CONFIDENTIAL. Limited circulation. For review only.
Further Analysis on Graph Rigidity
Minh Hoang Trinh† , Myoung-Chul Park† , Zhiyong Sun‡ , Brian. D. O. Anderson‡ , and Hyo-Sung Ahn†
Abstract— This paper presents novel results on the symmetric
rigidity matrix. First, we provide a physical meaning of eigenvectors of the symmetric rigidity matrix. Then, new concepts
in rigidity theory are proposed, including the worst-case, and
imbalance rigidity indices of planar frameworks. These new
indices are scale free and could be useful in synthesis problems
of rigid networks.
I. I NTRODUCTION
Recently, graph rigidity has come to play an important
role in formation control and network localization problems [1]–[4]. In rigidity theory, a framework is an abstract
mathematical model of a physical structure, and is defined
by a graph and a realization or embedding in some space,
typically here R2 . Frameworks can describe a large class of
physical structures, such as bridges, organic molecules, sensor network, formation of robotic systems, etc. The question
of determining whether a framework is rigid or flexible has
been studied extensively in the literature [5]–[7]. The answer
involves both combinatorial and geometrical aspects, and is
based on the rigidity matrix and for two dimensions at least,
Laman’s theorem [8].
While rigidity may be a requirement in many situations
where frameworks are relevant, the simple presence or
absence of rigidity will not usually be the sole criterion
for judging the efficacy of the framework. For example,
among sensor networks, which are normally required to be
globally rigid, there are some networks with better capacity
than others in terms of robustness and reliability to recover
their vertices positions in the presence of range measurement
noise. Thus, there is a need for some indices to quantitatively
measure formation rigidity, and to compare these quantities
among different networks. The authors of [9], [10] proposed
two indices called the worst-case rigidity index (WRI) and
the mean rigidity index (MRI) computable from the stiffness
matrix. In [10], the authors also considered an optimal
formation design problem in which links between agents
could vary continuously to maximize these indices. In a
two-dimensional space, the matrix M = RT R1 , termed the
symmetric rigidity matrix in [11], [12] or vertex rigidity
Gramian in [13], always has three zero eigenvalues when
the framework is infinitesimally rigid. The fourth smallest
eigenvalue of M , denoted by λ4 (M ), is another rigidity
index that has been widely used in the literature. Keeping
† School
of Mechatronics, Gwangju Institute
Technology (GIST), Gwangju, Republic of
of Science and
Korea. E-mail:
{trinhhoangminh,mcpark,hyosung}@gist.ac.kr
‡ College of Engineering and Computer Science, The Australian National University and National ICT Australia, ACT 0200, Australia. Email:
{zhiyong.sun,brian.anderson}@anu.edu.au
1 The
definitions will be clear in Section II
λ4 > 0 is crucial to the formation to maintain its infinitesimal
rigidity. For example, based on a weighted version of the
symmetric rigidity matrix M , the authors of [12] proposed
a decentralized rigidity maintenance control law for a multirobot system. The authors of [14] considered a rigidity optimization problem for mobile robotic systems. They proposed
a decentralized algorithm for robots to recursively update
their positions to maximize the eigenvalue λ4 . In [15], the
authors provided a distributed algorithm to maintain combinatorial rigidity of a multi-agent system. A local link addition
and deletion rule in [15] for each agent is based on redundant
rigidity theory. In [13], a heuristic design methodology for a
rigid network via submodular set function optimization was
proposed. The authors in [16] also proposed two measures
of generic rigidity and generalized redundancy in two-space
for both rigid and non-rigid graphs, which were termed the
rigidity index and the redundant index respectively.
Although the rigidity matrix R has been widely used, there
are fewer analytic results involving the matrices RT R and
RRT . This paper is devoted to studying the matrix RT R.
As the first contribution of this paper, we present a further
analysis on the eigenvalues and eigenvectors of the vertex
rigidity Gramian M = RT R. These results give insights
on the physical meaning of the nonzero eigenvalues of M ,
their dependence on the framework’s size and the addition or
removal of edges in the framework. The second contribution
proceeds from the observation that although various rigidity
indices have been proposed in the literature, for almost all
indices, the scale of the formation affects the value of the
index. For example, all equilateral triangles do not have the
same index, but an index which varies quadratically with the
length of the triangle side, a fact which to some degree is
counterintuitive. Thus, as the second and logically separate
contribution, we propose two new rigidity indices derived
from the matrix M , including the worst case Euclidean rigidity index and the imbalance index. Unlike existing indices in
the literature, these new indices are scale free and can be used
to effectively compare rigidity between different frameworks.
In formation control and network localization problems, the
new rigidity indices could be particularly useful in designing
rigid formations.
The rest of this paper is organized as follows. Section II
includes the background for this paper. Section III contains
further results on the symmetric rigidity matrix. New rigidity
indices are defined in Section IV. Finally, some concluding
remarks and suggestions for future works are given in Section
V.
Preprint submitted to 55th IEEE Conference on Decision and Control.
Received March 7, 2016.
CONFIDENTIAL. Limited circulation. For review only.
II. P RELIMINARIES
The following notations will be used in this article
n
• R : n−dimensional Euclidean space
m×n
• R
: the set of m by n real matrices
• [A]k : the k-th row of matrix A
• [A]ij : the entry in the i−th row and j−th column of
matrix A
• N (A): the null space of matrix A
• In : the n by n identity matrix
• A ⊗ B: the Kronecker product of matrices A and B
• kxk: the Euclidean norm of a vector x
m
• diag(Ai )i=1 = diag(A1 , . . . , Am ) : the block diagonal
matrix formed from m matrices A1 , . . . , Am
• |S|: the cardinality of a set S.
This section briefly summarizes some definitions from
algebraic graph theory and rigidity theory[6], [17]. Let
G = (V, E) be an undirected graph with a vertex set V =
{1, . . . , n}, and an edge set E ⊂ {(i, j) : i, j ∈ V, i 6= j}
with |E| = m edges. The neighbor set of a vertex i ∈ V is
defined as Ni := {j ∈ V : (i, j) ∈ E}. Further, Kn will be
used to denote the complete graph in n vertices.
For each vertex i in V, we associate i with a point pi =
[pxi , pyi ]T ∈ R2 . Then, the formation shape is determined by
a realization p = [pT1 , . . . , pTn ]T ∈ R2n . The graph G and a
realization p ∈ R2n together define a framework F = (G, p)
in the two dimensional plane. Two realizations p and p0 of a
graph G are similar if there exists a scale factor ζ > 0 such
that kpi − pj k = ζkp0i − p0j k, for all i, j ∈ V. Specifically,
when ζ = 1, p and p0 are called congruent.
Let H ∈ Rm×n be the incidence matrix of G with an
arbitrary edge ordering and orientations. The entries of H
are given as follows

the k-th edge sinks at node i
 1,
−1, the k-th edge leaves at node i
[H]ki =

0,
otherwise.
For any edge in E which is consistent with the construction
of H, let zk = pj − pi , k = 1, . . . , m. In this paper, zk
is sometimes used to refer to a corresponding edge in E
without ambiguity. If two edges zi , zj ∈ E share a common
vertex and have opposite (same) directions relative to their
shared vertex, we call zi , zj positively (negatively) adjacent
and denote zi ∼+ zj (resp., zi ∼− zj ). By convention, an
edge is not considered
adjacent
to itself.
T T
Define z = z1T , . . . , zm
∈ R2m ; then we have z =
2n
(H ⊗ I2 )p. Also, let fG : R 7→ Rm ,
has nonzero entries in columns 2i − 1, 2i, 2j − 1 and 2j, as
given below.
[R]k = · · · pTi − pTj · · · pTj − pTi · · · .
A framework F is infinitesimally rigid if and only if
rank(R) = 2n − 3 [8]. Note that R can be written as
R = D(z)T (H ⊗ I2 ),
diag(zi )m
i=1
(1)
2m×m
where D(z) =
∈R
. Now, the symmetric
rigidity matrix [12] is defined as follow
M := RT R ∈ R2n×2n .
The matrix M is symmetric and positively semidefinite.
Although the rigidity matrix R is widely used in infinitesimal
rigidity theory, there is a less use of the symmetric rigidity
matrix. The next section focuses on investigating some
properties of the matrix M .
III. T HE S YMMETRIC R IGIDITY M ATRIX
A. Physical meaning of eigenvectors of the symmetric rigidity matrix
Consider a framework F = (G, p) in the plane with the
rigidity matrix R. Let λk , k = 1, 2, . . . , 2n, denote the
k-th eigenvalue of M and v k , k = 1, 2, . . . , 2n, be its
corresponding eigenvector. Frther, let v k = [v1kT , . . . , vnkT ]T ,
where each vik is a 2-vector. For convenience, we further
assume that 2n eigenvalues of M are ordered as follows
λ1 ≤ λ2 ≤ . . . , λ2n . We have the following fact.
Theorem 1: [11] Consider a framework F = (G, p) in
the plane, with the rigidity matrix R ∈ Rm×2n and let
M = RT R ∈ R2n×2n . Then,
(i) rank(M ) ≤ 2n − 3, or M has at least three zero
eigenvalues;
(ii) N (M ) = N (R);
(iii) The framework F is infinitesimally rigid if and only
if rank(M ) = 2n − 3. Equivalently, the fourth smallest
eigenvalue λ4 of M is positive.
Three eigenvectors associating with three zero eigenvalues
of M span the null space of the rigidity matrix R and can
be found as [14]
v 1 = [1
0
1
0
...
1
0]T ,
v 2 = [0
1
0
1
...
0
1]T ,
px2
...
v3 =
[−py1
px1
−
py2
− pyn
pxn ]T .
fG (p) := (kpi − pj k2 )(i,j)∈E = [kz1 k2 , . . . , kzm k2 ]T
Observe that v 1 and v 2 are orthogonal, while v 3 is not
orthogonal to either v 1 or v 2 ; we construct from v 3 a vector
orthogonal to v 1 and v 2 by using the Gram-Schmidt process:
as the edge function of the framework F = (G, p); then, the
rigidity of frameworks can be defined as follows.
Definition 1: [6] A framework (G, p) is rigid in R2 if there
exists a neighborhood U of p in R2n such that fG−1 (fG (p)) ∩
−1
U = fK
(fK (p)) where K is the complete graph with the
same vertex set as G.
The rigidity matrix R ∈ Rm×2n is defined as the Jacobian
matrix R = 21 ∂fG (p)/∂p. Furthermore, suppose that the kth edge of E connects vertices i and j, the k-th row of R
v 3 = v 3 + p̄y v 1 − p̄x v 2 ,
Pn
Pn
where p̄x = n1 i=1 pxi and p̄y = n1 i=1 pyi are correspondingly x-, and y-coordinates of the formation’s centroid.
0
Intuitively, if we regard each vector v 1 , v 2 , v 3 as a
stacking of 2-vectors associated with n vertices and those 2vectors are velocities at the nodes in question, the framework
will translate along the x−, y− axes with unit speed, or rotate
around its centroid, respectively.
0
Preprint submitted to 55th IEEE Conference on Decision and Control.
Received March 7, 2016.
CONFIDENTIAL. Limited circulation. For review only.
Next, we will give a physical interpretation of the eigenvectors of the matrix M based on the study of statics
for frameworks [18]. Consider the framework as an actual
physical object whose edges are straight and comprise stiff
rods connected by articulated joints at the vertices.
For each eigenvector v k = [v1kT , . . . , vnkT ]T ∈ R2n of the
matrix M , we have,
M v k = RT Rv k = λk v k .
j∈Ni
Let w = Rv k ∈ Rm , it follows that
M v k = RT (Rv k ) = RT w = λk v k ,
k = 1, . . . , 2n. (2)
m
We can write w = [wij ](i,j)∈E ∈ R , and (2) can be
expressed as
X
wij (pi − pj ) = λk vik ,
(3)
j∈Ni
where i = 1, . . . , n and k = 1, . . . , 2n. Now, we regard each
vector v k as a stacking of 2-vectors force associated with
each of the n joints of the framework. More specifically,
wij (pi − pj ) is the force exerted by the rod on the vertex pi
along the rod (i, j). The scalar wij gives the magnitude of
the force per unit length. The force is called a tension in the
rod if wij < 0, and a compression if wij > 0.
Next, supposing that F is infinitesimally rigid, we focus on the nonzero eigenvalues λk , k = 4, . . . , 2n, and
their corresponding eigenvectors of M . A vector F =
[F1T , . . . , FnT ]T ∈ R2n is an equilibrium force for p if
n
X
Fi = 0,
(4)
pi × Fi = 0,
(5)
i=1
n
X
Theorem 2: Given an infinitesimally rigid framework F in
a plane. Then each vector F = −λk v k (k = 4, . . . , 2n−3) is
a resolvable force, where v k is the eigenvector corresponding
to a nonzero eigenvalue of the symmetric rigidity matrix M
of F.
Observe further that (7) can be rewritten as
X
− vik + λ−1
wij (pi − pj ) = 0, ∀i = 1, . . . , n, (8)
k
i=1
where × denotes the cross product. The two conditions
together imply that the sum of the forces Fi is zero and
the moments about any axis is zero. We say that F is a
resolvable force for F = G(p) if there exist scalars wij , one
for each edge (i, j) of F, such that
X
Fi +
wij (pi − pj ) = 0, ∀i = 1, . . . , n.
(6)
j∈Ni
There is a connection between the concepts of equilibirium
force and resolvable force.
Lemma 1: [7, Proposition 4.3] Suppose F = G(p) is a
framework in R2 where p1 , . . . , pn are not collinear. Then F
is infinitesimally rigid in R2 if and only if every equilibrium
force for p is a resolvable force for F. Moreover, each
equilibrium force is uniquely resolvable if and only if F
is infinitesimally rigid and stress free.
It follows from (3) that
X
− λk vik +
wij (pi − pj ) = 0, ∀i = 1, . . . , n. (7)
j∈Ni
Let Fi = −λk vik , i = 1, . . . , n and F = [F1T , . . . , FnT ]T =
−λk v k . Based on Lemma 1 and above analysis, we have
proved the following result.
where wij is depend on v k , i.e. wij = wij (v k ).
We define each eigenvector v k a prime force of the framework F. Then λk is the scalar that makes vk a resolvable
force. Loosely speaking, the bigger λk is, the smaller the
magnitude of the prime force needed to make the framework
static.
Now, considering the eigenvectors v k , k = 1, 2, 3, corresponding to a zero eigenvalue of M . Let us recall several
concepts from [7]. A stress 2 of a framework is a collection
of scalars wij , one for each edge (i, j) of G, such that
X
wij (pi − pj ) = 0, ∀i = 1, . . . , n.
(9)
j∈Ni
A stress is called trivial when wij = 0, ∀(i, j) ∈ E.
Moreover, a framework is termed stress free if it admits only
the trivial stress3 .
For each eigenvector corresponding to zero eigenvalue of
M , we have M v k = 0v k = 0, k = 1, 2, 3. Then, it follows
from N (M ) = N (R) that w = Rv k = 0. Therefore, w
immediately satisfies (9) and we have a trivial stress in this
case.
Next, considering an arbitrary external force vector v ∈
R2n . Since the nullspace and the row space of M are always
orthogonal and they together span whole R2n , v can always
be decomposed into two orthogonal vectors:
v = vnull + vrow
where vnull in N (M ) and vrow in the row space of M . Then,
vnull is associated with trivial stresses and congruent motions
of the framework, while vrow is an equilibrium force.
Example 1: Consider a K3 equilateral triangle framework
(each side’s length is 1). Since the graph is K3 , it is minimally infinitesimally rigid. Thus, M has six linearly independent eigenvectors. Three eigenvectors corresponding to zero
eigenvalues are in N (M ). We focus on three eigenvectors
v k = [v1kT , v2kT , v3kT ]T ∈ R6 corresponding to λ4 , λ5 , λ6 >
0. The force vectors Fi = −λk vik corresponding to pi ,
i = 1, 2, 3, are depicted in Fig. 1.
In equilateral triangle frameworks, we also have λ4 =
λ5 = 12 λ6 (see Fig. 2).
Hence, if we provide a force along the directions depicted
in Fig. 1c to break the framework, we need twice of
2 This term is also called “equilibirium stress” in [19]. The terminology
is not consistent with normal use of the term in physics or engineering, but
is consistent with standard literature [7], [18].
3 A framework is stress free if and only if rank(R) = m, the number of
edges of the framework [7]. Let F be an infinitesimally rigid framework,
so that we have rank(R) = 2n − 3. Thus, the condition for F to be stress
free is m = 2n − 3, i.e. that F is minimally infinitesimally rigid.
Preprint submitted to 55th IEEE Conference on Decision and Control.
Received March 7, 2016.
CONFIDENTIAL. Limited circulation. For review only.
(a) λ4
(b) λ5
(c) λ6
Fig. 1: The resolvable force assigned to vertices of an equilateral triangular framework corresponding to λk , k = 4, 5, 6.
magnitude than the direction of Fig. 1a and Fig. 1b. So,
the minimization of kλi − λj k, 4 ≤ i < j ≤ 2n, means that
the magnitudes of prime forces to break the realized rigid
graph are distributed in a more balanced way.
B. Influence of the framework’s scale, translation, orientation, and edge addition on the eigenvalues of the symmetric
rigidity matrix
The following Lemma characterizes the influence of the
eigenvalues of the matrix M
Lemma 2: The eigenvalues of the symmetric rigidity matrix M associated with a (G, p) are invariant to translation,
rotation and reflection of the framework.
The proof of 2 is trivial and will be omitted. We will now
look at the influence of the scale and the edge addition in
the remainder of the subsection.
Since a framework is defined based on a graph and a
realization in the space, the eigenvalues of M depends on
both the vertex positions and the associated graph. The
following fact characterizes the influence of framework’s
scale on the eigenvalues of the symmetric rigidity matrix.
Theorem 3: Consider two frameworks F1 = (G, p) and
F2 = (G, q), which have the same graph G = (V, E). Let
R1 , R2 be correspondingly rigidity matrices of F1 and F2 .
Denote the eigenvalues of M1 = R1T R1 and M2 = R2T R2
by λ1 ≤ . . . ≤ λ2n and λ01 ≤ . . . ≤ λ02n . If F1 is similar
with F2 by a scale factor ζ > 0, we have
λi = ζ 2 λ0i , ∀i = 1, . . . , 2n.
(10)
Proof: Since F2 is similar with F1 by a scale factor
ζ > 0, there exists a rotation matrix Ω ∈ R2×2 such that
zkT = ζΩzk0T , 1 ≤ k ≤ m.
(11)
From (1) and (11), we can write
R1 = diag(ziT )m
i=1 (H ⊗ I2 )
= ζ(Ω ⊗ Im )diag(zi0T )m
i=1 (H ⊗ I2 )
= ζ(Ω ⊗ Im )R2
T
(12)
2
Since Ω Ω = I2 , it follows that M1 = ζ M2 , the result
is then immediate.
Now, we analyze the influence of adding or removing bars
on the framework. In combinatorial rigidity theory, we know
that adding bars may convert a nonrigid framework to a
rigid one. Also, removing a bar from a rigid framework may
make it flexible. In order to quantify the effect of adding or
removing bars to the symmetric rigidity matrix, we have the
following Theorem.
Theorem 4: Consider an infinitesimally rigid framework
in the plane F = (G, p) with G = (V, E), |E| = m. Consider
the action of adding a new edge (i, j) to E, where (i, j) ∈
/
E, (i, j) ∈ V × V, and i 6= j. That action yields a new
framework F 0 = (G 0 , p) with G 0 = (V, E ∪ zm+1 ), |E| =
m + 1 having the same realization as p ∈ R2n . Assume that
the symmetric rigidity matrices of two frameworks M and
M 0 correspondingly have eigenvalues λ1 ≤ . . . ≤ λ2n and
λ01 ≤ . . . ≤ λ02n . The framework F 0 is more rigid than F,
in the sense that λ0i ≥ λi , ∀i = 1, . . . , 2n.
Proof: Since F is infinitesimally rigid, the framework
F 0 obtained by adding an edge (i, j) to E, is also infinitesimally rigid. Let H be the incidence matrix of F, then the
incidence matrix H 0 of F 0 can be written as
H
H0 = T
,
hm+1
where hm+1 ∈ Rn . The i−th and the j−th entries of hm+1
is 1 and −1 correspondingly, while other entries are all 0.
Then, the matrix M 0 can be expressed as
T
M 0 = M + (hm+1 ⊗ I2 )zm+1 zm+1
(hTm+1 ⊗ I2 ).
(13)
Equation (13) implies that the addition of a new edge is
equivalent to perturb the symmetric rigidity matrix M by
a rank-1 symmetric positive semidefinite matrix (hm+1 ⊗
T
(hTm+1 ⊗ I2 ). Since any symmetric matrix
I2 )zm+1 zm+1
perturbed by a positive semidefinite matrix will result in a
new matrix with non-descreasing eigenvalues [21], the claim
of Theorem 4 follows immediately.
The intuitive physical meaning of Theorem 4 is that adding
a new bar to a rigid framework makes the framework more
rigid. Also, the removal of a bar makes a rigid framework
become less rigid. The following result follows immediately
from Theorem 3 and Theorem 4.
Corollary 1: Consider a framework F = (G, p) and the
set of frameworks F 0 = (G 0 , p) with the same vertex set as
V, then the framework with G 0 = Kn is the most rigid in
this set.
Preprint submitted to 55th IEEE Conference on Decision and Control.
Received March 7, 2016.
CONFIDENTIAL. Limited circulation. For review only.
IV. T HE N EW R IGIDITY I NDEX
In the literature, there are several works on rigidity maintenance that considered the fourth smallest eigenvalue of M ,
λ4 , as an index of rigidity and tried to maximize λ4 . However,
it appears from one point of view at least that λ4 is not the
best variable to measure framework’s rigidity. This argument
will be made clear from the following example.
Let us consider two triangle frameworks which are similar
with the scale factor ζ. Theorem 2 suggests that by increasing
ζ, λ4 will be quadratically increased. If we consider λ4
as the measure of rigidity, it follows that a bigger triangle
framework is more rigid than a smaller one. Thus, λ4 alone
is not a good index to compare the degree of rigidity between
different frameworks, since λ4 depends on framework’s size.
Instead of λ4 , let us define a new measure of rigidity.
Definition 2: The worst-case rigidity index of a framework F = (G, p), denoted by χ , is the ratio between
the fourth smallest eigenvalue to the summation of all
eigenvalues of the symmetric rigidity matrix M , i.e.,
λ4
χ = P2n
λ4
.
(14)
tr(M )
i=1 λi
It follows from Definition 2, for any framework, χ = 0
means that the framework is not infinitesimally rigid. Moreover, χ is scale free as stated in the following Proposition.
=
Proposition 1: Consider two frameworks F1 = (G, p)
and F2 = (G, q) satisfying condition (10) in Theorem 3.
Let χ1 , χ2 be worst-case rigidity indices of F1 and F2
accordingly. Then χ1 = χ2 .
Proof: It follows from Theorem 3 that the eigenvalues
of M and M 0 satisfy λi = ζ 2 λ0i , ∀i = 1, . . . , 2n. Hence,
ζ 2 λ04
λ4
χ1 = P2n
i=1
λi
= P2n
i=1
ζ 2 λ0i
λ04
= P2n
i=1
λ0i
= χ2 ,
or i.e., all similar frameworks have the same degree of
rigidity.
P2n
P2n
It is worth
that Pi=1 λi =
i=4 λi =
P
Pn mentioning
m
2
2
tr(M ) =
d
d
=
2
.
As
discussed
in
ij
i
i=1
j∈Ni
i=1
Section 3, each eigenvector of M (prime force) represents
a prime direction to decompose an applied force and the
inverse of each eigenvalue of M represents the sensitivity
of M along each prime direction of the framework. Conseλ4
quently, χ = tr(M
) characterizes how rigid F is along the
most sensitive principal direction. Since χ is scale free, we
can compare the worst-case rigidity between frameworks.
Motivated from the observation in Example 1, it seems
reasonable to search for a graph that has balanced sensitivities along all prime directions of the framework. To this end,
we now propose the concept of imbalance index.
Definition 3: The ratio between the fourth eigenvalue and
the largest eigenvalue of the symmetric rigidity matrix M of
an infinitesimal rigid framework F, denoted by ξ, is defined
as the imbalance index of the framework.
ξ=
λ4
.
λ2n
(15)
The imbalance index ξ is also scale free. Furthermore,
ξ = 0 when F is not rigid and 0 < ξ < 1, otherwise.
Now, we try to find a bound for eigenvalues of M . For
that, we recall two useful lemmas.
Lemma 3: [20, Theorem 1.3.22] The symmetric rigidity
matrix N = RRT and the matrix M = RT R have the same
nonzero eigenvalues with the same multiplicities.
Lemma 4: [22, Theorem 2] Consider a symmetric matrix
A ∈ Rn×n . Furthermore, let λmin and λmax be the smallest
and largest eigenvalue of A, respectively. Then,
q
|λmax − λmin | ≥ max
([A]ii − [A]jj )2 + 4[A]2ij .
1≤i6=j≤n
Proposition 2: Consider an infinitesimally rigid framework F with n vertices and m ≥ 2n − 3 edges. Then the
largest eigenvalue of the matrix M , λ2n , satisfies
q
(d2i − d2j )2 + (di dj cos θij )2 , (16)
λ2n ≥ 2 max
1≤i6=j≤m
zT z
zT z
where cos θij = kziikkzjj k = dii djj .
Proof: Based on Lemma 3, the nonzero eigenvalues of
M = RT R are the same as the nonzero eigenvalues of the
matrix
N = RRT = D(z)(H ⊗ I2 )(H T ⊗ I2 )D(z)T
T
T m
= diag(zi )m
i=1 (HH ⊗ I2 )diag(zi )i=1 .
We have

2,
i = j,



+
1,
z
i ∼ zj ,
HH T ij =
−
−1, zi ∼ zj ,



0, otherwise.
Thus, the entries of N are given by

2z T zi ,
i = j,


 Ti
zi zj ,
zi ∼+ zj ,
[N ]ij =
−ziT zj , zi ∼− zj ,



0,
otherwise.
As a result, the matrix N ∈ Rm×m is symmetric and
positively semedefinite. Applying Lemma 4, since ziT zi =
d2i , kziT zj kq
= di dj cos θij , we have λmax (N ) − λmin (N ) ≥
(d2i − d2j )2 + (di dj cos θij )2 .
2 max
1≤i6=j≤m
Observe that when m > 2n − 3, since N has rank 2n − 3,
it has at least one zero eigenvalue 0. Thus, λmin (N ) = 0,
λmax (N ) = λmax (M ) = λ2n (M ), and it follows that
q
(d2i − d2j )2 + (di dj cos θij )2 .
λ2n (M ) ≥ 2 max
1≤i6=j≤m
To end this section, we will use Proposition 2 to roughly
estimate χ in two simple examples. Consider an equilateral
triangle framework with K3 underlying graph and each side’s
length is a as depicted in Fig. 2. Then, m = 3 = 2 × 3 − 3.
It follows from Proposition 2 that λ6 (M ) − λ4 (M ) ≥ a2 .
In this case, tr(M ) = 6a2 . Thus, 6a2 = λ6 (M ) + λ5 (M ) +
λ4 (M ) > a2 + 3λ4 (M ). It follows that λ4 (M ) < 5a2 /3 and
χ(M ) < 5/18. Calculation shows that χ(M ) = 1/4.
Preprint submitted to 55th IEEE Conference on Decision and Control.
Received March 7, 2016.
CONFIDENTIAL. Limited circulation. For review only.
Fig. 2: Eigenvalues and rigidity indices of a K3 equilateral
triangle framework side a.
Next, let the framework be a square (each side’s length is
a) with K4 underlying graph as in Fig 3. Since m = 6 √
> 2×
4 − 3 = 5, applying Proposition 2 yields, λ8 (M ) ≥ 2 2a2 .
Thus, √
χ = λ4 /tr(M ) ≤ (tr(M ) − λ8 (M ))/(4tr(M )) =
(8 − 2)/32. Exact calculation shows that in this case
χ(M ) = 1/8. We can also compare χ in two cases, where the
graph is K4 removing one edge and have the same realization
as depicted in Fig. 4–Fig. 5. It is easy to check that the
eigenvalues are decreased in these cases, which agrees with
Theorem 4.
Fig. 3: Eigenvalues and rigidity indices of a K4 square
framework side a.
Conjecture 1: If the framework F = (Kn , p) is infinitesimally rigid and its realization forms a convex polygon of n
vertices, χ and ξ are maximized when the realization is an
n-regular polygon.
Up to now, we have not found a method to to prove
Conjecture 1, other than for the case of triangular formations
treated below. The analysis requires treatment for all different
realizations p of F in the plane. The difficulties of the general
problem will partly illustrated be shown in the Appendix A,
when we provide a case study in K3 graphs. It is noted
that Conjecture 1 rules out all cases when the framework’s
realization is not a convex n-polygon. It is hard to use
intuition to even roughly estimate the values of χ and ξ.
V. C ONCLUSIONS AND FUTURE WORK
In this paper, we developed some properties of the rigidity
matrix M = RT R and proposed two new rigidity indices
from M , namely the worst-case rigidity index χ and the
imbalance index ξ. These indices are scale free and can be
used as a tool to compare rigidity between frameworks.
For future work, we are now working to unite the analysis
on the eigenvalues of the matrix M with these new rigidity
indices; in particular, we would like to understand how
the new indices vary when an edge is added to a graph.
All results in this paper are in two dimensions and could
be extended into three dimensions. Also, the constraint in
Proposition 2 is still not tight and need to be improved.
Conjecture 1 is difficult to prove completely, however, it
may be experimentally verifiable by exhaustive testing by
computers as in the case study of K3 frameworks. Finally,
applications of the new rigidity indices will be further
studied, especially in optimal network design problems.
ACKNOWLEDGEMENTS
Fig. 4: Eigenvalues and rigidity indices of a K4 removing
one edge square framework side a.
The work of B. D. O. Anderson and Z. Sun was supported
by NICTA, which is funded by the Australian Government
through the ICT Centre of Excellence program, and by
the Australian Research Council under grant DP130103610
and DP160104500. Z. Sun is also supported by the Prime
Minister’s Australia Asia Incoming Endeavour Postgraduate
Award from Australian Government. The work of M. H.
Trinh and H.-S. Ahn was supported by the National Research
Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (NRF-2015M2A8A4049953).
R EFERENCES
Fig. 5: Eigenvalues and rigidity indices of a K4 removing
one edge square framework side a.
We state here a conjecture for frameworks F = (Kn , p)
with a complete graph.
[1] R. Olfati-Saber and R. M. Murray, “Graph rigidity and distributed formation stabilization of multi-vehicle systems,” in in IEEE Conference
on Decision and Control, 2002, pp. 2965–2971.
[2] J. Aspnes, T. Eren, D. K. Goldenberg, A. S. Morse, W. Whiteley, Y. R.
Yang, B. D. Anderson, and P. N. Belhumeur, “A theory of network
localization,” IEEE Transactions on Mobile Computing, vol. 5, no. 12,
pp. 1663–1678, 2006.
[3] B. D. O. Anderson, C. Yu, B. Fidan, and J. M. Hendrickx, “Rigid
graph control architectures for autonomous formations,” IEEE Control
Systems Magazine, pp. 48–63, 2008.
[4] I. Shames, A. N. Bishop, and B. D. O. Anderson, “Analysis of noisy
bearing-only network localization,” IEEE Transactions on Automatic
Control, no. 1, pp. 247–252, 2013.
[5] G. Laman, “On graphs and rigidity of plane skeletal structures,”
Journal of Engineering Mathematics, vol. 4, no. 4, pp. 331–340, 1970.
Preprint submitted to 55th IEEE Conference on Decision and Control.
Received March 7, 2016.
CONFIDENTIAL. Limited circulation. For review only.
[6] L. Asimow and B. Roth, “The rigidity of graphs, ii,” Journal of
Mathematical Analysis and Applications, vol. 68, no. 1, pp. 171–190,
1979.
[7] B. Roth, “Rigid and flexible frameworks,” The American Mathematical
Monthly, vol. 88, no. 1, pp. 6–21, 1981.
[8] J. Graver, B. Servatlus, and H. Servatius, Combinatorial rigidity.
Graduate Studies in Mathematics, Vol. 2, American Mathematical
Society, 1993.
[9] G. Zhu and J. Hu, “Stiffness matrix and quantitative measure of
formation rigidity,” in Proc. of the Joint 48th IEEE CDC and 28th
CCC, Shanghai, P.R. China, 2009, pp. 3057–3062.
[10] ——, “Link resource allocation for maximizing the rigidity of multiagent formations,” in Proc. of the 2011 50th IEEE Conference on
Decision and Control and European Control Conference (CDC-ECC),
Orlando, FL, USA, 2011, pp. 2920–2925.
[11] D. Zelazo, A. Franchi, F. Allgower, H. H. Bulthoff, and P. R.
Giordano, “Rigidity maintenance control for multi-robot systems,” in
2012 Robotics: Science and Systems Conference, Sydney, Australia,
2012.
[12] D. Zelazo, A. Franchi, H. H. Bullthoff, and P. R. Giordano, “Decentralized rigidity maintenance control with range measurements for
multi-robot systems,” The International Journal of Robotics Research,
vol. 34, no. 1, pp. 105–128, 2015.
[13] I. Shames and T. H. Summers, “Rigid network design via submodular
set function optimization,” to appear in the IEEE Transactions on
Network Science and Engineering, 2015.
[14] Z. Sun, C. Yu, and B. D. O. Anderson, “Distributed optimization on
proximity network rigidity via robotic movements,” in Proceedings
of the 34th Chinese Control Conference, Hangzhou, China, 2015, pp.
6954–6960.
[15] R. K. Williams, A. Gasparri, A. Priolo, and G. S. Sukhatme, “Distributed combinatorial rigidity control in multi-agent networks,” in
52nd IEEE Conference on Decision and Control, December 10-13,
2013. Florence, Italy, 2013, pp. 44–49.
[16] T. Eren, “Graph invariants for unique localizability in cooperative
localization of wireless sensor networks: rigidity index and redundancy
index,” arXiv:1502.01699 [cs.SY], 2015.
[17] N. Biggs, Algebraic Graph Theory, 2nd ed. Cambridge University
Press, 1993.
[18] W. Whiteley, “Some matroids from discrete applied geometry,” in
Matroid theory (Seattle, WA, 1995), ser. Contemp. Math. Amer. Math.
Soc., Providence, RI, 1996, vol. 197, pp. 171–311.
[19] R. Connelly, “Generic global rigidity,” Discrete & Computational
Geometry, vol. 33, no. 4, p. 549–563, 2005.
[20] R. A. Horn and C. R. Johnson, Matrix Analysis. Cambridge University
Press, 1990.
[21] J. R. Bunch, C. P. Nielsen, and D. C. Sorensen, “Rank-one modification of the symmetric eigenproblem,” Numer. Math., vol. 31, pp.
31–48, 1978.
[22] L. Mirsky, “Inequalities for normal and hermitian matrices,” Duke
Math. J., vol. 24, no. 4, pp. 591–599, 1957.
Fig. 6: A K3 graph with an assigned edge orientation.
Based on Lemma 3, the nonzero eigenvalues of M = RRT
are the same as the eigenvalues of the matrix


2z1T z1 −z1T z2 −z1T z3
N = RT R =  −z2T z1 2z2T z2 −z2T z3 
−z3T z1 −z3T z2 2z3T z3


d21 +d23 −d22
d21 +d22 −d23
2d21
2
2
 2 2 2
d23 +d22 −d21 
=  d1 +d22 −d3
.
2d22
2
d21 +d23 −d22
2
d23 +d22 −d21
2
2d23
There are now two observations concerning the values assumed by χ and ξ over the set of triangular F:
• Observation 1: In all triangles A1 A2 A3 with a fixed
vertex angle α at A3 , ξ and χ are simultaneously
maximized when A1 A2 A3 is an isosceles triangle with
A1 A3 = A2 A3 .
• Observation 2: In all isosceles triangles A1 A2 A3 with
A1 A2 = A1 A3 , ξ and χ is maximized when A1 A2 A3
is an equilateral triangle.
These observations lead to the conclusion that ξ and χ are
maximized when the framework’s realization is an equilateral
triangle. This conclusion agrees with Conjecture 10. Our
setups and calculation results for these observations are as
follows.
A PPENDIX
A. A case study of the new rigidity indices in K3 frameworks
Consider an infinitesimally rigid framework F = (K3 , p)
in the plane. Then for each realization p = [pT1 , pT2 , pT3 ]T ∈
R6 , we have a triangle with three vertices located at pi . Let
Ai ≡ pi , i = 1, 2, 3, for convenience. Further, we assign an
orientation to edges as in Fig. 6. The rigidity matrix in this
case is given by
z1T

0
R=
−z3T

−z1T
z2T
0

0
−z2T 
z3T
Fig. 7: When A3 is located in the arc A1 mA2 , the triangle
has a vertex angle α at A3 a fixed side A1 A2 .
Consider a triangular realization A1 A2 A3 of F in which
the vertex angle at A3 is α. Without loss of generality,
let the circumcircle (O, r) of A1 A2 A3 be centered at the
origin O, and A1 and A2 be symmetric about the y-axis
for convenience. Then, A3 = A3 (ω), where ω is the angle
between OA3 and the x-axis as depicted in Fig. 7. Let ω1
(ω2 ) be angles between OA1 (OA2 ) and the x-axis. From
basic geometry, when A1 and A2 are fixed, if A3 is located
on the arc A1 mA2 := {A3 (ω) ∈ (O, r) : ω1 ≤ ω ≤ ω2 }, the
vertex angle at A3 is always α. Define S1 := {A1 A2 A3 :
Preprint submitted to 55th IEEE Conference on Decision and Control.
Received March 7, 2016.
CONFIDENTIAL. Limited circulation. For review only.
(a) α = 30o
(b) α = 45o
(c) α = 90o
(d) α = 120o
Fig. 8: The rigidity indices χ, ξ versus ω in four cases of vertex angle α. In all cases, χ and ξ are maximized when ω = 90o ,
i.e., the framework are isosceles triangles.
A3 ∈ A1 mA2 }. Every realization p of F having a vertex
angle α has a similar realization in S1 .
Based on Theorem 3, we only need to study χ and ξ in the
set S1 . The changes of ξ, χ versus ω, when α = 30o , 450 , 90o
and 120o are given in Fig. 8. Observe that in all cases ξ
and χ are simultaneously maximized when when ω = 900 ,
i.e., A1 A2 A3 is an isosceles triangle. Now, let us study the
then A3 = A3 (ω). Every isosceles triangular realizations of
F has a similar realization in S2 Based on Theorem 3, we
only need to study ξ and χ in the set S2 .
The changes of χ and ξ versus ω, 0 ≤ ω ≤ 180o are
given in Fig. 10. We can observe max χ = 1/4 and max ξ =
1/2. These maximal values are simultaneously attained when
when ω = 600 , i.e., A1 A2 A3 is an equilateral triangle.
Fig. 9: When A3 moves in the circle (O; r = 1), A1 A2 A3
is an isosceles triangle.
set of all isosceles triangular realizations. Without loss of
generality, let us fix A1 at the origin O, A2 at [1, 0]T , and
let A3 be any point in the circle centered at O ≡ A1 with
radius r = A1 A2 = 1. Then, A1 A2 A3 is a isosceles triangle.
Define S2 =:= {A1 A2 A3 : A3 ∈ (O; r = 1)}. Let ω be the
angle between OA3 and the x-axis as depicted in Fig. 9,
Fig. 10: The evolution of rigidity indices χ, ξ versus vertex
angle ω. χ and ξ are maximized when ω = 60o .
Preprint submitted to 55th IEEE Conference on Decision and Control.
Received March 7, 2016.