Spectrally arbitrary star sign patterns

Linear Algebra and its Applications 400 (2005) 99 119 wwwelseviercom/locate/laa Spectrally arbitrary star sign patterns G MacGillivray, RM Tifenbach, P van den Driessche Department of Mathematics and Statistics, University of Victoria, PO Box 3045, Victoria BC, Canada V8W 3P4 Received 30 August 2004; accepted 8 November 2004 Available online 20 January 2005 Submitted by RA Brualdi Abstract An n n sign pattern S n is spectrally arbitrary if, for any given real monic polynomial g(x) of degree n, there is a real matrix having sign pattern S n and characteristic polynomial g(x) All n n star sign patterns that are spectrally arbitrary, and all minimal such patterns, are characterized This subsequently leads to an explicit characterization of all n n star sign patterns that are potentially nilpotent It is shown that any super-pattern of a spectrally arbitrary star sign pattern is also spectrally arbitrary 2004 Elsevier Inc All rights reserved AMS classification: 05C50; 15A18 Keywords: Inertially arbitrary; Potentially nilpotent; Potentially stable; Spectrally arbitrary; Star sign pattern Research supported in part by NSERC through an USRA for RM Tifenbach and Discovery Grants for G MacGillivray and P van den Driessche Corresponding author E-mail addresses: gmacgill@mathuvicca (G MacGillivray), ryant@mathuvicca (RM Tifenbach), pvdd@mathuvicca (P van den Driessche) 0024-3795/$ - see front matter 2004 Elsevier Inc All rights reserved doi:101016/jlaa200411021

100 G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 1 Introduction Our goal is to characterize spectrally arbitrary star sign patterns To explain this goal, we start with some definitions and then put our problem in the context of other results on sign patterns An n n array whose entries are chosen from the set {+,, 0} is called a sign pattern Let S n = [s ij ] be an n n sign pattern (n 2) A real n n matrix M n = [m ij ] has sign pattern S n if sgn(m ij ) = s ij for all i and j If, for any given real monic polynomial g(x) of degree n, there is a real matrix having sign pattern S n and characteristic polynomial g(x), then S n is a spectrally arbitrary sign pattern (SAP) Thus, S n is a SAP if, given any self-conjugate spectrum, there is a real matrix having sign pattern S n and that spectrum Spectrally arbitrary sign patterns are normally identified up to equivalence, since the property of being a SAP is clearly preserved under negation, transposition, signature similarity and permutation similarity If there is a real matrix having sign pattern S n and characteristic polynomial g(x) = x n, then S n is potentially nilpotent; in particular each SAP must be potentially nilpotent The inertia of an n n matrix M n is the ordered triple i(m n ) = (i + (M n ), i (M n ), i 0 (M n )) where i + (M n ), i (M n ) and i 0 (M n ) are the number of eigenvalues of M n with positive, negative and zero real parts, respectively The sign pattern S n is an inertially arbitrary sign pattern (IAP) if, for any ordered triple of non-negative integers (i 1, i 2, i 3 ) such that i 1 + i 2 + i 3 = n, there is a real matrix M n having sign pattern S n such that i(m n ) = (i 1, i 2, i 3 ) The sign pattern S n is potentially stable if there is a real matrix M n having sign pattern S n and inertia i(m n ) = (0, n, 0) From the definitions, every SAP is an IAP and every IAP is potentially stable A super-pattern of S n is a sign pattern R n = [r ij ] such that if s ij /= 0, then r ij = s ij If R n is a super-pattern of S n, then S n is a sub-pattern of R n If sign pattern S n is a SAP (IAP) but none of its proper sub-patterns is a SAP (IAP), then S n is a minimal SAP (IAP) A sign pattern S n = [s ij ] is combinatorially symmetric if s ij /= 0 whenever s ji /= 0 The graph G(S n ) of a combinatorially symmetric sign pattern has vertices 1, 2,, n and an edge joining vertices i and j if and only if s ij /= 0 Note that loops are allowed A star is the graph with vertices 1, 2,, n and an edge joining a fixed centre vertex i and each leaf vertex j for all j /= i (and no other edges) In what follows, it is assumed that 1 is the centre vertex A combinatorially symmetric sign pattern S n is a star sign pattern if the graph obtained from G(S n ) by deleting all loops is a star, and more generally, is a tree sign pattern if this graph is a tree Spectrally arbitrary tree sign patterns, especially those whose graph (excluding loops) is a path, are considered in [1] A method, based on the implicit function theorem, for proving that a pattern (and all super-patterns) is a SAP is developed there A class of (full) spectrally arbitrary patterns is constructed in [2] by using a Soules matrix The implicit function theorem method is used in [3] to show that some Hessenberg sign patterns are minimal SAPs, the first such families for all orders to be presented Other spectrally arbitrary sign pattern classes are constructed in [4]

G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 101 also by using the implicit function theorem method All potentially stable star sign patterns are characterized in [5]; we make use of this characterization in Section 5 The inertias of matrices having a symmetric star sign pattern are characterized in [6] Potentially nilpotent star sign patterns are considered in [7], in which explicit characterizations are given for patterns of orders two and three, and a recursive characterization for patterns of general order n is proved Further aspects of these sign pattern problems can be found in references of the papers cited above We begin with some preliminary results (Section 2) and use these to prove that two families of star sign patterns, Z np and Y n, are spectrally arbitrary (Section 3) We then use the implicit function theorem method to show that any super-pattern of these is also a SAP (Section 4) We show that Z np and Y n are the only minimal inertially arbitrary star sign patterns (Section 5) We characterize all star sign patterns that are potentially nilpotent (Section 6), and conclude (Section 7) with a summary of our results for star sign patterns 2 Preliminary results Throughout this paper, we use the following conventions The lower-case roman letters a, b, c and d (with or without subscripts) denote positive real numbers In all matrices and arrays, entries that are not specified are equal to zero Let M n be the real n n matrix α 1 γ 2 γ 3 γ n β 2 α 2 M n = β 3 α 3 β n α n with characteristic polynomial det(xi n M n ), denoted by f n (x) Expanding this determinant about the first row gives the following expression for f n (x) Lemma 21 The characteristic polynomial of M n (as above) is n n f n (x) = (x α j ) β i γ i (x α j ) i=2 2 j n Note that if β i = 0 or γ i = 0 for some i, then α i is a zero of f n (x), and hence an eigenvalue of M n Thus, such a matrix has an eigenvalue of fixed sign, so its sign pattern cannot be an IAP or a SAP From Lemma 21, the off-diagonal entries β i and γ i enter into the characteristic polynomial of M n only as a product β i γ i In a spectrally arbitrary star sign pattern the entries (1, i) and (i, 1) for 2 i n are non-zero (the sign pattern is irreducible) It is therefore sufficient to consider only

102 G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 matrices with γ i = 1 for 2 i n and for star sign patterns it is sufficient to take entries (1, j) for 2 j n as + Lemma 22 Let n 2 If S n = [s ij ] is an inertially arbitrary star sign pattern, then s ii /= 0 for any leaf i Proof Let n 2 and let S n be an inertially arbitrary star sign pattern Let the real matrix α 1 1 1 1 β 2 α 2 M n = β 3 α 3 β n i=2 α n have sign pattern S n Then, n n det(m n ) = α j β i 2 j n α j By definition of a star sign pattern, β i /= 0 for all i Suppose that α i = 0 for exactly one i with 2 i n Then det(m n ) /= 0 So M n has only non-zero eigenvalues and, thus, the eigenvalues that have zero real parts must be conjugate pairs of purely imaginary numbers Therefore i 0 (M n ) is even for any real matrix M n with sign pattern S n, implying that S n is not inertially arbitrary, and giving a contradiction Now suppose that α i = 0 for more than one i with 2 i n Then det(m n ) = 0 But this implies that i 0 (M n ) 1 for all M n with sign pattern S n, again giving a contradiction Therefore, s ii /= 0 for all 2 i n Corollary 23 Let n 2 If S n = [s ij ] is a spectrally arbitrary star sign pattern, then s ii /= 0 for any leaf i Lemma 24 Let n 2 and M n (as in Lemma 22) be a real matrix having a star sign pattern If M n is nilpotent, then the non-zero entries among α 2, α 3,, α n are distinct Proof By Lemma 21, the characteristic polynomial of M n is n n f n (x) = (x α j ) β i (x α j ) i=2 2 j n If α i = α j for some i, j 2 with i /= j, then α i is a zero of f n (x) = x n, and consequently α i = 0 Thus, if α i /= 0, then α i /= α j for all i, j 2 with i /= j

G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 103 The following two results are used in our constructions of spectrally arbitrary star sign patterns Lemma 25 Let g(x) = x n + n 1 i=0 µ ix i be a real monic polynomial of degree n 1 Let t = max{ n 1 i=0 µ i, 1} Then t is greater than or equal to the absolute value of any zero of g(x) Furthermore, if γ > t and γ is real, then sgn(g(γ )) = sgn(γ n ) Proof Suppose γ > t = max{ n 1 i=0 µ i, 1} Then g(γ ) γ n = µ i γ i µ i γ i γ n 1 µ i < γ n i=0 i=0 Therefore g(γ ) /= 0 So, the absolute value of any zero of g(x) must be at most t Moreover, since g(x) is monic, g(x) as x and g(x) sgn( 1) n as x, thus sgn(g(γ )) = sgn(γ n ) for γ > t Lemma 26 Suppose a p < a p 1 < < a 1 and b 1 < b 2 < < b q are positive real numbers (where possibly p = 0 or q = 0) If p /= 0, then for each fixed i with 1 i p, q sgn ( a i + a j ) ( a i b k ) = sgn(( 1) p+q i ) 1 j p Further, if q /= 0, then for each fixed i with 1 i q, p sgn (b i + a j ) (b i b k ) = sgn(( 1) q i ) 1 j p 1 k q k /=i Proof q ( a i + a j ) ( a i b k ) i 1 = ( 1) (p i)+q (a j a i ) p j=i+1 (a i a j ) i=0 q (a i + b k ) Since each term in the product on the right is positive, the product on the left has the stated sign Similarly, p (b i + a j ) (b i b k ) 1 k q k /=i

104 G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 = ( 1) q i p i 1 (b i + a j ) (b i b k ) q k=i+1 (b k b i ) giving the second statement 3 Spectrally arbitrary star sign patterns As in the note after Lemma 21, the first row of a star sign pattern can be assumed to be [s 11 + + +] Let n 3 and 1 p n 2 Let Z np = [z ij ] be the n n star sign pattern with: z 11 = 0 { if 2 i p + 1 z ii = + if p + 2 i n { sgn( 1) i 1 if 2 i p + 1 z i1 = sgn( 1) n i+1 if p + 2 i n Thus, 0 + + + + Z np = sgn( 1) p sgn( 1) n p 1 + sgn( 1) n p 2 + + Note that Z np has p 1 negative entries and q = n p 1 1 positive entries among the entries z ii Theorem 31 The star sign pattern Z np (n 3 and 1 p n 2) is spectrally arbitrary Proof Let n 3 and 1 p n 2 be positive integers Define q = n p 1 and note that 1 q n 2 and p + q + 1 = n Let g(x) = x n + n 1 i=0 µ ix i be any real monic polynomial of degree n Find positive t greater than or equal to the absolute value of any zero of g(x); for example, as in Lemma 25, pick t = max{ n 1 i=0 µ i, 1} Pick positive numbers a 1, a 2,, a p and b 1, b 2,, b q such that

G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 105 t < a p < a p 1 < < a 1 and p a j q b k = µ n 1 t < b 1 < b 2 < < b q This is always possible: first pick the numbers so that they satisfy the inequalities, then increase either a 1 or b q until equality in the third condition is satisfied For each i with 1 i p, let ( 1) i+1 g( a i ) c i = 1 j p ( a i + a j ) q ( a i b k ) and for each i with 1 i q, let ( 1) q+i g(b i ) d i = p (b i + a j ) 1 k q (b i b k ) k /=i By Lemmas 25 and 26, sgn(c i ) = + and sgn(d i ) = + Consider the following matrix M n having sign pattern Z np : 0 1 1 c 1 a 1 c 2 a 2 M n = ( 1) p c p a p ( 1) q d 1 b 1 ( 1) q 1 d 2 b 2 d q b q Note that a i, b i, c i and d i are positive for all i, and for the entries in M n, c i is multiplied by ( 1) i and d i is multiplied by ( 1) q i+1 By Lemma 21, the characteristic polynomial of M n is f n (x) = x p q (x + a j ) (x b k ) p q + ( 1) i+1 c i (x + a j ) (x b k ) + q ( 1) q i d i 1 j p p (x + a j ) (x b k ) 1 k q k /=i

106 G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 This polynomial is monic of degree n, and the coefficient of x n 1 is p q a j b k = µ n 1 which is the coefficient of x n 1 in g(x) For each i with 1 i p q f n ( a i ) = ( 1) i+1 c i ( a i + a j ) ( a i b k ) = g( a i ) and for each i with 1 i q f n (b i ) = ( 1) q i d i 1 j p p (b i + a j ) (b i b k ) = g(b i ) 1 k q k /=i Since f n (x) and g(x) are both monic and have the same coefficients of x n 1, the polynomial f n (x) g(x) has degree at most n 2 Furthermore, since a 1, a 2,, a p, b 1, b 2, b q are distinct, f n (x) g(x) has n 1 distinct zeroes Therefore f n (x) g(x) = 0 The characteristic polynomial of M n is g(x) and M n has sign pattern Z np Thus, Z np is a spectrally arbitrary star sign pattern Let n 2 and Y n = [y ij ] be the n n star sign pattern with: y 11 = + y ii = if i /= 1 y i1 = sgn( 1) i 1 if i /= 1 + + + + Thus, Y n = + sgn( 1) n 1 Note that Y n is defined for n = 2 while Z np is not Although Z np and Y n are very similar, our proofs that these are SAPs are simpler if each pattern is treated individually Theorem 32 The star sign pattern Y n (n 2) is spectrally arbitrary Proof Let g(x) = x n + n 1 i=0 µ ix i be any real monic polynomial of degree n 2 Find positive t greater than or equal to the absolute value of any zero of g(x); for example, as in Lemma 25, pick t = max{ n 1 i=0 µ i, 1} Pick positive numbers a 0, a 1,, a n 1 such that

G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 107 t < a n 1 < a n 2 < < a 1 and a 0 + a j = µ n 1 First, pick the numbers so they satisfy the inequalities, then increase either a 1 or a 0 until equality in the second condition is satisfied For each i with 1 i n 1, let c i = ( 1)i+1 g( a i ) ( a i + a j ) 1 j n 1 By Lemma 25, and Lemma 26 with q = 0 and p = n 1 1 in the first statement, sgn(c i ) = + Consider the following matrix having sign pattern Y n : M n = ( 1) n 1 c n 1 a 0 1 1 c 1 a 1 c 2 a 2 a n 1 Note that a i and c i are positive for all i, and that for the entries in M n, c i is multiplied by ( 1) i By Lemma 21, the characteristic polynomial of M n is n 1 f n (x) = (x a 0 ) (x + a i ) + ( 1) i+1 c i (x + a j ) 1 j n 1 This polynomial is monic of degree n, and the coefficient of x n 1 is a 0 + a j = µ n 1 which is the coefficient of x n 1 in g(x) For each i with 1 i n 1, f n ( a i ) = ( 1) i+1 c i ( a i + a j ) = g( a i ) 1 j n 1 As in the proof of Theorem 31, f n (x) g(x) = 0 The characteristic polynomial of M n is g(x) and M n has sign pattern Y n Thus, Y n is a spectrally arbitrary star sign pattern Up to equivalence (as explained in Section 1), Y 2 is the unique spectrally arbitrary 2 2 sign pattern, and Z 31 and Y 3 are the only two minimal spectrally arbitrary 3 3 tree sign patterns; see [1] and [3] (in which Z 31 and Y 3 are denoted by T 3 and U 3, respectively)

108 G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 4 Super-patterns of spectrally arbitrary star sign patterns Some terminology is introduced for the statements in this section Let X = {x j j J } be a set of variables with finite index set J Let Σ 0 (X) = 1 and let Σ i (X) be the sum of the products of the elements of the i-element subsets of X for each i with 1 i J For each j J, let X j = X \ {x j } For example, Σ 1 (X j ) = x k, Σ 2 (X j ) = x k x l k J \{j} k,l J \{j} k /=l Lemma 41 Let n 2 and X = {x 1, x 2,, x n } For each i with 1 i n, if j, k J and j /= k, then Proof Σ i (X k ) Σ i (X j ) = (x j x k )Σ i 1 (X j X k ) Σ i (X k ) = Σ i (X j X k ) + x j Σ i 1 (X j X k ) Σ i (X j ) = Σ i (X j X k ) + x k Σ i 1 (X j X k ) and subtracting gives the result Lemma 42 Let n 2 and X = {x 1, x 2,, x n } If 1 1 1 Σ 1 (X 1 ) Σ 1 (X 2 ) Σ 1 (X n ) A n = Σ 2 (X 1 ) Σ 2 (X 2 ) Σ 2 (X n ) Σ n 1 (X 1 ) Σ n 1 (X 2 ) Σ n 1 (X n ) then det(a n ) = 1 i<j n (x i x j ) Proof (By induction) For n = 2, X = {x 1, x 2 } and det(a 2 ) = 1 1 x 2 x 1 = x 1 x 2 Now suppose that the statement is true for all m with 1 m n 1 Consider the matrix A n Subtracting the first column from each of the other columns and expanding about row 1, gives det(a n ) Σ 1 (X 2 ) Σ 1 (X 1 ) Σ 1 (X 3 ) Σ 1 (X 1 ) Σ 1 (X n ) Σ 1 (X 1 ) Σ 2 (X 2 ) Σ 2 (X 1 ) Σ 2 (X 3 ) Σ 2 (X 1 ) Σ 2 (X n ) Σ 2 (X 1 ) = Σ n 1 (X 2 ) Σ n 1 (X 1 ) Σ n 1 (X 3 ) Σ n 1 (X 1 ) Σ n 1 (X n ) Σ n 1 (X 1 )

G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 109 (x 1 x 2 )Σ 0 (X 1 X 2 ) (x 1 x 3 )Σ 0 (X 1 X 3 ) (x 1 x n )Σ 0 (X 1 X n ) (x 1 x 2 )Σ 1 (X 1 X 2 ) (x 1 x 3 )Σ 1 (X 1 X 3 ) (x 1 x n )Σ 1 (X 1 X n ) = (x 1 x 2 )Σ n 2 (X 1 X 2 ) (x 1 x 3 )Σ n 2 (X 1 X 3 ) (x 1 x n )Σ n 2 (X 1 X n ) by Lemma 41 Taking out common column factors gives det(a n ) 1 1 1 ( n ) Σ 1 (X 1 X 2 ) Σ 1 (X 1 X 3 ) Σ 1 (X 1 X n ) = (x 1 x i ) Σ 2 (X 1 X 2 ) Σ 2 (X 1 X 3 ) Σ 2 (X 1 X n ) i=2 Σ n 2 (X 1 X 2 ) Σ n 2 (X 1 X 3 ) Σ n 2 (X 1 X n ) 1 1 1 ( n ) Σ 1 (W 2 ) Σ 1 (W 3 ) Σ 1 (W n ) = (x 1 x i ) Σ 2 (W 2 ) Σ 2 (W 3 ) Σ 2 (W n ) i=2 Σ n 2 (W 2 ) Σ n 2 (W 3 ) Σ n 2 (W n ) where W = {x 2, x 3,, x n } Using the inductive hypothesis, n det(a n ) = (x 1 x i ) (x i x j ) = (x i x j ) i=2 2 i<j n 1 i<j n A different proof of Lemma 42 can be obtained using the theory of alternants, see [8] The following theorem and its proof using the implicit function theorem can be found in [1]; also see [3, Lemma 21] Theorem 43 Let S n be an n n sign pattern, and suppose that there exists a nilpotent matrix M n having sign pattern S n and at least n non-zero entries, m i1 j 1, m i2 j 2,, m in j n Let M n be the matrix obtained by replacing these entries in M n by variables u 1, u 2,, u n, and let the characteristic polynomial of M n be f n (x) = x n h 1 x n 1 + h 2 x n 2 + + ( 1) n 1 h n 1 x + ( 1) n h n If the Jacobian J = (h ([ 1, h 2,, h n ) (u 1, u 2,, u n ) = det hi u j is non-zero at (u 1, u 2,, u n ) = (m i1 j 1, m i2 j 2,, m in j n ), then every super-pattern of S n is spectrally arbitrary ])

110 G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 Theorem 44 Let n 2 and S n be a spectrally arbitrary star sign pattern Then every super-pattern of S n is spectrally arbitrary Proof Let S n be a spectrally arbitrary star sign pattern Then, S n is potentially nilpotent and so there is an n n nilpotent real matrix M n (of the form given in Lemma 22) with sign pattern S n where α i, β i /= 0 for i 2 and α i /= α j for i, j 2 and i /= j (by results in Section 2) Let α 1 1 1 1 u 1 v 1 M n = u 2 v 2 u n 1 v n 1 The conclusion will be shown to follow from an application of Theorem 43 with variables u 1, u 2,, u n 1, v n 1 By Lemma 21, the characteristic polynomial of M n is n 1 f n (x) = (x α 1 ) (x v j ) u i (x v j ) Letting 1 j n 1 f n (x) = x n h 1 x n 1 + h 2 x n 2 + + ( 1) n 1 h n 1 x + ( 1) n h n and X = {v 1, v 2,, v n 1 }, then h 1 = α 1 + v i = Σ 1 (X {α 1 }) h 2 = α 1 v i + 1 i 1 <i 2 n 1 v i1 v i2 = Σ 2 (X {α 1 }) u i Σ 0 (X i ) h 3 = α 1 1 i 1 <i 2 n 1 u i 1 j n 1 v i1 v i2 + v j u i 1 i 1 <i 2 <i 3 n 1 = Σ 3 (X {α 1 }) u i Σ 1 (X i ) v i1 v i2 v i3

G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 111 and in general for l 2, h l = Σ l (X {α 1 }) u i Σ l 2 (X i ) Taking the partial derivatives with respect to v n 1 and u i for 1 i n 1: h 1 h 1 h l = 1 = 0 = Σ l 2 (X i ) (l 2) v n 1 u i u i The partial derivatives h l v n 1 are not computed for l 2 because an examination of the following determinant shows that these terms do not appear in the calculation of the Jacobian (each such term is replaced by in J below) Now, let u n = v n 1 Then the Jacobian J = (h 1, h 2,, h n ) (u 1, u 2,, u n ) 0 0 0 1 1 1 1 Σ 1 (X 1 ) Σ 1 (X 2 ) Σ 1 (X n 1 ) = Σ 2 (X 1 ) Σ 2 (X 2 ) Σ 2 (X n 1 ) Σ n 2 (X 1 ) Σ n 2 (X 2 ) Σ n 2 (X n 1 ) 1 1 1 Σ 1 (X 1 ) Σ 1 (X 2 ) Σ 1 (X n 1 ) = ( 1) 2n 2 Σ 2 (X 1 ) Σ 2 (X 2 ) Σ 2 (X n 1 ) Σ n 2 (X 1 ) Σ n 2 (X 2 ) Σ n 2 (X n 1 ) = 1 i<j n 1 (v i v j ) by Lemma 42 Since the α i s are distinct for 2 i n, J is non-zero at (v 1, v 2,, v n 1, u 1, u 2,, u n 1 ) = (α 2, α 3,, α n, β 2, β 3,, β n ) Therefore, by Theorem 43, every super-pattern of S n is spectrally arbitrary Corollary 45 Every super-pattern of Y n or Z np (for appropriate n and p) is spectrally arbitrary, and thus inertially arbitrary Let Z + np and Z np be the super-patterns of Z np obtained by replacing the 0 in z 11 by + and, respectively By the above corollary, these star sign patterns are SAPs and IAPs

112 G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 5 Potentially stable and inertially arbitrary star sign patterns If sign pattern S n is inertially arbitrary, then both S n and S n are potentially stable Such sign patterns are considered in Theorem 52 below The proof of this theorem relies on the next result, which follows directly from theorems proved in the characterization of potentially stable star sign patterns in [5, Theorems 35 and 42] Theorem 51 Let n 2 and S n be a potentially stable star sign pattern where s 12, s 13,, s 1n are + and s 22, s 33,, s nn are non-zero If among s 22, s 33,, s nn exactly q terms are +, say s i1 i 1, s i2 i 2,, s iq i q are +, then among s i1 1, s i2 1,, s iq 1 exactly q 2 terms are + Theorem 52 Let n 2 and S n be a star sign pattern where s ii is non-zero for any leaf i If S n and S n are both potentially stable, then S n is equivalent to one of Y n, Z np, Z np + or Z np (for appropriate p) Proof Let n 2 and S n be a star sign pattern where s 22, s 33,, s nn are non-zero and S n and S n are both potentially stable For all i such that 2 i n, s 1i and s i1 are non-zero, otherwise S n would not be a star sign pattern Case 1: Among s 22, s 33,, s nn there are both + and entries This implies n 3 By permutation similarity, assume that s 22, s 33,, s p+1,p+1 are and s p+2,p+2, s p+3,p+3,, s nn are + for some p such that 1 p n 2 By Theorem 51, there are exactly n p 1 2 positive entries, and thus exactly n p 1 2 negative entries, among s p+2,1, s p+3,1,, s n1 This is exactly the number of positive and negative entries among z p+2,1, z p+3,1,, z n1 in Z np Since the off-diagonal terms in a matrix having a star sign pattern enter into its characteristic polynomial only as products, S n is equivalent to the pattern obtained from S n by taking the negative of the main diagonal So, similarly, there are exactly p 2 positive entries among s 21, s 31,,s p+1,1 and thus p 2 negative entries This is exactly the number of positive and negative entries among z 21, z 31,,z p+1,1 in Z np So, by permutation similarity, s ij = z ij in Z np for all (i, j) /= (1, 1) Therefore, S n is equivalent to one of Z np, Z np + or Znp (depending on the sign of s 11) Case 2: The entries s 22, s 33,, s nn are all the same sign Since S n is equivalent to the star sign pattern obtained by taking the negative of the main diagonal in S n, assume s 22, s 33,, s nn are all Thus, s 11 is +, otherwise S n would have a strictly positive main diagonal Furthermore, similarly to Case 1, there are exactly n 1 2 positive and n 1 2 negative entries among s 21, s 31,, s n1 This is exactly the number of positive and negative entries among y 21, y 31,, y n1 in Y n Therefore, S n is equivalent to Y n

G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 113 Note that Theorem 52 is sufficient for our purposes, since IAPs and SAPs are subclasses of the patterns S n for which S n and S n are both potentially stable However, using further results from [5], a characterization of all star sign patterns S n such that S n and S n are both potentially stable can be obtained Theorem 53 Let n 2 and S n be an inertially arbitrary star sign pattern Then S n is equivalent to one of Y n, Z np, Z + np or Z np (for appropriate p) Furthermore, Y n and Z np are minimal inertially arbitrary sign patterns and the only minimal inertially arbitrary star sign patterns Proof Let S n be an inertially arbitrary star sign pattern By Lemma 22, the entries s 22, s 33,, s nn are non-zero, thus the first statement follows directly from Theorem 52 Clearly, Z np + and Z np are not minimal As in Section 2, if s i1 or s 1i is replaced with 0 where 2 i n, any matrix with the resulting sign pattern has an eigenvalue with sign equal to s ii and so the sign pattern is not an IAP If s ii is replaced with 0 where 2 i n, any matrix with the resulting sign pattern has the sign of its determinant fixed, and so the sign pattern is not an IAP In Z np, the entry z 11 is equal to 0, so Z np is minimal In Y n, the entry y 11 is the only positive entry on the main diagonal, and so cannot be replaced by 0, so Y n is minimal Although the sign patterns Y n and Z np are minimal inertially (and thus, spectrally) arbitrary star sign patterns, there are proper sub-patterns of these that are potentially nilpotent or potentially stable For example, 0 + + + 0 0 0 0 is potentially nilpotent and 0 + + 0 0 0 is potentially stable 6 Potentially nilpotent star sign patterns Theorem 61 Let n 2 and S n = [s ij ] be a star sign pattern If S n is potentially nilpotent and has s ii /= 0 for any leaf i, then, up to equivalence, S n is one of Y n, Z np, Z np + or Z np (for appropriate p)

114 G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 Proof Let n 2 and S n be a potentially nilpotent star sign pattern with no zeroes among s 22, s 33,, s nn Let M n, as in Lemma 22, be a nilpotent matrix having sign pattern S n Thus, by Lemma 24, for i 2, the α i entries are distinct By permutation similarity, it can be assumed that α 2 < α 3 < < α n Case 1: Among α 2, α 3,, α n, there are both positive and negative entries This implies that n 3 Let p and q be the number of negative and positive entries, respectively (and note that 1 p, q n 2 and that p + q + 1 = n) Then α 2 < α 3 < < α p+1 < 0 < α p+2 < α p+3 < < α n Relabel α 2, α 3,, α n, β 2, β 3,, β n to give α 1 1 1 1 γ 1 a 1 γ 2 a 2 M n = γ p a p δ 1 b 1 δ 2 b 2 δ q where 0 < a p < a p 1 < < a 1 and 0 < b 1 < b 2 < < b q By Lemma 21, the characteristic polynomial of M n is p q f n (x) = (x α 1 ) (x + a j ) (x b k ) 1 j p p q γ i (x + a j ) (x b k ) q δ i p (x + a j ) (x b k ) 1 k q k /=i Since M n is nilpotent, f n (x) = x n So, for each i with 1 i p, q ( a i ) n = γ i ( a i + a j ) ( a i b k ) 1 j p By Lemma 26, sgn(γ i ) = sgn( 1) n 1 (p+q i) = sgn( 1) i (since p + q + 1 = n) Moreover, for each i with 1 i q, b q

G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 115 b n i = δ i p (b i + a j ) (b i b k ) 1 k p k /=i which implies, by Lemma 26, that sgn(δ i ) = sgn( 1) q i+1 So, for (i, j) /= (1, 1), the (i, j)th entry of M n has sign equal to z ij in Z np Thus, S n is equivalent to one of Z np, Z + np or Z np Case 2: For i 2, the α i entries all have the same sign; without loss of generality, assume that they are all negative (if M n is nilpotent, so is M n ) Order the leaf entries so that α 2 < α 3 < < α n and then relabel the matrix to give M n = α 1 1 1 1 γ 1 a 1 γ 2 a 2 γ n 1 a n 1 Since M n is nilpotent, α 1 = n 1 a j is positive Furthermore, by Lemma 21, the characteristic polynomial of M n is n 1 f n (x) = (x α 1 ) (x + a i ) γ i 1 j n 1 (x + a j ) Since M n is nilpotent, f n (x) = x n, so for each i with 1 i n 1, ( a i ) n = γ i 1 j n 1 ( a i + a j ) implying that γ i has sign sgn( 1) i So, by permutation similarity, for all (i, j), the sign of the (i, j)th entry of M n is equal to y ij in Y n Therefore, S n is equivalent to Y n Theorem 61, Corollary 45 and results of [7], which are introduced next, can be used to characterize potentially nilpotent star sign patterns Let n 2 and S n be a star sign pattern By permutation similarity, it can be assumed that s 22, s 33,, s mm are non-zero and s m+1,m+1, s m+2,m+2,, s nn are zero (either collection could be empty) Let R 1 be the m m star sign pattern obtained by deleting the rows and columns numbered m + 1, m + 2,, n in S n Let R 2 be the (n m + 1) (n m + 1) star sign pattern obtained by deleting rows and columns numbered 2, 3,, m in S n and replacing s 11 with 0 For example, if

116 G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 s 11 + + + s 21 s 22 s 31 s 33 S n = s m1 s mm s m+1,1 0 s m+2,1 0 s n1 0 then s 11 + + + s 21 s 22 R 1 = s 31 s 33 s m1 s mm 0 + + + s m+1,1 0 R 2 = s m+2,1 0 s n1 0 Theorem 62 [7, Theorems 2 and 3] Let n 2 and S n be a star sign pattern Then S n is potentially nilpotent if and only if R 1 and R 2 (as above) obtained from S n are potentially nilpotent In particular, R 2 is potentially nilpotent if and only if there is at least one positive and one negative entry among s m+1,1, s m+2,1,, s n1 Theorem 61 and Corollary 45 give necessary and sufficient conditions for the star sign pattern R 1 to be potentially nilpotent if R 1 has order at least 2 Theorem 62 gives necessary and sufficient conditions for the star sign pattern R 2 to be potentially nilpotent if the order of R 2 is at least 2; moreover, the only 1 1 potentially nilpotent sign pattern is S 1 = [0] Theorem 63 Let n 2 and S n be a star sign pattern Then S n is potentially nilpotent if and only if, up to equivalence, (a) S n is one of Y n, Z np, Z np +, Z np (for appropriate p); or, (b) for some m such that 1 m n 2, [ ] S11 S S n = 12 S 21 S 22

G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 117 where + + + 0 0 0 S 12 = 0 0 0 0 0 0 is m (n m); s m+1,1 0 0 0 s m+2,1 0 0 0 S 21 = s n1 0 0 0 is (n m) m and has both positive and negative entries among s m+1,1, s m+2,1,, s n1 ; S 22 = [0] (n m) (n m) ; and, (i) if m = 1, then S 11 = [0], or (ii) if m 2, then S 11 is one of Y m, Z mp, Z mp +, Z mp (for appropriate p) Theorem 64 Let n 2 and S n be a star sign pattern that is both potentially nilpotent and potentially stable Then S n has s ii /= 0 for any leaf i In particular, S n is equivalent to one of Y n, Z np, Z np + or Z np (for appropriate p) Proof Suppose n 2 and S n is a potentially nilpotent and potentially stable star sign pattern Let m be the number of non-zero entries among s 22, s 33,, s nn If m n 3, then there are at least two zeroes among s 22, s 33,, s nn, so, any matrix having sign pattern S n has determinant equal to zero Thus, m n 2, since S n is potentially stable If m = n 2, then there is exactly one zero among s 22, s 33,, s nn, implying that any matrix having sign pattern S n has a non-zero determinant Thus, since S n is potentially nilpotent, m = n 1 So, s ii /= 0 for all i 2 Therefore, by Theorem 63, S n is equivalent to one of Y n, Z np, Z np + or Z np (for appropriate p) 7 Summary The following theorem summarizes our results for star sign patterns Theorem 71 If n 2 and S n is a star sign pattern, then the following are equivalent: 1 S n is equivalent to one of Y n, Z np, Z np + or Z np (for appropriate p) 2 S n is spectrally arbitrary

118 G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 3 S n is inertially arbitrary 4 S n is potentially nilpotent and has a loop at each of the n 1 leaves in its graph 5 S n and S n are both potentially stable and S n has a loop at each of the n 1 leaves in its graph 6 S n is potentially nilpotent and potentially stable The equivalences in Theorem 71 are illustrated in the strongly connected directed graph given above, with the implications following from the definitions of the appropriate terms along with previous theorems, corollaries and lemmas given by number The equivalence of (2) and (3) holds for any 3 3 sign pattern; see [3] However, in general (3) does not imply (2), as illustrated by an example with n = 4 in [4] where the graph of the sign pattern is not a tree The same example shows that (6) does not necessarily imply (2) for general sign patterns However, for tree sign patterns with n 4, it is unknown whether (3) implies (2), or whether (6) implies (2) or (3)

G MacGillivray et al / Linear Algebra and its Applications 400 (2005) 99 119 119 Acknowledgment The authors thank DD Olesky for a careful reading of this manuscript and T Britz for discussions at the beginning of this project References [1] JH Drew, CR Johnson, DD Olesky, P van den Driessche, Spectrally arbitrary patterns, Linear Algebra Appl 308 (2000) 121 137 [2] JJ McDonald, DD Olesky, MJ Tsatsomeros, P van den Driessche, On the spectra of striped sign patterns, Linear and Multilinear Algebra 51 (2003) 39 48 [3] T Britz, JJ McDonald, DD Olesky, P van den Driessche, Minimal spectrally arbitrary sign patterns, SIAM J Matrix Anal Appl 26 (2004) 257 271 [4] MS Cavers, KN Vander Meulen, Spectrally and inertially arbitrary sign patterns, Linear Algebra Appl 394 (2005) 53 72 [5] Y Gao, J Li, On the potential stability of star sign pattern matrices, Linear Algebra Appl 327 (2001) 61 68 [6] Y Shao, L Sun, Y Gao, Inertia sets of two classes of symmetric sign patterns, Linear Algebra Appl 384 (2004) 85 95 [7] L Yeh, Sign pattern matrices that allow a nilpotent matrix, Bull Austral Math Soc 53 (1996) 189 196 [8] T Muir, A Treatise on the Theory of Determinants, Dover, New York, 1960