Semidefinite Farkas Lemma and Dimensional Rigidity of Bar Frameworks - - PowerPoint PPT Presentation
Semidefinite Farkas Lemma and Dimensional Rigidity of Bar Frameworks A. Y. Alfakih University of Windsor, ON Canada Workshop on Making Models: Stimulating research in rigidity Theory and Spatial-Visual Reasoning Fields Institute, Aug 2014
University of Windsor, ON Canada
Workshop on Making Models: Stimulating research in rigidity Theory and Spatial-Visual Reasoning Fields Institute, Aug 2014
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given an m × n matrix A and m-vector b, Farkas Lemma states that exactly one of the following two statements holds:
there exists x ∈ Rn such that Ax = b, x ≥ 0, there exists y ∈ Rm such that ATy ≤ 0, bTy > 0.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given an m × n matrix A and m-vector b, Farkas Lemma states that exactly one of the following two statements holds:
there exists x ∈ Rn such that Ax = b, x ≥ 0, there exists y ∈ Rm such that ATy ≤ 0, bTy > 0.
Farkas Lemma is at the heart of optimization theory.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given an m × n matrix A and m-vector b, Farkas Lemma states that exactly one of the following two statements holds:
there exists x ∈ Rn such that Ax = b, x ≥ 0, there exists y ∈ Rm such that ATy ≤ 0, bTy > 0.
Farkas Lemma is at the heart of optimization theory. In one direction the proof is easy. In the other direction, the proof is based on the separation theorm. S a
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given an m × n matrix A and m-vector b, Farkas Lemma states that exactly one of the following two statements holds:
there exists x ∈ Rn such that Ax = b, x ≥ 0, there exists y ∈ Rm such that ATy ≤ 0, bTy > 0.
Farkas Lemma is at the heart of optimization theory. In one direction the proof is easy. In the other direction, the proof is based on the separation theorm. S a
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
The linear programming problem: p∗ = min cTx (P) : subject to Ax = b x ≥ 0
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
The linear programming problem: p∗ = min cTx d∗ = max bTy (P) : subject to Ax = b (D) : subject to ATy ≤ c x ≥ 0
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
The linear programming problem: p∗ = min cTx d∗ = max bTy (P) : subject to Ax = b (D) : subject to ATy ≤ c x ≥ 0 LP strong duality: If both (P) or (D) are feasible, then both (P) and (D) have optimal solutions x∗ and y ∗ respectively, and p∗ = cTx∗ = d∗ = bTy ∗.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
The linear programming problem: p∗ = min cTx d∗ = max bTy (P) : subject to Ax = b (D) : subject to ATy ≤ c x ≥ 0 LP strong duality: If both (P) or (D) are feasible, then both (P) and (D) have optimal solutions x∗ and y ∗ respectively, and p∗ = cTx∗ = d∗ = bTy ∗. LP strict complementarity: If both (P) or (D) are feasible, then ∃ optimal pair x∗ ≥ 0 and s∗ = c − ATy ∗ ≥ 0 such that x∗
i s∗ i = 0 and x∗ i + s∗ i > 0 for all i.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
The linear programming problem: p∗ = min cTx d∗ = max bTy (P) : subject to Ax = b (D) : subject to ATy ≤ c x ≥ 0 LP strong duality: If both (P) or (D) are feasible, then both (P) and (D) have optimal solutions x∗ and y ∗ respectively, and p∗ = cTx∗ = d∗ = bTy ∗. LP strict complementarity: If both (P) or (D) are feasible, then ∃ optimal pair x∗ ≥ 0 and s∗ = c − ATy ∗ ≥ 0 such that x∗
i s∗ i = 0 and x∗ i + s∗ i > 0 for all i.
Semidefinite Programming (SDP): p∗ = min trace(CX) d∗ = max
trace(AiX) = bi
X 0
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Farkas Lemma: Ax = b, x ≥ 0, ATy ≤ 0, bTy > 0.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Farkas Lemma: Ax = b, x ≥ 0, ATy ≤ 0, bTy > 0. Potential SD Farkas Lemma: trace(AiX) = bi, X 0,
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Farkas Lemma: Ax = b, x ≥ 0, ATy ≤ 0, bTy > 0. False SD Farkas Lemma: trace(AiX) = bi, X 0,
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Farkas Lemma: Ax = b, x ≥ 0, ATy ≤ 0, bTy > 0. False SD Farkas Lemma: trace(AiX) = bi, X 0,
Correct SD Farkas Lemma: Given symmetric matrices A1, . . . , Am and m-vector b. Then exactly one of the following two statements holds:
∃ X ≻ 0: trace (AiX) = bi, for i = 1, . . . , m, ∃ Ω = y1A1 + · · · + ymAm: Ω 0, = 0,
i biyi ≤ 0.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Farkas Lemma: Ax = b, x ≥ 0, ATy ≤ 0, bTy > 0. False SD Farkas Lemma: trace(AiX) = bi, X 0,
Correct SD Farkas Lemma: Given symmetric matrices A1, . . . , Am and m-vector b. Then exactly one of the following two statements holds:
∃ X ≻ 0: trace (AiX) = bi, for i = 1, . . . , m, ∃ Ω = y1A1 + · · · + ymAm: Ω 0, = 0,
i biyi ≤ 0.
This Farkas Lemma is used to prove SD strong duality under Slater Condition.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
SD Farkas Lemma: Given symmetric matrices A1, . . . , Am and m-vector b and assume that there exists X ∗ 0 such that trace (AiX ∗) = bi for i = 1, . . . , m. Then exactly one of the following two statements holds:
∃ X ≻ 0: trace (AiX) = bi, for i = 1, . . . , m, ∃ Ω = y1A1 + · · · + ymAm: Ω 0, = 0, trace (ΩX ∗) = 0.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
SD Farkas Lemma: Given symmetric matrices A1, . . . , Am and m-vector b and assume that there exists X ∗ 0 such that trace (AiX ∗) = bi for i = 1, . . . , m. Then exactly one of the following two statements holds:
∃ X ≻ 0: trace (AiX) = bi, for i = 1, . . . , m, ∃ Ω = y1A1 + · · · + ymAm: Ω 0, = 0, trace (ΩX ∗) = 0.
Let F = {X 0 :trace (AiX) = bi, for i = 1, . . . , m}. Then F has no interior iff F is contained in the hyperplane: {X : trace (ΩX) = 0}.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given symmetric matrices A1, . . . , Am and m-vector b and assume that there exists X ∗ 0 such that trace (AiX ∗) = bi for i = 1, . . . , m and rank X ∗ = r. Then exactly one of the following two statements holds: ∃ X 0: trace (AiX) = bi, for i = 1, . . . , m, rank X ≥ r + 1, ∃ non-zero matrices Ω0, Ω1, . . . , Ωk, k ≤ n − r − 1, such that:
Ωj =
i yj i Ai,
Ω0 0, UT
1 Ω1U1 0, · · · , UT k ΩkUk 0,
trace (ΩjX ∗) = 0 for all j, rank Ω0 + rank (UT
1 Ω1U1) + · · · + rank (UT k ΩkUk) = n − r,
where U1, . . . , Uk and W0, W1, . . . , Wk are full column rank matrices defined as follows: for i = 0, 1, . . . , k, R(Wi) = N(U T
i ΩiUi), and Ui+1 = UiWi, U0 = I.
back
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Sn denotes the space of n × n symmetric matrices. Sn
+ denotes the cone of n × n symmetric PSD matrices. It is
closed and convex.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Sn denotes the space of n × n symmetric matrices. Sn
+ denotes the cone of n × n symmetric PSD matrices. It is
closed and convex. A subset F ⊂ Sn
+ is a face of Sn + if:
∀X, Y ∈ Sn
+ : (X + Y ) ∈ F =
⇒ X ∈ F and Y ∈ F.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Sn denotes the space of n × n symmetric matrices. Sn
+ denotes the cone of n × n symmetric PSD matrices. It is
closed and convex. A subset F ⊂ Sn
+ is a face of Sn + if:
∀X, Y ∈ Sn
+ : (X + Y ) ∈ F =
⇒ X ∈ F and Y ∈ F. The faces of Sn
+ are in one-to-one correspondence with the
subspaces of Rn. i.e., F is a face of Sn
+ iff:
F = {X ∈ Sn
+ : L ⊆ N(X)}, for some subspace L,
relint(F) = {X ∈ Sn
+ : L = N(X)},
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Sn denotes the space of n × n symmetric matrices. Sn
+ denotes the cone of n × n symmetric PSD matrices. It is
closed and convex. A subset F ⊂ Sn
+ is a face of Sn + if:
∀X, Y ∈ Sn
+ : (X + Y ) ∈ F =
⇒ X ∈ F and Y ∈ F. The faces of Sn
+ are in one-to-one correspondence with the
subspaces of Rn. i.e., F is a face of Sn
+ iff:
F = {X ∈ Sn
+ : L ⊆ N(X)}, for some subspace L,
relint(F) = {X ∈ Sn
+ : L = N(X)},
The faces of Sn
+ are uniquely characterized by their relative
interior.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let A ∈ Sn
+, face(A) denotes the minimal face containing A.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let A ∈ Sn
+, face(A) denotes the minimal face containing A.
Let A ∈ Sn
+ of rank r and let A = [W U]
Λ W T UT
spectral decomposition of A. Then face (A) = {X ∈ Sn
+ : XU = 0},
= {X ∈ Sn
+ : X = WYW T for some Y ∈ Sr +}.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let A ∈ Sn
+, face(A) denotes the minimal face containing A.
Let A ∈ Sn
+ of rank r and let A = [W U]
Λ W T UT
spectral decomposition of A. Then face (A) = {X ∈ Sn
+ : XU = 0},
= {X ∈ Sn
+ : X = WYW T for some Y ∈ Sr +}.
Sn
+ = face (I).
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let A ∈ Sn
+, face(A) denotes the minimal face containing A.
Let A ∈ Sn
+ of rank r and let A = [W U]
Λ W T UT
spectral decomposition of A. Then face (A) = {X ∈ Sn
+ : XU = 0},
= {X ∈ Sn
+ : X = WYW T for some Y ∈ Sr +}.
Sn
+ = face (I).
Note that if A, B ∈ Sn
+ and R(B) ⊂ R(A), then face (B) ⊂
face (A).
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given symmetric matrices A1, . . . , Am and m-vector b, let F = {X 0: trace (AiX) = bi, for i = 1, . . . , m}. Assume that X ∗ ∈ F: rank X ∗ = r. Then there does not exist X ∈ F: rank X ≥ r + 1 if and only if face (F) = face (X ∗).
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given symmetric matrices A1, . . . , Am and m-vector b, let F = {X 0: trace (AiX) = bi, for i = 1, . . . , m}. Assume that X ∗ ∈ F: rank X ∗ = r. Then there does not exist X ∈ F: rank X ≥ r + 1 if and only if face (F) = face (X ∗). Recall that face (F) ⊂ Sn
+ = face (I).
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given symmetric matrices A1, . . . , Am and m-vector b, let F = {X 0: trace (AiX) = bi, for i = 1, . . . , m}. Assume that X ∗ ∈ F: rank X ∗ = r. Then there does not exist X ∈ F: rank X ≥ r + 1 if and only if face (F) = face (X ∗). Recall that face (F) ⊂ Sn
+ = face (I).
Borwein and Wolkowicz ’81 devised a facial reduction algorithm for finding face (F).
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given symmetric matrices A1, . . . , Am and m-vector b, let F = {X 0: trace (AiX) = bi, for i = 1, . . . , m}. Assume that X ∗ ∈ F: rank X ∗ = r. Then there does not exist X ∈ F: rank X ≥ r + 1 if and only if face (F) = face (X ∗). Recall that face (F) ⊂ Sn
+ = face (I).
Borwein and Wolkowicz ’81 devised a facial reduction algorithm for finding face (F). At each step, a smaller dimensional face of Sn
+ is found. Thus,
this algorithm will find matrices U1, . . . , Uk such that: face (F) = face (Uk+1U T
k+1) ⊂ · · · ⊂ face (U1U T 1 ) ⊂ face (I).
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
A bar framework (G, p) is a simple graph G whose nodes are points p1, . . . , pn; and whose edges are line segments.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
A bar framework (G, p) is a simple graph G whose nodes are points p1, . . . , pn; and whose edges are line segments. (G, p) is r-dimensional if the dimension of the affine span of p1, . . . , pn is equal to r.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
A bar framework (G, p) is a simple graph G whose nodes are points p1, . . . , pn; and whose edges are line segments. (G, p) is r-dimensional if the dimension of the affine span of p1, . . . , pn is equal to r. The configuration p = (p1, . . . , pn) will be represented by the n × n Gram matrix B = (bij) where bij = (pi)Tpj.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
A bar framework (G, p) is a simple graph G whose nodes are points p1, . . . , pn; and whose edges are line segments. (G, p) is r-dimensional if the dimension of the affine span of p1, . . . , pn is equal to r. The configuration p = (p1, . . . , pn) will be represented by the n × n Gram matrix B = (bij) where bij = (pi)Tpj. B is PSD of rank r.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
A bar framework (G, p) is a simple graph G whose nodes are points p1, . . . , pn; and whose edges are line segments. (G, p) is r-dimensional if the dimension of the affine span of p1, . . . , pn is equal to r. The configuration p = (p1, . . . , pn) will be represented by the n × n Gram matrix B = (bij) where bij = (pi)Tpj. B is PSD of rank r. We fix the centroid of p at the origin, i.e., we require Be = 0 where e is the vector of all 1’s.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
A bar framework (G, p) is a simple graph G whose nodes are points p1, . . . , pn; and whose edges are line segments. (G, p) is r-dimensional if the dimension of the affine span of p1, . . . , pn is equal to r. The configuration p = (p1, . . . , pn) will be represented by the n × n Gram matrix B = (bij) where bij = (pi)Tpj. B is PSD of rank r. We fix the centroid of p at the origin, i.e., we require Be = 0 where e is the vector of all 1’s. B ⊂ face (VV T), where V Te = 0 and V has full column rank.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
(G, p) and (G, p′) are equivalent if the Euclidean length of the corresponding edges in both frameworks are the same.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
(G, p) and (G, p′) are equivalent if the Euclidean length of the corresponding edges in both frameworks are the same. An r-dimensional bar framework (G, p) is dimensionally rigid if there does not exist an s-dimensional bar framework (G, p′), that is equivalent to (G, p), for any s ≥ r + 1.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
(G, p) and (G, p′) are equivalent if the Euclidean length of the corresponding edges in both frameworks are the same. An r-dimensional bar framework (G, p) is dimensionally rigid if there does not exist an s-dimensional bar framework (G, p′), that is equivalent to (G, p), for any s ≥ r + 1. 1 2 3 4 (G1, p1) 1 2 3 4 1 2 3 4
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
(G, p) and (G, p′) are equivalent if the Euclidean length of the corresponding edges in both frameworks are the same. An r-dimensional bar framework (G, p) is dimensionally rigid if there does not exist an s-dimensional bar framework (G, p′), that is equivalent to (G, p), for any s ≥ r + 1. 1 2 3 4 (G1, p1) 1 2 3 4 1 2 3 4 1 2 3 (G2, p2) 1 2 3
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given r-dimensional framework (G, p) in Rr The matrix P = (p1)T . . . (pn)T is called the configuration matrix
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given r-dimensional framework (G, p) in Rr The matrix P = (p1)T . . . (pn)T is called the configuration matrix
a symmetric matrix Ω is called a stress matrix of (G, p) if:
PTΩ = 0, eTΩ = 0, Ωij = 0 for all {i, j} ∈ E(G).
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Given r-dimensional framework (G, p) in Rr The matrix P = (p1)T . . . (pn)T is called the configuration matrix
a symmetric matrix Ω is called a stress matrix of (G, p) if:
PTΩ = 0, eTΩ = 0, Ωij = 0 for all {i, j} ∈ E(G).
a symmetric matrix Ω is quasi-stress matrix of (G, p) if:
PTΩP = 0, eTΩ = 0, Ωij = 0 for all {i, j} ∈ E(G).
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
A Gale matrix of an r-dimensional framework (G, p) in Rr is the matrix Z whose columns form a basis of the null space of : PT eT
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
A Gale matrix of an r-dimensional framework (G, p) in Rr is the matrix Z whose columns form a basis of the null space of : PT eT
In Polytope theory, the rows of Z are called Gale transforms of p1, . . . , pn.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
A Gale matrix of an r-dimensional framework (G, p) in Rr is the matrix Z whose columns form a basis of the null space of : PT eT
In Polytope theory, the rows of Z are called Gale transforms of p1, . . . , pn. The Gale matrix Z encodes the affine dependency among the points p1, . . . , pn.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let Ω and Z be, respectively, a stress matrix and a Gale matrix of (G, p). Then Ω = ZΨZ T for some symmetric matrix Ψ. On the other hand, let Ψ′ be any symmetric matrix such that zi TΨ′zj = 0 for all {i, j} ∈ E(G), where zi is the ith row of Z. Then ZΨ′Z T is a stress matrix of (G, p).
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let Ω and Z be, respectively, a stress matrix and a Gale matrix of (G, p). Then Ω = ZΨZ T for some symmetric matrix Ψ. On the other hand, let Ψ′ be any symmetric matrix such that zi TΨ′zj = 0 for all {i, j} ∈ E(G), where zi is the ith row of Z. Then ZΨ′Z T is a stress matrix of (G, p). Thus results in terms of stress matrices an be equivalently stated in terms of Gale transform.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let (G, p) be an r-dimensional framework (G, p) on n nodes in Rr. Then (G, p) is dimensionally rigid if (G, p) admits a positive semidefinite stress matrix Ω with rank n − r − 1.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let (G, p) be an r-dimensional framework (G, p) on n nodes in Rr. Then (G, p) is dimensionally rigid if (G, p) admits a positive semidefinite stress matrix Ω with rank n − r − 1. This sufficient condition is not necessary.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let (G, p) be an r-dimensional framework (G, p) on n nodes in Rr. Then (G, p) is dimensionally rigid if (G, p) admits a positive semidefinite stress matrix Ω with rank n − r − 1. This sufficient condition is not necessary. A framework (G, p) is generic if all coordinates of p1, . . . , pn are algebraically independent
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let (G, p) be an r-dimensional framework (G, p) on n nodes in Rr. Then (G, p) is dimensionally rigid if (G, p) admits a positive semidefinite stress matrix Ω with rank n − r − 1. This sufficient condition is not necessary. A framework (G, p) is generic if all coordinates of p1, . . . , pn are algebraically independent
Let (G, p) be a generic r-dimensional framework (G, p) on n nodes in Rr. Assume that (G, p) is dimensionally rigid. Then (G, p) admits a positive semidefinite stress matrix Ω with rank n − r − 1.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let F = {B 0 : Be = 0, tr(BF ij) = ||pi − pj||2∀{i, j} ∈ E(G)}, where F ij = (ei − ej)(ei − ej)T and ei is the ith unit vector.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let F = {B 0 : Be = 0, tr(BF ij) = ||pi − pj||2∀{i, j} ∈ E(G)}, where F ij = (ei − ej)(ei − ej)T and ei is the ith unit vector. A framework (G, p′), with Gram matrix B′, is equivalent to (G, p) iff B′ ∈ F.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let F = {B 0 : Be = 0, tr(BF ij) = ||pi − pj||2∀{i, j} ∈ E(G)}, where F ij = (ei − ej)(ei − ej)T and ei is the ith unit vector. A framework (G, p′), with Gram matrix B′, is equivalent to (G, p) iff B′ ∈ F. An r-dimensional framework (G, p) is dimensionally rigid iff there does not exist B′ ∈ F : rank(B′) ≥ r + 1. (∗)
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
Let F = {B 0 : Be = 0, tr(BF ij) = ||pi − pj||2∀{i, j} ∈ E(G)}, where F ij = (ei − ej)(ei − ej)T and ei is the ith unit vector. A framework (G, p′), with Gram matrix B′, is equivalent to (G, p) iff B′ ∈ F. An r-dimensional framework (G, p) is dimensionally rigid iff there does not exist B′ ∈ F : rank(B′) ≥ r + 1. (∗) Condition (*) can be easily put in the form of Statement 1 in the new SD Farkas Lemma.
to Lemma
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
This Theorem is a refined version of a result by Bob and Gortler ’14
Let (G, p) be an r-dimensional bar framework on n vertices in Rr, r ≤ (n − 2). Then (G, p) is dimensionally rigid iff ∃ non-zero quasi-stress matrices Ω0, Ω1, . . . , Ωk, k ≤ n − r − 2, such that: Ω0 0, U T
1 Ω1U1 0, · · · , U T k ΩkUk 0,
rank Ω0 + rank (U T
1 Ω1U1) + · · · + rank (U T k ΩkUk) = n − r − 1,
PTΩ1ρ1 = 0, . . . , PTΩkρk = 0, where ρ1, U1, . . . , Uk and ξ1, . . . , ξk are full column rank matrices defined as follows: R(ρ1) = N( Ω0 PT eT ), R(ρi) = N(ρT
i Ωiρi), and
Ui = [P ρi] for i = 1, . . . , k and ρi+1 = ρiξi for i = 1, . . . , k − 1.
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18
(University of Windsor, ON Canada) Semidefinite Farkas’ Lemma and Dimensional Rigidity of Bar Frameworks Workshop on Making Models: Stimulating / 18