| {"paper_meta":{"paper_id":"arxiv:0710.3519","title":"0710.3519","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0710.3519v1 [cs.CC] 18 Oct 2007\nP-matrix recognition is co-NP-complete\nJan Foniok\nETH Zurich, Institute for Operations Research\nR ̈amistrasse 101, 8092 Zurich, Switzerland\nfoniok@math.ethz.ch\n18 October 2007\nThis is a summary of the proof by G.E. Coxson [1] that P-matrix recognition is co-NP-\ncomplete. The result follows by a reduction from the MAX CUT problem using results\nof S. Poljak and J. Rohn [5].\n1 Considered problems\nOur main interest is the complexity of deciding whether an input matrix is a P-matrix.\nA P-matrix is a square matrix M ∈Rn×n such that all its principal minors are positive.\nSuch matrices were first studied by Fiedler and Pt ́ak [2].\nP-MATRIX\nInstance:\nA square matrix M ∈Qn×n.\nQuestion:\nAre all the principal minors of M positive?\nTo start with, we use a well-known combinatorial problem.\nSIMPLE MAX CUT\nInstance:\nA graph G = (V, E), a positive integer K.\nQuestion:\nIs there a partition of the vertex set V into sets V1 and V2 such\nthat the number of edges with one end in V1 and the other end\nin V2 is at least K?\nGarey, Johnson and Stockmeyer [4] showed that the SIMPLE MAX CUT problem is\nNP-complete.\nThe reduction from SIMPLE MAX CUT to P-MATRIX uses two intermediate steps.\nThe first of them is the computation of the r-norm of a matrix.\n1\n\nFor an arbitrary matrix A ∈Rn×n, let\nr(A) = max\nn\nzTAy : z, y ∈{−1, 1}no\n.\nRemark. The function r is a matrix norm.\nProof. For an arbitrary square matrix A, we have r(A) ≥0 because zTAy = −(−z)TAy.\nMoreover if r(A) = 0, then zTAy = 0 for all choices of z, y ∈{−1, 1}n, hence A = 0. If\nk ∈R, then zT(kA)y = k · zTAy, so r(kA) = |k| · r(A).\nLet A, B ∈Rn×n. Then\nr(A+B) = max{zT(A+B)y : y, z ∈{−1, 1}n} = max{zTAy+zTBy : y, z ∈{−1, 1}n}\n≤max{zTAy : y, z ∈{−1, 1}n} + max{zTBy : y, z ∈{−1, 1}n}\n= r(A) + r(B).\nThus r is also subadditive.\nThe decision problem corresponding to r-norm computation is defined as follows.\nMATRIX R-NORM\nInstance:\nA matrix A ∈Qn×n and a rational number K.\nQuestion:\nIs r(A) ≥K?\nFor the last of the decision problems considered here, we need the notion of matrix\ninterval. If A−and A+ are n × n real matrices such that A−≤A+ (that is, for each\nr and s we have (A−)r,s ≤(A+)r,s), then the matrix interval∗[A−, A+] is the set of all\nmatrices A satisfying A−≤A ≤A+.\nA matrix interval is singular if it contains a singular matrix; otherwise it is non-\nsingular.\nThe decision problem we consider consists in testing whether a given matrix interval\nis singular.\nWe will see that this is a computationally hard problem even when the\ndifference A+ −A−has rank 1.\nRK1-MATRIX-INTERVAL SINGULARITY\nInstance:\nA non-singular matrix A ∈Qn×n and a non-negative matrix ∆∈\nQn×n of rank 1.\nQuestion:\nIs the matrix interval [A −∆, A + ∆] singular?\nThe rest of this exposition contains three polynomial reductions of these problems,\nultimately proving that P-MATRIX is co-NP-complete.\n∗This object is usually called an interval matrix. Since it is actually an interval and not a matrix, I\nbeg the reader to pardon my decision to call it an uncommon but appropriate name.\n2\n\n2 Reduction from SIMPLE MAX CUT to MATRIX R-NORM\nLet G = (V, E) be an undirected graph with n = |V | and let l= 2|E| + 1. If A(G) is\nthe adjacency matrix of G, define A = l· In −A(G). Thus\nAu,v =\n \n \n \n \n \nl\nif u = v,\n−1\nif uv ∈E,\n0\notherwise.\nObserve that for y, z ∈{−1, 1}n we have zTAy ≤yTAy because of the choice of l.\nHence r(A) = yTAy for some y ∈{−1, 1}n.\nLet S ⊆V be defined by S = {u : yu = 1} and let m′ be the number of edges of G\nwith one end in S and the other end in V \\ S. In this way, m′ is the size of the cut\ndefined by S and V \\ S.\nThen\nyTAy = nl+ 4m′ −2|E|\nand therefore there is a cut in G of size at least K if and only if r(A) ≥nl−2|E| + 4K.\nThe described reduction (by Poljak and Rohn [5]) establishes the hardness of comput-\ning the r-norm.\nTheorem 1. MATRIX R-NORM is NP-complete, even if input is restricted to non-\nsingular matrices.\nProof. It follows from the reduction above that MATRIX R-NORM is NP-hard. Observe\nthat by the choice of lthe matrix A in the reduction is strictly diagonally dominant and\nthus non-singular.\nA non-deterministic Turing machine can guess the values of y, z ∈{−1, 1}n and check\nin polynomial time that zTAy ≥K, so the problem is in the class NP.\n3 Reduction from MATRIX R-NORM to\nRK1-MATRIX-INTERVAL SINGULARITY\nFor a matrix A ∈Rn×n define\nρ0(A) = max{|λ| : λ is a real eigenvalue of A}\nand set ρ0(A) = 0 if A has no real eigenvalue.\nFurther for a vector y ∈Rn define D(y) to be the diagonal n×n matrix with diagonal\nvector y.\nThe following fact was proved by Rohn [6].\nLemma 2. Let A be a real non-singular n × n matrix and let ∆be a real non-negative\nn × n matrix.\nThen the matrix interval [A −∆, A + ∆] is singular if and only if\nρ0(A−1D(y)∆D(z)) ≥1 for some y, z ∈{−1, 1}n.\n3\n\nProof. For y, z ∈{−1, 1}n let ∆y,z denote the matrix D(y)∆D(z).\nFirst suppose that A−1∆y,z has a real eigenvalue λ such that |λ| ≥1 and A−1∆y,zx =\nλx for some y, z ∈{−1, 1}n and a non-zero vector x. Then\n 1 −1\nλA−1∆y,z\n \nx = 0,\n A −1\nλ∆y,z\n \nx = 0.\nHence A −(1/λ)∆y,z is a singular matrix in the interval [A −∆, A + ∆] because\n 1\nλ∆y,z\n =\n 1\nλD(y)∆D(z)\n ≤∆.\nTherefore the interval [A −∆, A + ∆] is singular.\nTo prove the converse, suppose that B is a singular matrix, B ∈[A −∆, A + ∆]. Let\nx be a non-zero vector for which Bx = 0.\nFor i = 1, 2, . . . , n set\nti = (Ax)i\n(∆|x|)i\n.\nWe claim that t ∈[0, 1]n.\nIndeed, |Ax| = |(A −B)x| ≤∆|x| because Bx = 0 and\nB ∈[A −∆, A + ∆].\nMoreover, set z = sgn x. Then D(z)x = |x| and\n(A −∆t,z)x = Ax −D(t)∆D(z)x = Ax −D(t)∆|x| = 0\nby the definition of t. Thus the matrix A −∆t,z is a singular matrix in the interval\n[A −∆, A + ∆].\nDefine ψ(s) = det(A−∆s,z). The function ψ is affine in each of the variables s1, . . . , sn.\nSince ψ(t) = det(A−∆t,z) = 0, either there exists y ∈{−1, 1}n such that det(A−∆y,z) =\n0, or there exist y, y′ ∈{−1, 1}n such that det(A −∆y,z) · det(A −∆y′,z) < 0.\nIn the latter case, without loss of generality we may assume that det A·det(A−∆y,z) <\n0. The function φ defined by φ(α) = det(A −α∆y,z) is continuous and φ(0)φ(1) < 0, so\nφ has a root in (0, 1).\nIn either case, there exist y ∈{−1, 1}n and α ∈(0, 1] such that det(A −α∆y,z) = 0.\nThen\ndet\n 1\nαA −∆y,z\n \n= 0,\ndet\n 1\nαI −A−1∆y,z\n \n= 0,\nhence\n1\nα is a real eigenvalue of the matrix A−1D(y)∆D(z) and\n1\nα ≥1, as we were\nsupposed to prove.\nThis lemma provides a useful connection between singularity of matrix intervals and\na parameter ρ0 dependent on the two matrices A, ∆that define the interval. Next we\nestablish a connection between ρ0 and the r-norm of matrices.\nFrom now on let\n1 be the all-one vector (1, 1, . . . , 1) ∈Rn and let J =\n1 ·\n1T be the\nall-one n × n matrix.\n4\n\nLemma 3. Let A ∈Rn×n be a non-singular matrix, let α be a positive real number and\nlet ∆= αJ. Then\nmax {ρ0(AD(y)∆D(z)) : y, z ∈{−1, 1}n} = α · r(A).\nProof. First observe that D(y)∆D(z) = α · D(y)1 ·\n1TD(z) = α · yzT for arbitrary\ny, z ∈{−1, 1}n. If λ is a non-zero real eigenvalue of α · AyzT and x is a non-zero vector\nsuch that\nα · AyzTx = λx ̸= 0,\nthen zTx ̸= 0 and\nα · zTAyzTx = λ · zTx,\nα · zTAy = λ.\nThus ρ0(AD(y)∆D(z)) = α · |zTAy|. Hence\nmax {ρ0(AD(y)∆D(z)) : y, z ∈{−1, 1}n}\n= α · max\nn\n|zTAy| : y, z ∈{−1, 1}no\n= α · r(A).\nNow everything is set for Poljak and Rohn’s reduction [5].\nTheorem 4. Let A ∈Rn×n be a non-singular matrix, let K be a positive real number and\nlet ∆= (1/K) · J. Then r(A) ≥K if and only if the matrix interval [A−1 −∆, A−1 + ∆]\nis singular.\nProof. By Lemma 2, the matrix interval [A−1 −∆, A−1 + ∆] is singular if and only if\nρ0(AD(y)∆D(z)) ≥1 for some y, z ∈{−1, 1}n. By Lemma 3, ρ0(AD(y)∆D(z)) ≥1 for\nsome y, z ∈{−1, 1}n if and only if r(A) ≥K.\nCorollary 5. RK1-MATRIX-INTERVAL SINGULARITY is NP-hard.\nRemark. Poljak and Rohn [5] show that RK1-MATRIX-INTERVAL SINGULARITY\nbelongs to the class NP by proving the existence of a singular matrix in every singular\nmatrix interval, with a polynomial bound on the size of all entries of that matrix.\n4 Reduction from RK1-MATRIX-INTERVAL SINGULARITY\nto P-MATRIX\nThe described reduction is by Coxson [1].\n5\n\nLet A, ∆∈Rn×n. Consider the matrix interval [A, A + ∆]. Let ∆i,j be the matrix\nwhose element in the ith row and jth column is ∆i,j and which has zeros elsewhere.\nThen each matrix M in the interval [A, A + ∆] can be uniquely expressed as\nM = A +\nn\nX\ni,j=1\npi,j∆i,j,\n(1)\nwhere pi,j ∈[0, 1] for all values of i, j.\nEach matrix ∆i,j is a rank-1 matrix (even if ∆has higher rank), and so ∆i,j = ri,jsT\ni,j\nfor some vectors ri,j, si,j ∈Rn. We can actually take ri,j to be ∆i,j in its ith entry and\nzero elsewhere, and si,j to be 1 in its jth entry and zero elsewhere.\nNow let R be the matrix whose columns are all the n2 vectors ri,j and let S be the\nmatrix whose columns are all the n2 vectors si,j. Thus ∆= RST. Moreover, if p ∈Rn2\nis the vector formed by the numbers pi,j, we can write (1) as\nM = A + RD(p)ST.\nSuppose that A is non-singular. Then the matrix interval [A, A + ∆] is non-singular\nif and only if\ndet(A + RD(p)ST) = det(A) det(In + A−1RD(p)ST) ̸= 0\n(2)\nfor each vector p ∈[0, 1]n2.\nSupposing that the matrix A is non-singular, inequality (2) holds if and only if\ndet(In + A−1RD(p)ST) ̸= 0.\n(3)\nIn this way we have proved that for a non-singular matrix A, singularity of the matrix\ninterval [A, A + ∆] is equivalent to the existence of a vector p ∈[0, 1]n2 that does not\nsatisfy inequality (3). Since the expression in (3) is a multi-affine function of p, we can\nactually derive another condition.\nLemma 6. Let ψ(p) = det(In + A−1RD(p)ST).\nThen inequality (3) holds for each\np ∈[0, 1]n2 if and only if ψ(p) > 0 for each p ∈{0, 1}n2.\nProof. First observe that ψ(p) = det(In + A−1RD(p)ST) is a multi-affine function of p,\nthat is, for each i we have ψ(p) = c1 + c2pi, where c1, c2 depend on i and pj for j ̸= i.\nWe claim that any multi-affine function φ : [0, 1]k →R is non-zero on the whole\ndomain if and only if its values on the vertices {0, 1}k have all the same sign. Assuming\nthis claim holds, we notice that ψ(0) = det In = 1 > 0, so ψ is non-zero on [0, 1]n2 if and\nonly if it is positive on {0, 1}k.\nTo prove the claim, first suppose that φ is non-zero on [0, 1]k but there are two vertices\nu, v ∈{0, 1}k such that φ(u) < 0 and φ(v) > 0. Following the path along the edges\nof {0, 1}, we will find two vertices u′, v′ ∈{0, 1} that differ in exactly one coordinate\nand such that φ(u′) < 0 and φ(v′) > 0. Without loss of generality we may assume\nthat u′\n1 = 0 and v′\n1 = 1, while u′\ni = v′\ni for i ≥2.\nLet x ∈[0, 1]k be defined by\nx1 = φ(u′)/(φ(u′) −φ(v′)) and xi = u′\ni for i ≥2. Then φ(x) = 0, a contradiction.\nConversely, if φ is positive (negative) on all the vertices, it is easy to prove by induction\non face dimension that φ is positive (negative) in every internal point of each face.\n6\n\nLemma 6 together with the discussion that precedes it imply the following character-\nisation.\nLemma 7. Let A be a non-singular matrix and let R, S be defined as above. Then the\nmatrix interval [A, A + ∆] is singular if and only if\ndet(In + A−1RD(p)ST) ≤0\nfor some p ∈{0, 1}n2.\nIn order to get D(p) from the middle of the product to the beginning, we use the\nfollowing lemma, whose proof we present in the Appendix.\nLemma 8. Let F ∈Rk×n and G ∈Rn×k. Then det(Ik + FG) = det(In + GF).\nThis fact can be exploited to prove the following equivalence.\nTheorem 9. Let A be a non-singular matrix and let R, S be defined as in Lemma 7.\nThen the matrix interval [A, A + ∆] is singular if and only if the matrix M = In2 +\nSTA−1R is not a P-matrix.\nProof. Because of Lemma 8,\nψ(p) = det(In2 + A−1RD(p)ST) = det(In2 + D(p)STA−1R).\nIf p ∈{0, 1}n2 and p ̸= 0, the expression det(In2 + D(p)STA−1R) is equal to the\nprincipal minor of the matrix M obtained by selecting exactly those rows and columns\nthat correspond to the 1-entries of the vector p. Thus ψ(p) is non-positive for some\np ∈{0, 1}n2 if and only if the matrix M is not a P-matrix.\nThe proof is now completed by applying Lemma 7.\nCorollary 10. The problem P-MATRIX is co-NP-complete.\nProof. NP-hardness follows from Corollary 5 and Theorem 9.\nThe problem belongs to co-NP because after guessing the rows and columns, the\ncorresponding principal minor, which certifies the negative answer, can be computed in\npolynomial time.\nAppendix: Proof of Lemma 8\nOne of the basic facts about determinants is that adding a multiple of a row to another\nrow does not change the determinant. The following lemma (Theorem 3 in Section 2.5 of\nGantmacher’s book [3]) is a block version of this fact. Even though it holds for matrices\nwith an arbitrary number of blocks, we state it just for 2 × 2 blocks. This variant is\nsufficient for the proof of Lemma 8.\n7\n\nLemma 11. Let A ∈Rm×n be a matrix with block structure\nA =\n \nm1{\nn1\nz}|{\nA1,1\nn2\nz}|{\nA1,2\nm2{\nA2,1\nA2,2\n \nand let X ∈Rm1×m2, Y ∈Rn1×n2. Then\ndet A = det\n A1,1 + XA2,1\nA1,2 + XA2,2\nA2,1\nA2,2\n \n= det\n A1,1\nA1,2 + A1,1Y\nA2,1\nA2,2 + A2,1Y\n \n.\nProof. Since\n A1,1 + XA2,1\nA1,2 + XA2,2\nA2,1\nA2,2\n \n=\n Im1\nX\n0\nIm2\n \nA,\nwe have\ndet\n A1,1 + XA2,1\nA1,2 + XA2,2\nA2,1\nA2,2\n \n= det\n Im1\nX\n0\nIm2\n \n· det A = det A.\nSimilarly\ndet\n A1,1\nA1,2 + A1,1Y\nA2,1\nA2,2 + A2,1Y\n \n= det A · det\n In1\nY\n0\nIn2\n \n= det A.\nFinally comes the proof of Lemma 8.\nProof of Lemma 8. Applying Lemma 11 twice, we get\ndet(Ik + FG) = det\n Ik + FG\n0\nG\nIn\n \n(∗)\n= det\n Ik\n−F\nG\nIn\n \n(†)\n= det\n Ik\n0\nG\nIn + GF\n \n= det(In + GF).\nHere (∗) follows by applying Lemma 11 to rows with X = F and (†) follows by applying\nit to columns with Y = F.\nReferences\n[1] G. E. Coxson.\nThe P-matrix problem is co-NP-complete.\nMath. Programming,\n64(2):173–178, 1994.\n[2] M. Fiedler and V. Pt ́ak. On matrices with non-positive off-diagonal elements and\npositive principal minors. Czechoslovak Math. J., 12 (87):382–400, 1962.\n[3] F. R. Gantmacher. The Theory of Matrices, volume I. Chelsea, New York, 1959.\n[4] M. R. Garey, D. S. Johnson, and L. Stockmeyer. Some simplified NP-complete graph\nproblems. Theoret. Comput. Sci., 1(3):237–267, 1976.\n8\n\n[5] S. Poljak and J. Rohn. Checking robust nonsingularity is NP-hard. Math. Control\nSignals Systems, 6(1):1–9, 1993.\n[6] J. Rohn. Systems of linear interval equations. Linear Algebra Appl., 126:39–78, 1989.\n9","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0710.3519v1 [cs.CC] 18 Oct 2007\nP-matrix recognition is co-NP-complete\nJan Foniok\nETH Zurich, Institute for Operations Research\nR ̈amistrasse 101, 8092 Zurich, Switzerland\nfoniok@math.ethz.ch\n18 October 2007\nThis is a summary of the proof by G.E. Coxson [1] that P-matrix recognition is co-NP-\ncomplete. The result follows by a reduction from the MAX CUT problem using results\nof S. Poljak and J. Rohn [5].\n1 Considered problems\nOur main interest is the complexity of deciding whether an input matrix is a P-matrix.\nA P-matrix is a square matrix M ∈Rn×n such that all its principal minors are positive.\nSuch matrices were first studied by Fiedler and Pt ́ak [2].\nP-MATRIX\nInstance:\nA square matrix M ∈Qn×n.\nQuestion:\nAre all the principal minors of M positive?\nTo start with, we use a well-known combinatorial problem.\nSIMPLE MAX CUT\nInstance:\nA graph G = (V, E), a positive integer K.\nQuestion:\nIs there a partition of the vertex set V into sets V1 and V2 such\nthat the number of edges with one end in V1 and the other end\nin V2 is at least K?\nGarey, Johnson and Stockmeyer [4] showed that the SIMPLE MAX CUT problem is\nNP-complete.\nThe reduction from SIMPLE MAX CUT to P-MATRIX uses two intermediate steps.\nThe first of them is the computation of the r-norm of a matrix.\n1"},{"paragraph_id":"p2","order":2,"text":"For an arbitrary matrix A ∈Rn×n, let\nr(A) = max\nn\nzTAy : z, y ∈{−1, 1}no\n.\nRemark. The function r is a matrix norm.\nProof. For an arbitrary square matrix A, we have r(A) ≥0 because zTAy = −(−z)TAy.\nMoreover if r(A) = 0, then zTAy = 0 for all choices of z, y ∈{−1, 1}n, hence A = 0. If\nk ∈R, then zT(kA)y = k · zTAy, so r(kA) = |k| · r(A).\nLet A, B ∈Rn×n. Then\nr(A+B) = max{zT(A+B)y : y, z ∈{−1, 1}n} = max{zTAy+zTBy : y, z ∈{−1, 1}n}\n≤max{zTAy : y, z ∈{−1, 1}n} + max{zTBy : y, z ∈{−1, 1}n}\n= r(A) + r(B).\nThus r is also subadditive.\nThe decision problem corresponding to r-norm computation is defined as follows.\nMATRIX R-NORM\nInstance:\nA matrix A ∈Qn×n and a rational number K.\nQuestion:\nIs r(A) ≥K?\nFor the last of the decision problems considered here, we need the notion of matrix\ninterval. If A−and A+ are n × n real matrices such that A−≤A+ (that is, for each\nr and s we have (A−)r,s ≤(A+)r,s), then the matrix interval∗[A−, A+] is the set of all\nmatrices A satisfying A−≤A ≤A+.\nA matrix interval is singular if it contains a singular matrix; otherwise it is non-\nsingular.\nThe decision problem we consider consists in testing whether a given matrix interval\nis singular.\nWe will see that this is a computationally hard problem even when the\ndifference A+ −A−has rank 1.\nRK1-MATRIX-INTERVAL SINGULARITY\nInstance:\nA non-singular matrix A ∈Qn×n and a non-negative matrix ∆∈\nQn×n of rank 1.\nQuestion:\nIs the matrix interval [A −∆, A + ∆] singular?\nThe rest of this exposition contains three polynomial reductions of these problems,\nultimately proving that P-MATRIX is co-NP-complete.\n∗This object is usually called an interval matrix. Since it is actually an interval and not a matrix, I\nbeg the reader to pardon my decision to call it an uncommon but appropriate name.\n2"},{"paragraph_id":"p3","order":3,"text":"2 Reduction from SIMPLE MAX CUT to MATRIX R-NORM\nLet G = (V, E) be an undirected graph with n = |V | and let l= 2|E| + 1. If A(G) is\nthe adjacency matrix of G, define A = l· In −A(G). Thus\nAu,v ="},{"paragraph_id":"p4","order":4,"text":"l\nif u = v,\n−1\nif uv ∈E,\n0\notherwise.\nObserve that for y, z ∈{−1, 1}n we have zTAy ≤yTAy because of the choice of l.\nHence r(A) = yTAy for some y ∈{−1, 1}n.\nLet S ⊆V be defined by S = {u : yu = 1} and let m′ be the number of edges of G\nwith one end in S and the other end in V \\ S. In this way, m′ is the size of the cut\ndefined by S and V \\ S.\nThen\nyTAy = nl+ 4m′ −2|E|\nand therefore there is a cut in G of size at least K if and only if r(A) ≥nl−2|E| + 4K.\nThe described reduction (by Poljak and Rohn [5]) establishes the hardness of comput-\ning the r-norm.\nTheorem 1. MATRIX R-NORM is NP-complete, even if input is restricted to non-\nsingular matrices.\nProof. It follows from the reduction above that MATRIX R-NORM is NP-hard. Observe\nthat by the choice of lthe matrix A in the reduction is strictly diagonally dominant and\nthus non-singular.\nA non-deterministic Turing machine can guess the values of y, z ∈{−1, 1}n and check\nin polynomial time that zTAy ≥K, so the problem is in the class NP.\n3 Reduction from MATRIX R-NORM to\nRK1-MATRIX-INTERVAL SINGULARITY\nFor a matrix A ∈Rn×n define\nρ0(A) = max{|λ| : λ is a real eigenvalue of A}\nand set ρ0(A) = 0 if A has no real eigenvalue.\nFurther for a vector y ∈Rn define D(y) to be the diagonal n×n matrix with diagonal\nvector y.\nThe following fact was proved by Rohn [6].\nLemma 2. Let A be a real non-singular n × n matrix and let ∆be a real non-negative\nn × n matrix.\nThen the matrix interval [A −∆, A + ∆] is singular if and only if\nρ0(A−1D(y)∆D(z)) ≥1 for some y, z ∈{−1, 1}n.\n3"},{"paragraph_id":"p5","order":5,"text":"Proof. For y, z ∈{−1, 1}n let ∆y,z denote the matrix D(y)∆D(z).\nFirst suppose that A−1∆y,z has a real eigenvalue λ such that |λ| ≥1 and A−1∆y,zx =\nλx for some y, z ∈{−1, 1}n and a non-zero vector x. Then\n 1 −1\nλA−1∆y,z"},{"paragraph_id":"p6","order":6,"text":"x = 0,\n A −1\nλ∆y,z"},{"paragraph_id":"p7","order":7,"text":"x = 0.\nHence A −(1/λ)∆y,z is a singular matrix in the interval [A −∆, A + ∆] because\n 1\nλ∆y,z\n =\n 1\nλD(y)∆D(z)\n ≤∆.\nTherefore the interval [A −∆, A + ∆] is singular.\nTo prove the converse, suppose that B is a singular matrix, B ∈[A −∆, A + ∆]. Let\nx be a non-zero vector for which Bx = 0.\nFor i = 1, 2, . . . , n set\nti = (Ax)i\n(∆|x|)i\n.\nWe claim that t ∈[0, 1]n.\nIndeed, |Ax| = |(A −B)x| ≤∆|x| because Bx = 0 and\nB ∈[A −∆, A + ∆].\nMoreover, set z = sgn x. Then D(z)x = |x| and\n(A −∆t,z)x = Ax −D(t)∆D(z)x = Ax −D(t)∆|x| = 0\nby the definition of t. Thus the matrix A −∆t,z is a singular matrix in the interval\n[A −∆, A + ∆].\nDefine ψ(s) = det(A−∆s,z). The function ψ is affine in each of the variables s1, . . . , sn.\nSince ψ(t) = det(A−∆t,z) = 0, either there exists y ∈{−1, 1}n such that det(A−∆y,z) =\n0, or there exist y, y′ ∈{−1, 1}n such that det(A −∆y,z) · det(A −∆y′,z) < 0.\nIn the latter case, without loss of generality we may assume that det A·det(A−∆y,z) <\n0. The function φ defined by φ(α) = det(A −α∆y,z) is continuous and φ(0)φ(1) < 0, so\nφ has a root in (0, 1).\nIn either case, there exist y ∈{−1, 1}n and α ∈(0, 1] such that det(A −α∆y,z) = 0.\nThen\ndet\n 1\nαA −∆y,z"},{"paragraph_id":"p8","order":8,"text":"= 0,\ndet\n 1\nαI −A−1∆y,z"},{"paragraph_id":"p9","order":9,"text":"= 0,\nhence\n1\nα is a real eigenvalue of the matrix A−1D(y)∆D(z) and\n1\nα ≥1, as we were\nsupposed to prove.\nThis lemma provides a useful connection between singularity of matrix intervals and\na parameter ρ0 dependent on the two matrices A, ∆that define the interval. Next we\nestablish a connection between ρ0 and the r-norm of matrices.\nFrom now on let\n1 be the all-one vector (1, 1, . . . , 1) ∈Rn and let J =\n1 ·\n1T be the\nall-one n × n matrix.\n4"},{"paragraph_id":"p10","order":10,"text":"Lemma 3. Let A ∈Rn×n be a non-singular matrix, let α be a positive real number and\nlet ∆= αJ. Then\nmax {ρ0(AD(y)∆D(z)) : y, z ∈{−1, 1}n} = α · r(A).\nProof. First observe that D(y)∆D(z) = α · D(y)1 ·\n1TD(z) = α · yzT for arbitrary\ny, z ∈{−1, 1}n. If λ is a non-zero real eigenvalue of α · AyzT and x is a non-zero vector\nsuch that\nα · AyzTx = λx ̸= 0,\nthen zTx ̸= 0 and\nα · zTAyzTx = λ · zTx,\nα · zTAy = λ.\nThus ρ0(AD(y)∆D(z)) = α · |zTAy|. Hence\nmax {ρ0(AD(y)∆D(z)) : y, z ∈{−1, 1}n}\n= α · max\nn\n|zTAy| : y, z ∈{−1, 1}no\n= α · r(A).\nNow everything is set for Poljak and Rohn’s reduction [5].\nTheorem 4. Let A ∈Rn×n be a non-singular matrix, let K be a positive real number and\nlet ∆= (1/K) · J. Then r(A) ≥K if and only if the matrix interval [A−1 −∆, A−1 + ∆]\nis singular.\nProof. By Lemma 2, the matrix interval [A−1 −∆, A−1 + ∆] is singular if and only if\nρ0(AD(y)∆D(z)) ≥1 for some y, z ∈{−1, 1}n. By Lemma 3, ρ0(AD(y)∆D(z)) ≥1 for\nsome y, z ∈{−1, 1}n if and only if r(A) ≥K.\nCorollary 5. RK1-MATRIX-INTERVAL SINGULARITY is NP-hard.\nRemark. Poljak and Rohn [5] show that RK1-MATRIX-INTERVAL SINGULARITY\nbelongs to the class NP by proving the existence of a singular matrix in every singular\nmatrix interval, with a polynomial bound on the size of all entries of that matrix.\n4 Reduction from RK1-MATRIX-INTERVAL SINGULARITY\nto P-MATRIX\nThe described reduction is by Coxson [1].\n5"},{"paragraph_id":"p11","order":11,"text":"Let A, ∆∈Rn×n. Consider the matrix interval [A, A + ∆]. Let ∆i,j be the matrix\nwhose element in the ith row and jth column is ∆i,j and which has zeros elsewhere.\nThen each matrix M in the interval [A, A + ∆] can be uniquely expressed as\nM = A +\nn\nX\ni,j=1\npi,j∆i,j,\n(1)\nwhere pi,j ∈[0, 1] for all values of i, j.\nEach matrix ∆i,j is a rank-1 matrix (even if ∆has higher rank), and so ∆i,j = ri,jsT\ni,j\nfor some vectors ri,j, si,j ∈Rn. We can actually take ri,j to be ∆i,j in its ith entry and\nzero elsewhere, and si,j to be 1 in its jth entry and zero elsewhere.\nNow let R be the matrix whose columns are all the n2 vectors ri,j and let S be the\nmatrix whose columns are all the n2 vectors si,j. Thus ∆= RST. Moreover, if p ∈Rn2\nis the vector formed by the numbers pi,j, we can write (1) as\nM = A + RD(p)ST.\nSuppose that A is non-singular. Then the matrix interval [A, A + ∆] is non-singular\nif and only if\ndet(A + RD(p)ST) = det(A) det(In + A−1RD(p)ST) ̸= 0\n(2)\nfor each vector p ∈[0, 1]n2.\nSupposing that the matrix A is non-singular, inequality (2) holds if and only if\ndet(In + A−1RD(p)ST) ̸= 0.\n(3)\nIn this way we have proved that for a non-singular matrix A, singularity of the matrix\ninterval [A, A + ∆] is equivalent to the existence of a vector p ∈[0, 1]n2 that does not\nsatisfy inequality (3). Since the expression in (3) is a multi-affine function of p, we can\nactually derive another condition.\nLemma 6. Let ψ(p) = det(In + A−1RD(p)ST).\nThen inequality (3) holds for each\np ∈[0, 1]n2 if and only if ψ(p) > 0 for each p ∈{0, 1}n2.\nProof. First observe that ψ(p) = det(In + A−1RD(p)ST) is a multi-affine function of p,\nthat is, for each i we have ψ(p) = c1 + c2pi, where c1, c2 depend on i and pj for j ̸= i.\nWe claim that any multi-affine function φ : [0, 1]k →R is non-zero on the whole\ndomain if and only if its values on the vertices {0, 1}k have all the same sign. Assuming\nthis claim holds, we notice that ψ(0) = det In = 1 > 0, so ψ is non-zero on [0, 1]n2 if and\nonly if it is positive on {0, 1}k.\nTo prove the claim, first suppose that φ is non-zero on [0, 1]k but there are two vertices\nu, v ∈{0, 1}k such that φ(u) < 0 and φ(v) > 0. Following the path along the edges\nof {0, 1}, we will find two vertices u′, v′ ∈{0, 1} that differ in exactly one coordinate\nand such that φ(u′) < 0 and φ(v′) > 0. Without loss of generality we may assume\nthat u′\n1 = 0 and v′\n1 = 1, while u′\ni = v′\ni for i ≥2.\nLet x ∈[0, 1]k be defined by\nx1 = φ(u′)/(φ(u′) −φ(v′)) and xi = u′\ni for i ≥2. Then φ(x) = 0, a contradiction.\nConversely, if φ is positive (negative) on all the vertices, it is easy to prove by induction\non face dimension that φ is positive (negative) in every internal point of each face.\n6"},{"paragraph_id":"p12","order":12,"text":"Lemma 6 together with the discussion that precedes it imply the following character-\nisation.\nLemma 7. Let A be a non-singular matrix and let R, S be defined as above. Then the\nmatrix interval [A, A + ∆] is singular if and only if\ndet(In + A−1RD(p)ST) ≤0\nfor some p ∈{0, 1}n2.\nIn order to get D(p) from the middle of the product to the beginning, we use the\nfollowing lemma, whose proof we present in the Appendix.\nLemma 8. Let F ∈Rk×n and G ∈Rn×k. Then det(Ik + FG) = det(In + GF).\nThis fact can be exploited to prove the following equivalence.\nTheorem 9. Let A be a non-singular matrix and let R, S be defined as in Lemma 7.\nThen the matrix interval [A, A + ∆] is singular if and only if the matrix M = In2 +\nSTA−1R is not a P-matrix.\nProof. Because of Lemma 8,\nψ(p) = det(In2 + A−1RD(p)ST) = det(In2 + D(p)STA−1R).\nIf p ∈{0, 1}n2 and p ̸= 0, the expression det(In2 + D(p)STA−1R) is equal to the\nprincipal minor of the matrix M obtained by selecting exactly those rows and columns\nthat correspond to the 1-entries of the vector p. Thus ψ(p) is non-positive for some\np ∈{0, 1}n2 if and only if the matrix M is not a P-matrix.\nThe proof is now completed by applying Lemma 7.\nCorollary 10. The problem P-MATRIX is co-NP-complete.\nProof. NP-hardness follows from Corollary 5 and Theorem 9.\nThe problem belongs to co-NP because after guessing the rows and columns, the\ncorresponding principal minor, which certifies the negative answer, can be computed in\npolynomial time.\nAppendix: Proof of Lemma 8\nOne of the basic facts about determinants is that adding a multiple of a row to another\nrow does not change the determinant. The following lemma (Theorem 3 in Section 2.5 of\nGantmacher’s book [3]) is a block version of this fact. Even though it holds for matrices\nwith an arbitrary number of blocks, we state it just for 2 × 2 blocks. This variant is\nsufficient for the proof of Lemma 8.\n7"},{"paragraph_id":"p13","order":13,"text":"Lemma 11. Let A ∈Rm×n be a matrix with block structure\nA ="},{"paragraph_id":"p14","order":14,"text":"m1{\nn1\nz}|{\nA1,1\nn2\nz}|{\nA1,2\nm2{\nA2,1\nA2,2"},{"paragraph_id":"p15","order":15,"text":"and let X ∈Rm1×m2, Y ∈Rn1×n2. Then\ndet A = det\n A1,1 + XA2,1\nA1,2 + XA2,2\nA2,1\nA2,2"},{"paragraph_id":"p16","order":16,"text":"= det\n A1,1\nA1,2 + A1,1Y\nA2,1\nA2,2 + A2,1Y"},{"paragraph_id":"p17","order":17,"text":".\nProof. Since\n A1,1 + XA2,1\nA1,2 + XA2,2\nA2,1\nA2,2"},{"paragraph_id":"p18","order":18,"text":"=\n Im1\nX\n0\nIm2"},{"paragraph_id":"p19","order":19,"text":"A,\nwe have\ndet\n A1,1 + XA2,1\nA1,2 + XA2,2\nA2,1\nA2,2"},{"paragraph_id":"p20","order":20,"text":"= det\n Im1\nX\n0\nIm2"},{"paragraph_id":"p21","order":21,"text":"· det A = det A.\nSimilarly\ndet\n A1,1\nA1,2 + A1,1Y\nA2,1\nA2,2 + A2,1Y"},{"paragraph_id":"p22","order":22,"text":"= det A · det\n In1\nY\n0\nIn2"},{"paragraph_id":"p23","order":23,"text":"= det A.\nFinally comes the proof of Lemma 8.\nProof of Lemma 8. Applying Lemma 11 twice, we get\ndet(Ik + FG) = det\n Ik + FG\n0\nG\nIn"},{"paragraph_id":"p24","order":24,"text":"(∗)\n= det\n Ik\n−F\nG\nIn"},{"paragraph_id":"p25","order":25,"text":"(†)\n= det\n Ik\n0\nG\nIn + GF"},{"paragraph_id":"p26","order":26,"text":"= det(In + GF).\nHere (∗) follows by applying Lemma 11 to rows with X = F and (†) follows by applying\nit to columns with Y = F.\nReferences\n[1] G. E. Coxson.\nThe P-matrix problem is co-NP-complete.\nMath. Programming,\n64(2):173–178, 1994.\n[2] M. Fiedler and V. Pt ́ak. On matrices with non-positive off-diagonal elements and\npositive principal minors. Czechoslovak Math. J., 12 (87):382–400, 1962.\n[3] F. R. Gantmacher. The Theory of Matrices, volume I. Chelsea, New York, 1959.\n[4] M. R. Garey, D. S. Johnson, and L. Stockmeyer. Some simplified NP-complete graph\nproblems. Theoret. Comput. Sci., 1(3):237–267, 1976.\n8"},{"paragraph_id":"p27","order":27,"text":"[5] S. Poljak and J. Rohn. Checking robust nonsingularity is NP-hard. Math. Control\nSignals Systems, 6(1):1–9, 1993.\n[6] J. Rohn. Systems of linear interval equations. Linear Algebra Appl., 126:39–78, 1989.\n9"}],"pages":[{"page":1,"text":"arXiv:0710.3519v1 [cs.CC] 18 Oct 2007\nP-matrix recognition is co-NP-complete\nJan Foniok\nETH Zurich, Institute for Operations Research\nR ̈amistrasse 101, 8092 Zurich, Switzerland\nfoniok@math.ethz.ch\n18 October 2007\nThis is a summary of the proof by G.E. Coxson [1] that P-matrix recognition is co-NP-\ncomplete. The result follows by a reduction from the MAX CUT problem using results\nof S. Poljak and J. Rohn [5].\n1 Considered problems\nOur main interest is the complexity of deciding whether an input matrix is a P-matrix.\nA P-matrix is a square matrix M ∈Rn×n such that all its principal minors are positive.\nSuch matrices were first studied by Fiedler and Pt ́ak [2].\nP-MATRIX\nInstance:\nA square matrix M ∈Qn×n.\nQuestion:\nAre all the principal minors of M positive?\nTo start with, we use a well-known combinatorial problem.\nSIMPLE MAX CUT\nInstance:\nA graph G = (V, E), a positive integer K.\nQuestion:\nIs there a partition of the vertex set V into sets V1 and V2 such\nthat the number of edges with one end in V1 and the other end\nin V2 is at least K?\nGarey, Johnson and Stockmeyer [4] showed that the SIMPLE MAX CUT problem is\nNP-complete.\nThe reduction from SIMPLE MAX CUT to P-MATRIX uses two intermediate steps.\nThe first of them is the computation of the r-norm of a matrix.\n1"},{"page":2,"text":"For an arbitrary matrix A ∈Rn×n, let\nr(A) = max\nn\nzTAy : z, y ∈{−1, 1}no\n.\nRemark. The function r is a matrix norm.\nProof. For an arbitrary square matrix A, we have r(A) ≥0 because zTAy = −(−z)TAy.\nMoreover if r(A) = 0, then zTAy = 0 for all choices of z, y ∈{−1, 1}n, hence A = 0. If\nk ∈R, then zT(kA)y = k · zTAy, so r(kA) = |k| · r(A).\nLet A, B ∈Rn×n. Then\nr(A+B) = max{zT(A+B)y : y, z ∈{−1, 1}n} = max{zTAy+zTBy : y, z ∈{−1, 1}n}\n≤max{zTAy : y, z ∈{−1, 1}n} + max{zTBy : y, z ∈{−1, 1}n}\n= r(A) + r(B).\nThus r is also subadditive.\nThe decision problem corresponding to r-norm computation is defined as follows.\nMATRIX R-NORM\nInstance:\nA matrix A ∈Qn×n and a rational number K.\nQuestion:\nIs r(A) ≥K?\nFor the last of the decision problems considered here, we need the notion of matrix\ninterval. If A−and A+ are n × n real matrices such that A−≤A+ (that is, for each\nr and s we have (A−)r,s ≤(A+)r,s), then the matrix interval∗[A−, A+] is the set of all\nmatrices A satisfying A−≤A ≤A+.\nA matrix interval is singular if it contains a singular matrix; otherwise it is non-\nsingular.\nThe decision problem we consider consists in testing whether a given matrix interval\nis singular.\nWe will see that this is a computationally hard problem even when the\ndifference A+ −A−has rank 1.\nRK1-MATRIX-INTERVAL SINGULARITY\nInstance:\nA non-singular matrix A ∈Qn×n and a non-negative matrix ∆∈\nQn×n of rank 1.\nQuestion:\nIs the matrix interval [A −∆, A + ∆] singular?\nThe rest of this exposition contains three polynomial reductions of these problems,\nultimately proving that P-MATRIX is co-NP-complete.\n∗This object is usually called an interval matrix. Since it is actually an interval and not a matrix, I\nbeg the reader to pardon my decision to call it an uncommon but appropriate name.\n2"},{"page":3,"text":"2 Reduction from SIMPLE MAX CUT to MATRIX R-NORM\nLet G = (V, E) be an undirected graph with n = |V | and let l= 2|E| + 1. If A(G) is\nthe adjacency matrix of G, define A = l· In −A(G). Thus\nAu,v =\n \n \n \n \n \nl\nif u = v,\n−1\nif uv ∈E,\n0\notherwise.\nObserve that for y, z ∈{−1, 1}n we have zTAy ≤yTAy because of the choice of l.\nHence r(A) = yTAy for some y ∈{−1, 1}n.\nLet S ⊆V be defined by S = {u : yu = 1} and let m′ be the number of edges of G\nwith one end in S and the other end in V \\ S. In this way, m′ is the size of the cut\ndefined by S and V \\ S.\nThen\nyTAy = nl+ 4m′ −2|E|\nand therefore there is a cut in G of size at least K if and only if r(A) ≥nl−2|E| + 4K.\nThe described reduction (by Poljak and Rohn [5]) establishes the hardness of comput-\ning the r-norm.\nTheorem 1. MATRIX R-NORM is NP-complete, even if input is restricted to non-\nsingular matrices.\nProof. It follows from the reduction above that MATRIX R-NORM is NP-hard. Observe\nthat by the choice of lthe matrix A in the reduction is strictly diagonally dominant and\nthus non-singular.\nA non-deterministic Turing machine can guess the values of y, z ∈{−1, 1}n and check\nin polynomial time that zTAy ≥K, so the problem is in the class NP.\n3 Reduction from MATRIX R-NORM to\nRK1-MATRIX-INTERVAL SINGULARITY\nFor a matrix A ∈Rn×n define\nρ0(A) = max{|λ| : λ is a real eigenvalue of A}\nand set ρ0(A) = 0 if A has no real eigenvalue.\nFurther for a vector y ∈Rn define D(y) to be the diagonal n×n matrix with diagonal\nvector y.\nThe following fact was proved by Rohn [6].\nLemma 2. Let A be a real non-singular n × n matrix and let ∆be a real non-negative\nn × n matrix.\nThen the matrix interval [A −∆, A + ∆] is singular if and only if\nρ0(A−1D(y)∆D(z)) ≥1 for some y, z ∈{−1, 1}n.\n3"},{"page":4,"text":"Proof. For y, z ∈{−1, 1}n let ∆y,z denote the matrix D(y)∆D(z).\nFirst suppose that A−1∆y,z has a real eigenvalue λ such that |λ| ≥1 and A−1∆y,zx =\nλx for some y, z ∈{−1, 1}n and a non-zero vector x. Then\n 1 −1\nλA−1∆y,z\n \nx = 0,\n A −1\nλ∆y,z\n \nx = 0.\nHence A −(1/λ)∆y,z is a singular matrix in the interval [A −∆, A + ∆] because\n 1\nλ∆y,z\n =\n 1\nλD(y)∆D(z)\n ≤∆.\nTherefore the interval [A −∆, A + ∆] is singular.\nTo prove the converse, suppose that B is a singular matrix, B ∈[A −∆, A + ∆]. Let\nx be a non-zero vector for which Bx = 0.\nFor i = 1, 2, . . . , n set\nti = (Ax)i\n(∆|x|)i\n.\nWe claim that t ∈[0, 1]n.\nIndeed, |Ax| = |(A −B)x| ≤∆|x| because Bx = 0 and\nB ∈[A −∆, A + ∆].\nMoreover, set z = sgn x. Then D(z)x = |x| and\n(A −∆t,z)x = Ax −D(t)∆D(z)x = Ax −D(t)∆|x| = 0\nby the definition of t. Thus the matrix A −∆t,z is a singular matrix in the interval\n[A −∆, A + ∆].\nDefine ψ(s) = det(A−∆s,z). The function ψ is affine in each of the variables s1, . . . , sn.\nSince ψ(t) = det(A−∆t,z) = 0, either there exists y ∈{−1, 1}n such that det(A−∆y,z) =\n0, or there exist y, y′ ∈{−1, 1}n such that det(A −∆y,z) · det(A −∆y′,z) < 0.\nIn the latter case, without loss of generality we may assume that det A·det(A−∆y,z) <\n0. The function φ defined by φ(α) = det(A −α∆y,z) is continuous and φ(0)φ(1) < 0, so\nφ has a root in (0, 1).\nIn either case, there exist y ∈{−1, 1}n and α ∈(0, 1] such that det(A −α∆y,z) = 0.\nThen\ndet\n 1\nαA −∆y,z\n \n= 0,\ndet\n 1\nαI −A−1∆y,z\n \n= 0,\nhence\n1\nα is a real eigenvalue of the matrix A−1D(y)∆D(z) and\n1\nα ≥1, as we were\nsupposed to prove.\nThis lemma provides a useful connection between singularity of matrix intervals and\na parameter ρ0 dependent on the two matrices A, ∆that define the interval. Next we\nestablish a connection between ρ0 and the r-norm of matrices.\nFrom now on let\n1 be the all-one vector (1, 1, . . . , 1) ∈Rn and let J =\n1 ·\n1T be the\nall-one n × n matrix.\n4"},{"page":5,"text":"Lemma 3. Let A ∈Rn×n be a non-singular matrix, let α be a positive real number and\nlet ∆= αJ. Then\nmax {ρ0(AD(y)∆D(z)) : y, z ∈{−1, 1}n} = α · r(A).\nProof. First observe that D(y)∆D(z) = α · D(y)1 ·\n1TD(z) = α · yzT for arbitrary\ny, z ∈{−1, 1}n. If λ is a non-zero real eigenvalue of α · AyzT and x is a non-zero vector\nsuch that\nα · AyzTx = λx ̸= 0,\nthen zTx ̸= 0 and\nα · zTAyzTx = λ · zTx,\nα · zTAy = λ.\nThus ρ0(AD(y)∆D(z)) = α · |zTAy|. Hence\nmax {ρ0(AD(y)∆D(z)) : y, z ∈{−1, 1}n}\n= α · max\nn\n|zTAy| : y, z ∈{−1, 1}no\n= α · r(A).\nNow everything is set for Poljak and Rohn’s reduction [5].\nTheorem 4. Let A ∈Rn×n be a non-singular matrix, let K be a positive real number and\nlet ∆= (1/K) · J. Then r(A) ≥K if and only if the matrix interval [A−1 −∆, A−1 + ∆]\nis singular.\nProof. By Lemma 2, the matrix interval [A−1 −∆, A−1 + ∆] is singular if and only if\nρ0(AD(y)∆D(z)) ≥1 for some y, z ∈{−1, 1}n. By Lemma 3, ρ0(AD(y)∆D(z)) ≥1 for\nsome y, z ∈{−1, 1}n if and only if r(A) ≥K.\nCorollary 5. RK1-MATRIX-INTERVAL SINGULARITY is NP-hard.\nRemark. Poljak and Rohn [5] show that RK1-MATRIX-INTERVAL SINGULARITY\nbelongs to the class NP by proving the existence of a singular matrix in every singular\nmatrix interval, with a polynomial bound on the size of all entries of that matrix.\n4 Reduction from RK1-MATRIX-INTERVAL SINGULARITY\nto P-MATRIX\nThe described reduction is by Coxson [1].\n5"},{"page":6,"text":"Let A, ∆∈Rn×n. Consider the matrix interval [A, A + ∆]. Let ∆i,j be the matrix\nwhose element in the ith row and jth column is ∆i,j and which has zeros elsewhere.\nThen each matrix M in the interval [A, A + ∆] can be uniquely expressed as\nM = A +\nn\nX\ni,j=1\npi,j∆i,j,\n(1)\nwhere pi,j ∈[0, 1] for all values of i, j.\nEach matrix ∆i,j is a rank-1 matrix (even if ∆has higher rank), and so ∆i,j = ri,jsT\ni,j\nfor some vectors ri,j, si,j ∈Rn. We can actually take ri,j to be ∆i,j in its ith entry and\nzero elsewhere, and si,j to be 1 in its jth entry and zero elsewhere.\nNow let R be the matrix whose columns are all the n2 vectors ri,j and let S be the\nmatrix whose columns are all the n2 vectors si,j. Thus ∆= RST. Moreover, if p ∈Rn2\nis the vector formed by the numbers pi,j, we can write (1) as\nM = A + RD(p)ST.\nSuppose that A is non-singular. Then the matrix interval [A, A + ∆] is non-singular\nif and only if\ndet(A + RD(p)ST) = det(A) det(In + A−1RD(p)ST) ̸= 0\n(2)\nfor each vector p ∈[0, 1]n2.\nSupposing that the matrix A is non-singular, inequality (2) holds if and only if\ndet(In + A−1RD(p)ST) ̸= 0.\n(3)\nIn this way we have proved that for a non-singular matrix A, singularity of the matrix\ninterval [A, A + ∆] is equivalent to the existence of a vector p ∈[0, 1]n2 that does not\nsatisfy inequality (3). Since the expression in (3) is a multi-affine function of p, we can\nactually derive another condition.\nLemma 6. Let ψ(p) = det(In + A−1RD(p)ST).\nThen inequality (3) holds for each\np ∈[0, 1]n2 if and only if ψ(p) > 0 for each p ∈{0, 1}n2.\nProof. First observe that ψ(p) = det(In + A−1RD(p)ST) is a multi-affine function of p,\nthat is, for each i we have ψ(p) = c1 + c2pi, where c1, c2 depend on i and pj for j ̸= i.\nWe claim that any multi-affine function φ : [0, 1]k →R is non-zero on the whole\ndomain if and only if its values on the vertices {0, 1}k have all the same sign. Assuming\nthis claim holds, we notice that ψ(0) = det In = 1 > 0, so ψ is non-zero on [0, 1]n2 if and\nonly if it is positive on {0, 1}k.\nTo prove the claim, first suppose that φ is non-zero on [0, 1]k but there are two vertices\nu, v ∈{0, 1}k such that φ(u) < 0 and φ(v) > 0. Following the path along the edges\nof {0, 1}, we will find two vertices u′, v′ ∈{0, 1} that differ in exactly one coordinate\nand such that φ(u′) < 0 and φ(v′) > 0. Without loss of generality we may assume\nthat u′\n1 = 0 and v′\n1 = 1, while u′\ni = v′\ni for i ≥2.\nLet x ∈[0, 1]k be defined by\nx1 = φ(u′)/(φ(u′) −φ(v′)) and xi = u′\ni for i ≥2. Then φ(x) = 0, a contradiction.\nConversely, if φ is positive (negative) on all the vertices, it is easy to prove by induction\non face dimension that φ is positive (negative) in every internal point of each face.\n6"},{"page":7,"text":"Lemma 6 together with the discussion that precedes it imply the following character-\nisation.\nLemma 7. Let A be a non-singular matrix and let R, S be defined as above. Then the\nmatrix interval [A, A + ∆] is singular if and only if\ndet(In + A−1RD(p)ST) ≤0\nfor some p ∈{0, 1}n2.\nIn order to get D(p) from the middle of the product to the beginning, we use the\nfollowing lemma, whose proof we present in the Appendix.\nLemma 8. Let F ∈Rk×n and G ∈Rn×k. Then det(Ik + FG) = det(In + GF).\nThis fact can be exploited to prove the following equivalence.\nTheorem 9. Let A be a non-singular matrix and let R, S be defined as in Lemma 7.\nThen the matrix interval [A, A + ∆] is singular if and only if the matrix M = In2 +\nSTA−1R is not a P-matrix.\nProof. Because of Lemma 8,\nψ(p) = det(In2 + A−1RD(p)ST) = det(In2 + D(p)STA−1R).\nIf p ∈{0, 1}n2 and p ̸= 0, the expression det(In2 + D(p)STA−1R) is equal to the\nprincipal minor of the matrix M obtained by selecting exactly those rows and columns\nthat correspond to the 1-entries of the vector p. Thus ψ(p) is non-positive for some\np ∈{0, 1}n2 if and only if the matrix M is not a P-matrix.\nThe proof is now completed by applying Lemma 7.\nCorollary 10. The problem P-MATRIX is co-NP-complete.\nProof. NP-hardness follows from Corollary 5 and Theorem 9.\nThe problem belongs to co-NP because after guessing the rows and columns, the\ncorresponding principal minor, which certifies the negative answer, can be computed in\npolynomial time.\nAppendix: Proof of Lemma 8\nOne of the basic facts about determinants is that adding a multiple of a row to another\nrow does not change the determinant. The following lemma (Theorem 3 in Section 2.5 of\nGantmacher’s book [3]) is a block version of this fact. Even though it holds for matrices\nwith an arbitrary number of blocks, we state it just for 2 × 2 blocks. This variant is\nsufficient for the proof of Lemma 8.\n7"},{"page":8,"text":"Lemma 11. Let A ∈Rm×n be a matrix with block structure\nA =\n \nm1{\nn1\nz}|{\nA1,1\nn2\nz}|{\nA1,2\nm2{\nA2,1\nA2,2\n \nand let X ∈Rm1×m2, Y ∈Rn1×n2. Then\ndet A = det\n A1,1 + XA2,1\nA1,2 + XA2,2\nA2,1\nA2,2\n \n= det\n A1,1\nA1,2 + A1,1Y\nA2,1\nA2,2 + A2,1Y\n \n.\nProof. Since\n A1,1 + XA2,1\nA1,2 + XA2,2\nA2,1\nA2,2\n \n=\n Im1\nX\n0\nIm2\n \nA,\nwe have\ndet\n A1,1 + XA2,1\nA1,2 + XA2,2\nA2,1\nA2,2\n \n= det\n Im1\nX\n0\nIm2\n \n· det A = det A.\nSimilarly\ndet\n A1,1\nA1,2 + A1,1Y\nA2,1\nA2,2 + A2,1Y\n \n= det A · det\n In1\nY\n0\nIn2\n \n= det A.\nFinally comes the proof of Lemma 8.\nProof of Lemma 8. Applying Lemma 11 twice, we get\ndet(Ik + FG) = det\n Ik + FG\n0\nG\nIn\n \n(∗)\n= det\n Ik\n−F\nG\nIn\n \n(†)\n= det\n Ik\n0\nG\nIn + GF\n \n= det(In + GF).\nHere (∗) follows by applying Lemma 11 to rows with X = F and (†) follows by applying\nit to columns with Y = F.\nReferences\n[1] G. E. Coxson.\nThe P-matrix problem is co-NP-complete.\nMath. Programming,\n64(2):173–178, 1994.\n[2] M. Fiedler and V. Pt ́ak. On matrices with non-positive off-diagonal elements and\npositive principal minors. Czechoslovak Math. J., 12 (87):382–400, 1962.\n[3] F. R. Gantmacher. The Theory of Matrices, volume I. Chelsea, New York, 1959.\n[4] M. R. Garey, D. S. Johnson, and L. Stockmeyer. Some simplified NP-complete graph\nproblems. Theoret. Comput. Sci., 1(3):237–267, 1976.\n8"},{"page":9,"text":"[5] S. Poljak and J. Rohn. Checking robust nonsingularity is NP-hard. Math. Control\nSignals Systems, 6(1):1–9, 1993.\n[6] J. Rohn. Systems of linear interval equations. Linear Algebra Appl., 126:39–78, 1989.\n9"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"A graph G = (V, E), a positive integer K.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"r(A) = max","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"Proof. For an arbitrary square matrix A, we have r(A) ≥0 because zTAy = −(−z)TAy.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"Moreover if r(A) = 0, then zTAy = 0 for all choices of z, y ∈{−1, 1}n, hence A = 0. If","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"k ∈R, then zT(kA)y = k · zTAy, so r(kA) = |k| · r(A).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"r(A+B) = max{zT(A+B)y : y, z ∈{−1, 1}n} = max{zTAy+zTBy : y, z ∈{−1, 1}n}","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"= r(A) + r(B).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"Let G = (V, E) be an undirected graph with n = |V | and let l= 2|E| + 1. If A(G) is","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"the adjacency matrix of G, define A = l· In −A(G). Thus","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"Au,v =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"if u = v,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"Hence r(A) = yTAy for some y ∈{−1, 1}n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"Let S ⊆V be defined by S = {u : yu = 1} and let m′ be the number of edges of G","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"yTAy = nl+ 4m′ −2|E|","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"ρ0(A) = max{|λ| : λ is a real eigenvalue of A}","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"and set ρ0(A) = 0 if A has no real eigenvalue.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"First suppose that A−1∆y,z has a real eigenvalue λ such that |λ| ≥1 and A−1∆y,zx =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"x = 0,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"x = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"x be a non-zero vector for which Bx = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"For i = 1, 2, . . . , n set","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"ti = (Ax)i","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"Indeed, |Ax| = |(A −B)x| ≤∆|x| because Bx = 0 and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"Moreover, set z = sgn x. Then D(z)x = |x| and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"(A −∆t,z)x = Ax −D(t)∆D(z)x = Ax −D(t)∆|x| = 0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"Define ψ(s) = det(A−∆s,z). The function ψ is affine in each of the variables s1, . . . , sn.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"Since ψ(t) = det(A−∆t,z) = 0, either there exists y ∈{−1, 1}n such that det(A−∆y,z) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"0. The function φ defined by φ(α) = det(A −α∆y,z) is continuous and φ(0)φ(1) < 0, so","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"In either case, there exist y ∈{−1, 1}n and α ∈(0, 1] such that det(A −α∆y,z) = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"1 be the all-one vector (1, 1, . . . , 1) ∈Rn and let J =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"let ∆= αJ. Then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"max {ρ0(AD(y)∆D(z)) : y, z ∈{−1, 1}n} = α · r(A).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq33","equation_number":null,"raw_text":"Proof. First observe that D(y)∆D(z) = α · D(y)1 ·","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq34","equation_number":null,"raw_text":"1TD(z) = α · yzT for arbitrary","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq35","equation_number":null,"raw_text":"α · AyzTx = λx ̸= 0,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq36","equation_number":null,"raw_text":"then zTx ̸= 0 and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq37","equation_number":null,"raw_text":"α · zTAyzTx = λ · zTx,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq38","equation_number":null,"raw_text":"α · zTAy = λ.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq39","equation_number":null,"raw_text":"Thus ρ0(AD(y)∆D(z)) = α · |zTAy|. Hence","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq40","equation_number":null,"raw_text":"= α · max","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq41","equation_number":null,"raw_text":"= α · r(A).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq42","equation_number":null,"raw_text":"let ∆= (1/K) · J. Then r(A) ≥K if and only if the matrix interval [A−1 −∆, A−1 + ∆]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq43","equation_number":null,"raw_text":"M = A +","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq44","equation_number":null,"raw_text":"i,j=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq45","equation_number":null,"raw_text":"Each matrix ∆i,j is a rank-1 matrix (even if ∆has higher rank), and so ∆i,j = ri,jsT","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq46","equation_number":null,"raw_text":"matrix whose columns are all the n2 vectors si,j. Thus ∆= RST. Moreover, if p ∈Rn2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq47","equation_number":null,"raw_text":"M = A + RD(p)ST.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq48","equation_number":null,"raw_text":"det(A + RD(p)ST) = det(A) det(In + A−1RD(p)ST) ̸= 0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq49","equation_number":null,"raw_text":"det(In + A−1RD(p)ST) ̸= 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq50","equation_number":null,"raw_text":"Lemma 6. Let ψ(p) = det(In + A−1RD(p)ST).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq51","equation_number":null,"raw_text":"Proof. First observe that ψ(p) = det(In + A−1RD(p)ST) is a multi-affine function of p,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq52","equation_number":null,"raw_text":"that is, for each i we have ψ(p) = c1 + c2pi, where c1, c2 depend on i and pj for j ̸= i.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq53","equation_number":null,"raw_text":"this claim holds, we notice that ψ(0) = det In = 1 > 0, so ψ is non-zero on [0, 1]n2 if and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq54","equation_number":null,"raw_text":"1 = 0 and v′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq55","equation_number":null,"raw_text":"1 = 1, while u′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq56","equation_number":null,"raw_text":"i = v′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq57","equation_number":null,"raw_text":"x1 = φ(u′)/(φ(u′) −φ(v′)) and xi = u′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq58","equation_number":null,"raw_text":"i for i ≥2. Then φ(x) = 0, a contradiction.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq59","equation_number":null,"raw_text":"Lemma 8. Let F ∈Rk×n and G ∈Rn×k. Then det(Ik + FG) = det(In + GF).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq60","equation_number":null,"raw_text":"Then the matrix interval [A, A + ∆] is singular if and only if the matrix M = In2 +","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq61","equation_number":null,"raw_text":"ψ(p) = det(In2 + A−1RD(p)ST) = det(In2 + D(p)STA−1R).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq62","equation_number":null,"raw_text":"If p ∈{0, 1}n2 and p ̸= 0, the expression det(In2 + D(p)STA−1R) is equal to the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq63","equation_number":null,"raw_text":"A =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq64","equation_number":null,"raw_text":"det A = det","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq65","equation_number":null,"raw_text":"= det","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq66","equation_number":null,"raw_text":"= det","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq67","equation_number":null,"raw_text":"· det A = det A.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq68","equation_number":null,"raw_text":"= det A · det","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq69","equation_number":null,"raw_text":"= det A.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq70","equation_number":null,"raw_text":"det(Ik + FG) = det","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq71","equation_number":null,"raw_text":"= det","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq72","equation_number":null,"raw_text":"= det","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq73","equation_number":null,"raw_text":"= det(In + GF).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq74","equation_number":null,"raw_text":"Here (∗) follows by applying Lemma 11 to rows with X = F and (†) follows by applying","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq75","equation_number":null,"raw_text":"it to columns with Y = F.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":14168,"parse_confidence":0.5,"equation_parse_rate_proxy":1.0,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}} |