Theorem 5.4 For any data s > 0, 0 6= l ≥ 0, problem (Pλn−λ2) has optimal value 0 if
and only if G = Kn. In this case,
wij = sisj P k<h l2 khsksh for 1 ≤ i < j ≤ n, is optimal.
Proof. Any feasible solution w with λn= λ2 =: λ satisfies w 6= 0 and λ > 0.
Because µ only serves to shift the trivial eigenvalue 0 of the Laplacian, we shift it appro- priately, i. e., in such a way that all the eigenvalues of DLwD + µD−111TD−1 are the same
(thus µ > 0). Because of Courant-Fischer this is equivalent to λI DLwD + µD−111TD−1 λI
which is equivalent to
Lw+ µD−211TD−2 = λD−2.
For 1 ≤ i < j ≤ n this forces wij = µsisj 6= 0, therefore the graph must be complete.
Furthermore the i-th diagonal element must satisfy X j∈N,j6=i wij + µs2i = µsi X j∈N sj = λsi, thus µ = Pλ i∈N si = λ kD−11k2. The constraint 1 = X i<j l2ijwij = µ X i<j l2ijsisj = λ kD−11k2 X i<j l2ijsisj determines λ.
Next we describe some optimal dual realizations for G = Kn.
Example 5.5 (Complete Graphs) Let G be the complete graph Kn with given data s >
0 and 0 6= l ≥ 0. By Theorem 5.4 and strong duality, any optimal solution ξ, U = [u1, . . . , un] and V = [v1, . . . , vn] of (Eλn−λ2) fulfills ξ = 0. Because all wij are positive,
An optimal d-dimensional realization of (Dλn−λ2) (1 ≤ d ≤ n − 1) is given by taking M ⊆ N , where |M | = d + 1 and ξ = 0, Xkl = Ykl= 1 d¯s(M ) ¯ s(M \ {k}) for k, l ∈ M, k = l, −√sksl for k, l ∈ M, k 6= l, 0 otherwise. (5.6)
Note, the n − d − 1 nodes of N \ M are embedded in the origin.
For illustration, Figure 5.3 shows a one-, a two- and a three-dimensional realization of the tetrahedron with s = 1 and l = 1 constructed by (5.6) with node sets M1 = {1, 2},
M2 = {1, 2, 3} and M3 = {1, 2, 3, 4}, respectively. −0.6 −0.4 −0.2 0 0.2 0.4 0.6 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 1 3, 4 2 −0.5 0 0.5 −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5 3 4 1 2 −0.5 0 0.5 −0.5 0 0.5 −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5 4 1 2 3
Figure 5.3: A one-, a two- and a three-dimensional realization of the tetrahedron (see Example 5.5).
If the strictly active subgraph is not connected, the problem almost decomposes into sub- problems (Pλn) on the components. More precisely, the value of λ2 is zero and the min-
imization of the maximum eigenvalue leads to an identical maximum eigenvalue on each component consisting of at least two nodes. In order to state and prove this result in detail, we denote by LN0
w and DN
0
the principal submatrix of the weighted Laplacian Lw and of
D, respectively, with indices i ∈ N0 ⊆ N .
Proposition 5.6 Given G = (N, E 6= ∅) and data s > 0, 0 6= l ≥ 0, let λ2, λn, w
be optimal for (Pλn−λ2) and let the strictly active subgraph Gw consist of k connected
components Gh = (Nh, Eh), h = 1, . . . , k. Then
(i) k > 1 if and only if λ2 = 0,
(ii) there is an optimal ¯w ≥ w so that for each component h = 1, . . . , k, λn = λmax(DNhLwN¯hDNh) if and only if Eh 6= ∅.
Proof. (i) was already observed by Fiedler [25], but the argument is short: Suppose k ≥ 2, then q = D−1 1 ¯ s(N1) X i∈N1 ei− 1 ¯ s(N2) X i∈N2 ei !
is an eigenvector to λ2 = 0, because DLwDq = 0 and 1TD−1q = 0.
Let otherwise λ2 = 0 and assume for contradiction that G is connected. Then Lw is
irreducible as well as the matrix 2 max{[Lw]ii : i ∈ N }I − Lw. Moreover the previous
matrix is nonnegative and positive definite. Applying Perron-Frobenius yields that the spectral radius of 2 max{[Lw]ii : i ∈ N }I − Lw is a simple eigenvalue which is equal to
2 max{[Lw]ii : i ∈ N }−λ1(Lw). Thus λ1(Lw) is a simple eigenvalue of Lwwhich contradicts
λ2(Lw) = 0 thus G is not connected.
For (ii), we know λn > 0 by Theorem 5.4. Because Lw consists of independent principal
submatrices corresponding to the connected components, there is at least one block Nh
with λn = λmax(DNhLwNhDNh) and a wij > 0 with ij ∈ Eh having lij > 0, otherwise the
solution could be improved.
Suppose E¯h = ∅ for some ¯h, then the component is an isolated node, |N¯h| = 1, and 0 = LNw¯h,
so λn> λmax(DNh¯L N¯h w DN¯h).
Suppose now there is a connected component (N¯h, E¯h 6= ∅) with λn> λmax(DN¯hL N¯h w DN¯h).
If all lij = 0 for ij ∈ E¯h, the weights wij can be increased to some values ¯wij so that
λn = λmax(DN¯hL N¯h
¯
w DNh¯) on this component. Otherwise, slightly increasing the weight
of a wij with lij > 0 and ij ∈ E¯h and decreasing the weights of all components with
λn = λmax(DNhLNwhDNh) allows to preserve feasibility and to improve the solution at the
same time, so this contradicts optimality.
Remark 5.7 By Proposition 5.6 and its proof, the number of components of the strictly active subgraph with at least one edge is a lower bound on the dimension of the eigenspace corresponding to the maximum eigenvalue λn of DLw¯D for ¯w of the proof.
By summing the KKT conditions (5.4) over all nodes of the graph, or alternatively over the nodes of each connected component, it follows that with respect to optimal V the equilibrium constraint holds automatically for each connected component.
Proposition 5.8 Given G = (N, E 6= ∅) and data s > 0, 0 6= l ≥ 0, let ξ, U , V = [v1, . . . , vn] be optimal for (Eλn−λ2) and λ2, λn, w be optimal for (Pλn−λ2). For any con-
nected component (Nh, Eh) of the strictly active subgraph, of the active subgraph, or of the
graph itself, the weighted barycenter with respect to V is in the origin, i. e., P
i∈Nhsivi = 0.
In particular, nodes i ∈ N isolated in Gw satisfy vi = 0.
Considering the U -realization, the barycenter of the entire graph is explicitly forced to lie in the origin. This, however, does not extend to the connected components. In fact, whenever the strictly active subgraph is not connected, the optimal U -realizations of each component collapse to single points.
Proposition 5.9 Given G = (N, E 6= ∅) and data s > 0, 0 6= l ≥ 0, let U = [u1, . . . , un]
be optimal for (Eλn−λ2) and λ2, w be optimal for (Pλn−λ2). The strictly active subgraph
Gw = (N, Ew) is not connected if and only if ui = uj for ij ∈ Ew if and only if λ2 = 0.
Proof. The claim follows from semidefinite complementarity hX, DLwD + µD−111>D−1− λ2Ii =
X
ij∈E
wijkui− ujk2− λ2 = 0 (5.7)
and Proposition 5.6(i), i. e., λ2 = 0 if and only if Gw is not connected.
If G itself is not connected there exists an optimal one-dimensional U (independent of an optimal V ). To see this, split the graph into two disjoint node sets such that no edges connect nodes in distinct sets. Each set is mapped onto a separate coordinate so that the normalization constraint and the equilibrium constraint are satisfied.
If G is connected but its strictly active subgraph Gw is not, no optimal one-dimensional
realizations U need to exist for a given optimal V , because the distance constraints of inactive edges may cause problems. This is illustrated by the following example.
Example 5.10 Consider the graph of Figure 5.4 with data s = 1 and l = 1. In the optimal solution the edges in the dashed triangle on nodes {1, 2, 3} have optimal weight zero, all others wij = 19 giving λ2 = 0 and λn= 49 = ξ. The strictly active subgraph consists of three
stars with centers 1, 2 and 3.
2 3
1
Figure 5.4: There may be no one-dimensional realization U even if Gw is not connected.
In an optimal realization U , each star is mapped onto one point.
In an optimal V each star is embedded in a one-dimensional subspace with its center at distance 12 from the origin opposite to its three leaves which are embedded on top of each other at distance 16 from the origin, see Corollary 5.16 below. Arranging the three centers in an equilateral triangle with barycenter in the origin results in mutual squared distances kvi − vjk2 = 34 for i, j ∈ {12, 13, 23}.
Hence, kui− ujk2 ≤ 34 − 49 = 1136 (ij ∈ {12, 23, 13}). This is satisfied, e. g., by embedding
the ui for i = {1, 2, 3} in an equilateral triangle around the origin with kuik = 2√13, each
having its leaves embedded on top of it.
The mentioned optimal realizations are illustrated in Figure 5.5.
−0.2 −0.1 0 0.1 0.2 −0.25 −0.2 −0.15 −0.1 −0.05 0 0.05 0.1 0.15 0.2 0.25 1, 4, 5, 6 2, 7, 8, 9 3, 10, 11, 12 −0.4 −0.2 0 0.2 0.4 −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 1 2 3 4, 5, 6 7, 8, 9 10, 11, 12
Figure 5.5: Optimal two-dimensional realizations U and V of the graph of Figure 5.4 for (Eλn−λ2), cf. Example 5.10.
There is, however, no optimal one-dimensional realization ui = xih (i ∈ N ) with h ∈ Rn,
khk = 1 and xi ∈ R (i ∈ N), because it would have to satisfy the following infeasible
system,
equilibrium constraint 4x1 + 4x2+ 4x3 = 0,
normalization constraint 4x2
1 + 4x22+ 4x23 = 1,
distance constraints (xi − xj)2 ≤ 1136 (ij ∈ {12, 23, 13}).
The normalization constraint specifies a ball with radius 1
2 and the distance constraints
specify a polyhedron which lies completely in the ball, so they have no points in common. The equilibrium constraint is just an additional plane through the origin.
Figure 5.6 illustrates the situation after we have eliminated x3 by the equilibrium constraint.
The normalization constraint specifies the red ellipse and the distance constraints the green polyhedron.
The next proposition provides a bound on the length of vectors of optimal realizations of (Eλn−λ2).
Proposition 5.11 Given G = (N, E 6= ∅) and data s > 0, 0 6= l ≥ 0, let ξ, U = [u1, . . . , un], V = [v1, . . . , vn] be optimal for (Eλn−λ2) and put ˆl = (max{kui− ujk
2+ l2 ijξ : ij ∈ E})1/2. Then kuik < s −1/2 i and kvik < min{s −1/2 i , ˆl} for i ∈ N .
x2
x1
Figure 5.6: The reduced system of constraints is infeasible, cf. Example 5.10. Proof. The bound concerning the ui is a direct consequence of the normalization and equi-
librium constraint. The same argument works for the term s−1/2i of the bound concerning the vi using Proposition 5.8.
The proof for ˆl follows that of Proposition 4.5, so we only repeat the basic idea: We observe that ˆl > 0 because l 6= 0 and, by Theorem 5.4 and Example 5.5, ξ > 0 or kui− ujk2 > 0
for at least one ij ∈ E. We suppose for contradiction, that there is a node k ∈ N with kvkk = ˆl + ≥ ˆl. Then we may show that there is another feasible realization V0 (and
U ) with no smaller objective value having the barycenter of VN0 outside the origin, which contradicts the optimality of V by Proposition 5.8.
The following example illustrates that the bounds of Proposition 5.11 cannot be improved. Example 5.12 Let s = c1, l = 1 with real c > 0. For n > 2 consider the graph
G = (N, E) = ({1, . . . , n}, {{2, k} : k ∈ N \ {1, 2}}),
i. e., it consists of two components: an isolated node and a star. Let h ∈ Rn, khk = 1.
Optimal realizations of (Pλn−λ2) and (Eλn−λ2) are given by
λ2 = 0, λn= n − 1 c(n − 2), µ = 0, wij = 1 n − 2 (ij ∈ E) and ξ = n − 1 c(n − 2), ui = q n−1 cn h for i = 1, −q 1 cn(n−1)h otherwise, vi = 0 for i = 1, q n−2 c(n−1)h for i = 2, −q 1 c(n−2)(n−1)h otherwise.
Because ui = uj (ij ∈ E) the bounds are ˆl =
√
ξ > c−1/2 = s−1/2i (i ∈ N ).
For n → ∞ we obtain ˆl → c−1/2 = s−1/2i (i ∈ N ), ku1k → c−1/2 and kv2k → c−1/2. Thus,
Remark 5.13 The realizations of complete graphs described in Example 5.5 allow to con- struct a sequence of problems and solutions with the property that ˆl → 0 in Proposition 5.11. Indeed, the analysis of the realization yields kvik2 = kuik2 = d−1(s−1i − ¯s(M )
−1) and d−1(s−1i − ¯s(M )−1) < s−1i for i ∈ M . In addition, ˆ l2 = max{0, kuik2 (i ∈ M ), kui− ujk2 = d−1(s−1i + s −1 j ) (i, j ∈ M )} = max{d−1(s−1i + s−1j ) : i, j ∈ M }.
For s = c1 > 0 and d > 2 we have s−1i = c−1 > ˆl2 = 2(dc)−1 (i ∈ N ) and for, e. g., d = n − 1 and n → ∞ we obtain ˆl → 0.