This is a summary of the proof by G.E. Coxson that P-matrix recognition is
co-NP-complete. The result follows by a reduction from the MAX CUT problem
using results of S. Poljak and J. Rohn.Comment: 9 page

Foniok, Jan

English

arXiv

This is a summary of the proof by G.E. Coxson [1] that P-matrix recognition is co-NPcomplete. The result follows by a reduction from the MAX CUT problem using results of S. Poljak and J. Rohn [5]

Foniok, J

E-space: Manchester Metropolitan University's Research Repository

arXiv:0710.3519v1  [cs.CC]  18 Oct 2007P-matrix recognition is co-NP-completeJan FoniokETH Zurich, Institute for Operations ResearchRa¨mistrasse 101, 8092 Zurich, Switzerlandfoniok@math.ethz.ch18 October 2007This is a summary of the proof by G.E. Coxson [1] that P-matrix recognition is co-NP-complete. The result follows by a reduction from the MAX CUT problem using resultsof S. Poljak and J. Rohn [5].1 Considered problemsOur main interest is the complexity of deciding whether an input matrix is a P-matrix.A P-matrix is a square matrix M ∈ Rn×n such that all its principal minors are positive.Such matrices were first studied by Fiedler and Pta´k [2].P-MATRIXInstance: A square matrix M ∈ Qn×n.Question: Are all the principal minors of M positive?To start with, we use a well-known combinatorial problem.SIMPLE MAX CUTInstance: A graph G = (V,E), a positive integer K.Question: Is there a partition of the vertex set V into sets V1 and V2 suchthat the number of edges with one end in V1 and the other endin V2 is at least K?Garey, Johnson and Stockmeyer [4] showed that the SIMPLE MAX CUT problem isNP-complete.The reduction from SIMPLE MAX CUT to P-MATRIX uses two intermediate steps.The first of them is the computation of the r-norm of a matrix.1For an arbitrary matrix A ∈ Rn×n, letr(A) = max{zTAy : z, y ∈ {−1, 1}n}.Remark. The function r is a matrix norm.Proof. For an arbitrary square matrix A, we have r(A) ≥ 0 because zTAy = −(−z)TAy.Moreover if r(A) = 0, then zTAy = 0 for all choices of z, y ∈ {−1, 1}n, hence A = 0. Ifk ∈ R, then zT(kA)y = k · zTAy, so r(kA) = |k| · r(A).Let A,B ∈ Rn×n. Thenr(A+B) = max{zT(A+B)y : y, z ∈ {−1, 1}n} = max{zTAy+zTBy : y, z ∈ {−1, 1}n}≤ max{zTAy : y, z ∈ {−1, 1}n}+max{zTBy : y, z ∈ {−1, 1}n}= r(A) + r(B).Thus r is also subadditive.The decision problem corresponding to r-norm computation is defined as follows.MATRIX R-NORMInstance: A matrix A ∈ Qn×n and a rational number K.Question: Is r(A) ≥ K?For the last of the decision problems considered here, we need the notion of matrixinterval. If A− and A+ are n × n real matrices such that A− ≤ A+ (that is, for eachr and s we have (A−)r,s ≤ (A+)r,s), then the matrix interval∗ [A−, A+] is the set of allmatrices A satisfying A− ≤ A ≤ A+.A matrix interval is singular if it contains a singular matrix; otherwise it is non-singular.The decision problem we consider consists in testing whether a given matrix intervalis singular. We will see that this is a computationally hard problem even when thedifference A+ −A− has rank 1.RK1-MATRIX-INTERVAL SINGULARITYInstance: A non-singular matrix A ∈ Qn×n and a non-negative matrix ∆ ∈Qn×n of rank 1.Question: Is the matrix interval [A−∆, A+∆] singular?The rest of this exposition contains three polynomial reductions of these problems,ultimately proving that P-MATRIX is co-NP-complete.∗This object is usually called an interval matrix. Since it is actually an interval and not a matrix, Ibeg the reader to pardon my decision to call it an uncommon but appropriate name.22 Reduction from SIMPLE MAX CUT to MATRIX R-NORMLet G = (V,E) be an undirected graph with n = |V | and let ℓ = 2|E| + 1. If A(G) isthe adjacency matrix of G, define A = ℓ · In −A(G). ThusAu,v =ℓ if u = v,−1 if uv ∈ E,0 otherwise.Observe that for y, z ∈ {−1, 1}n we have zTAy ≤ yTAy because of the choice of ℓ.Hence r(A) = yTAy for some y ∈ {−1, 1}n.Let S ⊆ V be defined by S = {u : yu = 1} and let m′ be the number of edges of Gwith one end in S and the other end in V \ S. In this way, m′ is the size of the cutdefined by S and V \ S.ThenyTAy = nℓ+ 4m′ − 2|E|and therefore there is a cut in G of size at least K if and only if r(A) ≥ nℓ− 2|E|+4K.The described reduction (by Poljak and Rohn [5]) establishes the hardness of comput-ing the r-norm.Theorem 1. MATRIX R-NORM is NP-complete, even if input is restricted to non-singular matrices.Proof. It follows from the reduction above that MATRIX R-NORM is NP-hard. Observethat by the choice of ℓ the matrix A in the reduction is strictly diagonally dominant andthus non-singular.A non-deterministic Turing machine can guess the values of y, z ∈ {−1, 1}n and checkin polynomial time that zTAy ≥ K, so the problem is in the class NP.3 Reduction from MATRIX R-NORM toRK1-MATRIX-INTERVAL SINGULARITYFor a matrix A ∈ Rn×n defineρ0(A) = max{|λ| : λ is a real eigenvalue of A}and set ρ0(A) = 0 if A has no real eigenvalue.Further for a vector y ∈ Rn define D(y) to be the diagonal n×n matrix with diagonalvector y.The following fact was proved by Rohn [6].Lemma 2. Let A be a real non-singular n× n matrix and let ∆ be a real non-negativen × n matrix. Then the matrix interval [A − ∆, A + ∆] is singular if and only ifρ0(A−1D(y)∆D(z)) ≥ 1 for some y, z ∈ {−1, 1}n.3Proof. For y, z ∈ {−1, 1}n let ∆y,z denote the matrix D(y)∆D(z).First suppose that A−1∆y,z has a real eigenvalue λ such that |λ| ≥ 1 and A−1∆y,zx =λx for some y, z ∈ {−1, 1}n and a non-zero vector x. Then(1− 1λA−1∆y,z)x = 0,(A− 1λ∆y,z)x = 0.Hence A− (1/λ)∆y,z is a singular matrix in the interval [A−∆, A+∆] because∣∣ 1λ∆y,z∣∣ = ∣∣ 1λD(y)∆D(z)∣∣ ≤ ∆.Therefore the interval [A−∆, A+∆] is singular.To prove the converse, suppose that B is a singular matrix, B ∈ [A−∆, A+∆]. Letx be a non-zero vector for which Bx = 0.For i = 1, 2, . . . , n setti =(Ax)i(∆|x|)i.We claim that t ∈ [0, 1]n. Indeed, |Ax| = |(A − B)x| ≤ ∆|x| because Bx = 0 andB ∈ [A−∆, A+∆].Moreover, set z = sgnx. Then D(z)x = |x| and(A−∆t,z)x = Ax−D(t)∆D(z)x = Ax−D(t)∆|x| = 0by the definition of t. Thus the matrix A − ∆t,z is a singular matrix in the interval[A−∆, A+∆].Define ψ(s) = det(A−∆s,z). The function ψ is affine in each of the variables s1, . . . , sn.Since ψ(t) = det(A−∆t,z) = 0, either there exists y ∈ {−1, 1}n such that det(A−∆y,z) =0, or there exist y, y′ ∈ {−1, 1}n such that det(A−∆y,z) · det(A−∆y′,z) < 0.In the latter case, without loss of generality we may assume that detA·det(A−∆y,z) <0. The function φ defined by φ(α) = det(A− α∆y,z) is continuous and φ(0)φ(1) < 0, soφ has a root in (0, 1).In either case, there exist y ∈ {−1, 1}n and α ∈ (0, 1] such that det(A − α∆y,z) = 0.Thendet(1αA−∆y,z)= 0,det(1αI −A−1∆y,z)= 0,hence 1αis a real eigenvalue of the matrix A−1D(y)∆D(z) and 1α≥ 1, as we weresupposed to prove.This lemma provides a useful connection between singularity of matrix intervals anda parameter ρ0 dependent on the two matrices A, ∆ that define the interval. Next weestablish a connection between ρ0 and the r-norm of matrices.From now on let 1 be the all-one vector (1, 1, . . . , 1) ∈ Rn and let J = 1 · 1T be theall-one n× n matrix.4Lemma 3. Let A ∈ Rn×n be a non-singular matrix, let α be a positive real number andlet ∆ = αJ . Thenmax {ρ0(AD(y)∆D(z)) : y, z ∈ {−1, 1}n} = α · r(A).Proof. First observe that D(y)∆D(z) = α · D(y)1 · 1TD(z) = α · yzT for arbitraryy, z ∈ {−1, 1}n. If λ is a non-zero real eigenvalue of α ·AyzT and x is a non-zero vectorsuch thatα · AyzTx = λx 6= 0,then zTx 6= 0 andα · zTAyzTx = λ · zTx,α · zTAy = λ.Thus ρ0(AD(y)∆D(z)) = α · |zTAy|. Hencemax {ρ0(AD(y)∆D(z)) : y, z ∈ {−1, 1}n}= α ·max{|zTAy| : y, z ∈ {−1, 1}n}= α · r(A).Now everything is set for Poljak and Rohn’s reduction [5].Theorem 4. Let A ∈ Rn×n be a non-singular matrix, let K be a positive real number andlet ∆ = (1/K) ·J . Then r(A) ≥ K if and only if the matrix interval [A−1−∆, A−1 +∆]is singular.Proof. By Lemma 2, the matrix interval [A−1 − ∆, A−1 + ∆] is singular if and only ifρ0(AD(y)∆D(z)) ≥ 1 for some y, z ∈ {−1, 1}n. By Lemma 3, ρ0(AD(y)∆D(z)) ≥ 1 forsome y, z ∈ {−1, 1}n if and only if r(A) ≥ K.Corollary 5. RK1-MATRIX-INTERVAL SINGULARITY is NP-hard.Remark. Poljak and Rohn [5] show that RK1-MATRIX-INTERVAL SINGULARITYbelongs to the class NP by proving the existence of a singular matrix in every singularmatrix interval, with a polynomial bound on the size of all entries of that matrix.4 Reduction from RK1-MATRIX-INTERVAL SINGULARITYto P-MATRIXThe described reduction is by Coxson [1].5Let A,∆ ∈ Rn×n. Consider the matrix interval [A,A + ∆]. Let ∆i,j be the matrixwhose element in the ith row and jth column is ∆i,j and which has zeros elsewhere.Then each matrix M in the interval [A,A +∆] can be uniquely expressed asM = A+n∑i,j=1pi,j∆i,j, (1)where pi,j ∈ [0, 1] for all values of i, j.Each matrix ∆i,j is a rank-1 matrix (even if ∆ has higher rank), and so ∆i,j = ri,jsTi,jfor some vectors ri,j, si,j ∈ Rn. We can actually take ri,j to be ∆i,j in its ith entry andzero elsewhere, and si,j to be 1 in its jth entry and zero elsewhere.Now let R be the matrix whose columns are all the n2 vectors ri,j and let S be thematrix whose columns are all the n2 vectors si,j. Thus ∆ = RST. Moreover, if p ∈ Rn2is the vector formed by the numbers pi,j, we can write (1) asM = A+RD(p)ST.Suppose that A is non-singular. Then the matrix interval [A,A + ∆] is non-singularif and only ifdet(A+RD(p)ST) = det(A) det(In +A−1RD(p)ST) 6= 0 (2)for each vector p ∈ [0, 1]n2.Supposing that the matrix A is non-singular, inequality (2) holds if and only ifdet(In +A−1RD(p)ST) 6= 0. (3)In this way we have proved that for a non-singular matrix A, singularity of the matrixinterval [A,A + ∆] is equivalent to the existence of a vector p ∈ [0, 1]n2that does notsatisfy inequality (3). Since the expression in (3) is a multi-affine function of p, we canactually derive another condition.Lemma 6. Let ψ(p) = det(In + A−1RD(p)ST). Then inequality (3) holds for eachp ∈ [0, 1]n2if and only if ψ(p) > 0 for each p ∈ {0, 1}n2.Proof. First observe that ψ(p) = det(In +A−1RD(p)ST) is a multi-affine function of p,that is, for each i we have ψ(p) = c1 + c2pi, where c1, c2 depend on i and pj for j 6= i.We claim that any multi-affine function φ : [0, 1]k → R is non-zero on the wholedomain if and only if its values on the vertices {0, 1}k have all the same sign. Assumingthis claim holds, we notice that ψ(0) = det In = 1 > 0, so ψ is non-zero on [0, 1]n2 if andonly if it is positive on {0, 1}k .To prove the claim, first suppose that φ is non-zero on [0, 1]k but there are two verticesu, v ∈ {0, 1}k such that φ(u) < 0 and φ(v) > 0. Following the path along the edgesof {0, 1}, we will find two vertices u′, v′ ∈ {0, 1} that differ in exactly one coordinateand such that φ(u′) < 0 and φ(v′) > 0. Without loss of generality we may assumethat u′1 = 0 and v′1 = 1, while u′i = v′i for i ≥ 2. Let x ∈ [0, 1]k be defined byx1 = φ(u′)/(φ(u′)− φ(v′)) and xi = u′i for i ≥ 2. Then φ(x) = 0, a contradiction.Conversely, if φ is positive (negative) on all the vertices, it is easy to prove by inductionon face dimension that φ is positive (negative) in every internal point of each face.6Lemma 6 together with the discussion that precedes it imply the following character-isation.Lemma 7. Let A be a non-singular matrix and let R,S be defined as above. Then thematrix interval [A,A +∆] is singular if and only ifdet(In +A−1RD(p)ST) ≤ 0for some p ∈ {0, 1}n2.In order to get D(p) from the middle of the product to the beginning, we use thefollowing lemma, whose proof we present in the Appendix.Lemma 8. Let F ∈ Rk×n and G ∈ Rn×k. Then det(Ik + FG) = det(In +GF ).This fact can be exploited to prove the following equivalence.Theorem 9. Let A be a non-singular matrix and let R,S be defined as in Lemma 7.Then the matrix interval [A,A + ∆] is singular if and only if the matrix M = In2 +STA−1R is not a P-matrix.Proof. Because of Lemma 8,ψ(p) = det(In2 +A−1RD(p)ST) = det(In2 +D(p)STA−1R).If p ∈ {0, 1}n2and p 6= 0, the expression det(In2 + D(p)STA−1R) is equal to theprincipal minor of the matrix M obtained by selecting exactly those rows and columnsthat correspond to the 1-entries of the vector p. Thus ψ(p) is non-positive for somep ∈ {0, 1}n2if and only if the matrix M is not a P-matrix.The proof is now completed by applying Lemma 7.Corollary 10. The problem P-MATRIX is co-NP-complete.Proof. NP-hardness follows from Corollary 5 and Theorem 9.The problem belongs to co-NP because after guessing the rows and columns, thecorresponding principal minor, which certifies the negative answer, can be computed inpolynomial time.Appendix: Proof of Lemma 8One of the basic facts about determinants is that adding a multiple of a row to anotherrow does not change the determinant. The following lemma (Theorem 3 in Section 2.5 ofGantmacher’s book [3]) is a block version of this fact. Even though it holds for matriceswith an arbitrary number of blocks, we state it just for 2 × 2 blocks. This variant issufficient for the proof of Lemma 8.7Lemma 11. Let A ∈ Rm×n be a matrix with block structureA =(m1{n1︷︸︸︷A1,1n2︷︸︸︷A1,2m2{ A2,1 A2,2)and let X ∈ Rm1×m2 , Y ∈ Rn1×n2 . ThendetA = det(A1,1 +XA2,1 A1,2 +XA2,2A2,1 A2,2)= det(A1,1 A1,2 +A1,1YA2,1 A2,2 +A2,1Y).Proof. Since (A1,1 +XA2,1 A1,2 +XA2,2A2,1 A2,2)=(Im1 X0 Im2)A,we havedet(A1,1 +XA2,1 A1,2 +XA2,2A2,1 A2,2)= det(Im1 X0 Im2)· detA = detA.Similarlydet(A1,1 A1,2 +A1,1YA2,1 A2,2 +A2,1Y)= detA · det(In1 Y0 In2)= detA.Finally comes the proof of Lemma 8.Proof of Lemma 8. Applying Lemma 11 twice, we getdet(Ik + FG) = det(Ik + FG 0G In)(∗)= det(Ik −FG In)(†)= det(Ik 0G In +GF)= det(In +GF ).Here (∗) follows by applying Lemma 11 to rows with X = F and (†) follows by applyingit to columns with Y = F .References[1] G. E. Coxson. The P-matrix problem is co-NP-complete. Math. Programming,64(2):173–178, 1994.[2] M. Fiedler and V. Pta´k. On matrices with non-positive off-diagonal elements andpositive principal minors. Czechoslovak Math. J., 12 (87):382–400, 1962.[3] F. R. Gantmacher. The Theory of Matrices, volume I. Chelsea, New York, 1959.[4] M. R. Garey, D. S. Johnson, and L. Stockmeyer. Some simplified NP-complete graphproblems. Theoret. Comput. Sci., 1(3):237–267, 1976.8[5] S. Poljak and J. Rohn. Checking robust nonsingularity is NP-hard. Math. ControlSignals Systems, 6(1):1–9, 1993.[6] J. Rohn. Systems of linear interval equations. Linear Algebra Appl., 126:39–78, 1989.9

P-matrix recognition is co-NP-complete

https://e-space.mmu.ac.uk/618316/1/0710.3519v1.pdf

P-matrix recognition is co-NP-complete

Abstract

Similar works

Full text

Available Versions

E-space: Manchester Metropolitan University's Research Repository