗ is maximally monotone and dim E i∗ = n. This implies that dim{u ∈ R p | B T u ∈ domG} = n,
123
dim{u ∈ R p | AT u ∈ ranG} = n.
Maximally monotone linear subspace extensions
345
Let Mi ∈ R p× p with rankMi = n and {u ∈ R p | B T u ∈ domG} = ranM1 ,
(5.8)
T
{u ∈ R | A u ∈ ranG} = ranM2 . p
(5.9)
Corollary 3.8 shows that graE i = {(x, x ∗ ) | MiT Ax + MiT Bx ∗ = 0}. However, finding Mi from (5.8) and (5.9) may not be easy as it seems. Remark 5.4 Unfortunately, we do not know how to determine all maximally monotone linear subspace extensions of G if it is given in the form of (2.4). 6 Examples In the final section, we illustrate our maximally monotone extensions by considering rely three examples. In particular, they show that maximally monotone extensions G on the representation of G in terms of A, B and choices of N we shall use. However, the maximally monotone extensions E i are intrinsic, only depending on graG. Example 6.1 Consider Id 0 ∗ n n ∗ graG = (x, x ) ∈ R × R | x+ x =0 0 C where C ∈ Rn×n is symmetric and positive definite, and Id ∈ Rn×n . Clearly, graG =
0 . 0
We have α defined by (i) For every α ∈ [−1, 1] , G ' α = graG
n {(0, ( R )}, ) 1+α −1 (x, 1−α C x) | x ∈ Rn ,
if α = 1; otherwise
is a maximally monotone linear extension of G. 1 and E 2 = G −1 . (ii) E 1 = G α , we need eigenvectors of Proof (i): To find G A=
0 Id 0 C (Id 0) = (0 C T ) + . C 0 C0
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Counting multiplicity, the positive definite matrix C has eigen-pairs (λi , wi ) (i = 1, . . . , n) such that λi > 0, wi = 1 and wi , w j = 0 for i = j. As such, the matrix A has 2n eigen-pairs, namely wi λi , wi and wi −λi , −wi with i = 1, . . . , n. Put W = (w1 . . . wn ) ∈ Rn×n and write W W V = . W −W Then W T C W = D = diag(λ1 , λ2 , . . . , λn ). In Theorem 3.7, take Nα =
0 αId 0 Id
∈ R2n×2n
where Id ∈ Rn×n . We have rank Nα = n, 00 00 = NαT Idλ Nα = 0 (α 2 − 1)D 0 (α 2 − 1)W T C W being negative semidefinite, and V Nα =
0 (1 + α)W 0 (α − 1)W
.
α Then by Theorem 3.7, we have an maximally monotone linear extension G given by 0 ∗ n n =0 graG α = (x, x ) ∈ R × R | (1 + α)W T x + (α − 1)W T C x ∗ = (x, x ∗ ) ∈ Rn × Rn | (1 + α)x + (α − 1)C x ∗ = 0 ' {(0, Rn )}, if α = 1; ( ) = 1+α −1 n (x, 1−α C x) | x ∈ R , otherwise. Hence we get the result as desire.
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347
(ii): It is immediate from Theorem 5.2 and (i). Example 6.2 Consider ⎧ ⎨
⎛ ⎞ 1 −1 0 x graG = (x, x ∗ ) ∈ R2 × R2 | ⎝ 0 0 ⎠ 1 + ⎝ 0 x2 ⎩ 0 0 −1 ⎛
⎫ ⎞ 0 ∗ ⎬ x 1 ⎠ 1∗ = 0 . x2 ⎭ 1
Then i : R2 ⇒ R2 for i = 1, 2 given by (i) the linear operators G 1 = G
10 √ 0 −1+√ 2
2 = , G
2− 2
1 0
2 5 √
2 10
are two maximally monotone extensions of G. (ii) E 1 (x1 , 0) = (x1 , R) ∀x1 ∈ R. (iii) E 2 (x1 , y) = (x1 , 0) ∀x1 , y ∈ R. Proof We have ⎫ ⎧⎛ ⎞ x1 ⎪ ⎪ ⎪ ⎪ ⎬ ⎨⎜ ⎟ 0 ⎟ | x1 ∈ R graG = ⎜ ⎝ ⎠ ⎪ ⎪ ⎪ ⎪ x1 ⎭ ⎩ 0 is monotone. Since dim G = 1, G is not maximally monotone by Fact 2.5. The matrix ⎛ ⎞ −2 0 0 T T AB + B A = ⎝ 0 0 −1 ⎠ 0 −1 −2 has a positive eigenvalue −1 +
√ 2 with an eigenvector
⎛
⎞ 0 u = ⎝1 √ ⎠ 1− 2
so that
T
B AT
⎞ 0 √ ⎜2 − 2 ⎟ ⎟. u=⎜ ⎠ ⎝0 √ −1 + 2 ⎛
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Then by Corollary 3.4, ⎧⎛ ⎞ ⎫ ⎧⎛ ⎫ ⎞ 0 √ x1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎟ ⎬ ⎨⎜ ⎬ ⎟ 2 − 0 2 ⎟ x2 | x2 ∈ R 1 = ⎜ ⎟ | x1 ∈ R + ⎜ graG ⎝ ⎝ ⎠ ⎠ x1 ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ √ ⎩ ⎩ ⎭ ⎪ ⎭ 0 −1 + 2 ⎧⎛ ⎫ ⎞ x1 √ ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎬ ⎟ 2)x (2 − 2 ⎟ ⎜ | x = ⎝ . , x ∈ R 1 2 ⎠ x1 ⎪ ⎪ ⎪ ⎪ √ ⎩ ⎭ (−1 + 2)x2 Therefore, 1 = G
10 √ 0 −1+√ 2
2− 2
is a maximally monotone extension of G. Now we have √ ⎞ ⎛ ⎛ ⎞ 0 0 1 −1 + 2 0√ 0 1 1 Idλ = ⎝ 0 −1 − 2 0 ⎠ , V = ⎝ − −1+√2 − −1−√2 0 ⎠ . 0 0 −2 1 1 0
(6.1)
Take ⎛
⎞ 0 −1 1 N = ⎝ 0 2 −1 ⎠ . 0 1 1
(6.2)
We have rank N = 2 and ⎞ 0 0 √ 0√ 3 2 1 + 2⎠, N T Idλ N = ⎝ 0 −7 − √ 0 1+ 2 −4 ⎛
(6.3)
being negative semidefinite. By Theorem 3.7, with V, N given in (6.1) and (6.2), we use the NullSpace command in Maple to solve (V N )T Ax + (V N )T Bx ∗ = 0, and get ⎧ ⎛ ⎞ ⎛ √ ⎞⎫ −2 2 ⎪ 1 ⎪ ⎪ ⎬ ⎨ ⎜ ⎟ ⎜ √ ⎟⎪ 0 2 = span ⎜ ⎟ ⎜ 5 2 ⎟ . graG ⎝1⎠⎝0 ⎠⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0 1
123
Maximally monotone linear subspace extensions
2 = Thus G
√ −1 1 1 −2 √ 2 = 05 2 0
2 5 √
349
is another maximally monotone extension of
2 10
G. On the other hand, ⎧⎛ ⎞ ⎫ ⎛ ⎞ ⎧⎛ ⎞ ⎫ x1 0 x1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎟ ⎬ ⎜ ⎟ ⎨⎜ ⎟ ⎬ 0⎟ 0 ⎟ 0 ⎟ ⎜ ⎜ ⎜ graE 1 = ⎝ ⎠ | x1 ∈ R + ⎝ ⎠ = ⎝ ⎠ | x1 ∈ R 0 x1 x1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ 0 R R gives E 1 (x1 , 0) = (x1 , R) ∀x1 ∈ R. And ⎧⎛ ⎞ ⎫ x1 ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎟ ⎬ R⎟ ⎜ graE 2 = ⎝ ⎠ | x1 ∈ R . x1 ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0 gives E 2 (x1 , y) = (x1 , 0) ∀x1 , y ∈ R. In [5], the authors use autoconjugates to find maximally monotone extensions of monotone operators. In general, it is not clear whether the maximally monotone extensions of a linear relation is still a linear relation. As both monotone operators in Examples 6.2 and 6.1 are subset of {(x, x) | x ∈ Rn }, [5, Example 5.10] shows that the maximally monotone extension obtained by autoconjugates must be Id, which is different from the ones given here. Example 6.3 Consider graG = {(x, x ∗ ) ∈ R2 × R2 | Ax + Bx ∗ = 0} where ⎛
1 A = ⎝2 3
⎞ ⎛ 1 1 0⎠, B = ⎝1 1 0
⎞ ⎛ ⎞ 5 1115 7 ⎠ , thus (A B) = ⎝ 2 0 1 7 ⎠ . 2 3102
(6.4)
i : Rn ⇒ Rn for i = 1, 2 given by Then linear operators G ⎛ 1 = G
√ √ −117+17 √ 201 −107+7√ 201 2(−1+ 201) ⎝ 2(−1+ 201) √ √ − −23+3√ 201 − −21+√201 2(−1+ 201) 2(−1+ 201)
⎞
2 = ⎠, G
√ 201 201 13 6√ 4 − 6√ 201 201 29 9 − 20 + 30 − 20 + 30 33 4
−
√
are two maximally monotone linear extensions of G.
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Moreover, ⎧⎛ ⎫ ⎞ ⎛ ⎞ 0 ⎪ ⎪ ⎪ −1 ⎪ ⎨ ⎬ ⎜1 ⎟ ⎜0⎟ ⎜ ⎟ ⎜ ⎟ , x + | x , x ∈ R graE 1 = ⎝ x 1 2 1 2 ⎝1⎠ ⎪ −5 ⎠ ⎪ ⎪ ⎪ ⎩ ⎭ 1 1 ⎧⎛ ⎫ ⎞ ⎛ ⎞ −1 1 ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎬ ⎜5⎟ 1 ⎟ ⎜ ⎟ ⎜ ⎟ graE 2 = ⎝ . x + | x , x ∈ R x 1 2 1 2 ⎝0⎠ −5 ⎠ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 1 0 Proof We have rank(A B) = 3 and ⎛ Idλ = ⎝
13 +
√ ⎞ ⎛ 20 √ 0 201 0 0 1+ 201 ⎠, V = ⎝ 1 0 −6 0√ −1 0 0 13 − 201 1 1
20 √ 1− 201
1 1
⎞ ⎠,
(6.5)
and Vg =
0 20 √ 1− 201
−1 1 1 1
.
(6.6)
Clearly, here p = 3, n = 2 and AB T + B AT has exactly p − n = 3 − 2 = 1 positive eigenvalue. By Theorem 3.3(i)(v), G is monotone. Since AB T + B AT is not negative semidefinite, by Theorem 3.6(i)(iii), G is not maximally monotone. With Vg given in (6.6) and A, B in (6.4), use the NullSpace command in maple to 1 defined by solve Vg Ax + Vg Bx ∗ = 0 and obtain G ⎧⎛ ⎞ ⎛ ⎞⎫ √ √ −21+√201 −107+7√ 201 ⎪ ⎪ − − ⎪ ⎪ ⎪ 2(−1+ 2(−1+√201) ⎟⎪ ⎪ ⎪ √ 201) ⎟ ⎜ ⎬ ⎨⎜ ⎜ −23+3√ 201 ⎟ ⎜ −117+17 ⎟ √ 201 ⎟ ⎜ ⎜ ⎟ graG 1 = span ⎜ 2(−1+ 201) ⎟ , ⎜ 2(−1+ 201) ⎟ . ⎪ ⎪ ⎪ ⎝1 ⎠ ⎝0 ⎠⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 0 1 1 is a maximally monotone linear subspace extension of G. Then By Corollary 3.4, G ⎛ 1 = ⎝ G
√ √ −21+√201 − −107+7√ 201 2(−1+ 2(−1+√201) √ 201) −117+17 −23+3√ 201 √ 201 2(−1+ 201) 2(−1+ 201)
−
⎛
⎞−1 ⎠
=
√ √ −117+17 √ 201 −107+7√ 201 2(−1+ 201) ⎝ 2(−1+ 201) √ √ − −23+3√ 201 − −21+√201 2(−1+ 201) 2(−1+ 201)
⎞ ⎠.
Let N be defined by ⎛
⎞ 0 0 15 N = ⎝0 1 0 ⎠. 001
123
(6.7)
Maximally monotone linear subspace extensions
351
Then rankN = 2 and ⎛
0 0 N T Idλ N = ⎝ 0 −6 0 0
0 0√
⎞ ⎠.
338−24 201 25
is negative semidefinite. With N in (6.7), A, B in (6.4) and V in (6.5), use the NullSpace command in maple to solve (V N )T Ax +(V N )T Bx ∗ = 0. By Theorem 3.7, we get a maximally monotone 2 , defined by linear extension of G, G 2 = G
√
√
201 201 9 − 20 +√ 30 − 13 4 + √ 6 201 201 33 29 20 − 30 4 − 6
−1 =
√ 201 201 13 − 6√ 4 6√ 201 201 29 9 − 20 + 30 − 20 + 30 33 4
−
√
.
To find E 1 and E 2 , using the LinearSolve command in Maple, we get graG = C ran , where D C=
−1 , 1
D=
−5 . 1
It follows from Theorem 5.2 that ⎧⎛ ⎫ ⎞ ⎛ ⎞ −1 0 ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎬ ⎟ ⎜0⎟ 1 ⎟ x1 + ⎜ ⎟ x2 | x1 , x2 ∈ R , graE 1 = ⎜ ⎝ −5 ⎠ ⎝1⎠ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 1 1 ⎧⎛ ⎫ ⎞ ⎛ ⎞ −1 1 ⎪ ⎪ ⎪ ⎪ ⎨⎜ ⎬ ⎜5⎟ 1 ⎟ ⎜ ⎜ ⎟ ⎟ graE 2 = ⎝ . + | x , x ∈ R x x 1 2 1 2 ⎝0⎠ −5 ⎠ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 1 0 Acknowledgments The authors thank two anonymous referees for their constructive comments which have improved the presentation of the paper. The authors also thank Dr. Heinz Bauschke for bring their attentions of [10] and many valuable discussions.
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