Unspecified Journal Volume 00, Number 0, Pages 000 000 S????-????(XX)0000-0 DETERMINANT AND PFAFFIAN OF SUM OF SKEW SYMMETRIC MATRICES TIN-YAU TAM AND MARY CLAIR THOMPSON Abstract. We completely describe the determinants of the sum of orbits of two real skew symmetric matrices, under similarity action of orthogonal group the special orthogonal group respectively. We also study the Pfaffian case the complex case. 1. Introduction Let U(n) C n n be the unitary group. following result. Fiedler [3] obtained the Theorem 1.1. (Fiedler) Let A, B C n n be Hermitian matrices with eigenvalues α 1,..., α n β 1,..., β n. Then H(A, B) := det(uau 1 + V BV 1 ) : U, V U(n)} is the interval [min σ Sn n (α i+β σ(i) ), max σ Sn n (α i+β σ(i) )], where S n denotes the symmetric group on 1,..., n}. See [2, 6, 7] for related studies. The result remains the same if U(n) is replaced by the special unitary group SU(n). It is also the same if A, B R n n are real symmetric matrices U(n) is replaced by the orthogonal group O(n) or the spectral orthogonal group SO(n). If A, B R n n are skew symmetric, we consider the set D(A, B) := det(uau 1 + V BV 1 ) : U, V O(n)} = det(a + V BV 1 ) : V O(n)}. The case n = 2k + 1 is trivial since n n real skew symmetric matrices are singular so that D(A, B) = 0}. We only need to consider the even case n = 2k. Theorem 1.1 can only provide lower upper bounds for the compact set D(A, B) R via the Hermitian matrices ia ib. The orthogonal group O(n) is not connected has two 2000 Mathematics Subject Classification. Primary 15A15, 15A42 1 c 0000 (copyright holder)
2 T.Y. TAM AND M.C. THOMPSON components corresponding to +1 1 determinant, namely, the special orthogonal group SO(n) SO(n) ˆ := DSO(n), where D := diag (1,..., 1, 1). Though O(n) is disconnected, we show that D(A, B) is an interval when k 2. In Section 2 we determine D(A, B) D 0 (A, B) := det(uau 1 + V BV 1 ) : U, V SO(n)}. In Section 3 we study the case when the determinant function is replaced by the Pfaffian. In Section 4 we obtain corresponding result for the complex case. 2. Determinant sum of orbits Let A, B R 2k 2k be skew symmetric. Denote by (A, B) := UAU 1 + V BV 1 : U, V O(2k)} 0 (A, B) := UAU 1 + V BV 1 : U, V SO(2k)} the sums of orbits of A B under O(2k) SO(2k) respectively. Set D(A, B) := det M : M (A, B)} D 0 (A, B) := det M : M 0 (A, B)}. Notice that D 0 (A, B) is a closed interval since SO(2k) is compact connected. The spectral theorem for real skew symmetric matrix A under O(2k) [5] asserts that there is an orthogonal matrix O O(2k) such that OAO 1 is in the following canonical form (1) OAO 1 = α 1 J α k J = diag (α 1,..., α k ) J where ( ) 0 1 J :=. 1 0 Thus we may assume that A B are in their canonical forms when D(A, B) is studied. The eigenvalues of A are purely imaginary occur in pairs ±iα 1,..., ±iα k we may assume that α 1 α 1 α 2 α 2 α k α k ( 0) which are the singular values of A. However the spectral theorem for A under SO(2k) involves the sign of the Pfaffian of A, denoted by δ A = ±1. To quickly see it, consider J J which are similar via O(2) but not SO(2). Eigenvalues fail
DETERMINANT, PFAFFIAN AND SKEW SYMMETRIC MATRICES 3 to distinguish J J. However Pf J = 1 Pf ( J) = 1, where Pfaffian of A [4] is defined as Pf A = 1 sgn(σ) a 2 k σ(2i 1),σ(2i), k! σ S 2k sgn(σ) is the signature of σ. For example [ ] 0 a b c 0 a Pf = a, Pf a 0 d e a 0 b d 0 f = af be + dc. c e f 0 In particular, if A = diag (a 1,..., a k ) J, then Pf A = a 1 a k. It is well known that [8, 9] (Pf A) 2 = det A. If A R 2k 2k is skew symmetric, there is O SO(2k) such that (2) OAO 1 = diag (α 1,..., α k ) J if δ A = 1 diag (α 1,..., α k ) J if δ A = 1 Lemma 2.1. Let A, B R 2k 2k be real skew symmetric. Then (a) D 0 (A, B) R +. (b) D(A, B) = D 0 (A, B) D 0 (A D, B), where A D := DAD 1. Proof. (a) The determinant of each real skew symmetric matrix in R 2k 2k is nonnegative so D 0 (A, B) R +. (b) It is easy to see that (A, B) = 0 (A, B) 0 (A D, B) 0 (A, B D ) 0 (A D, B D ) where A D := DAD 1. Notice that D 0 (A, B) = D 0 (A D, B D ) D 0 (A D, B) = D 0 (A, B D ) since = D 0 (A, B) = det(uau 1 + V BV 1 ) : U, V SO(2k)} = det(dud 1 A D DU 1 D 1 + DV D 1 B D DV 1 D 1 ) : U, V SO(2k)} = D 0 (A D, B D ). Notice that D 0 (A, B) D 0 (A, B D ) are closed intervals. We will show that is the case for D(A, B) when k 2. Lemma 2.2. Let k 2 let A, B R 2k 2k be real skew symmetric with singular values α 1 α 1 α 2 α 2 α k α k β 1 β 1 β 2 β 2 β k β k. The following are equivalent.
4 T.Y. TAM AND M.C. THOMPSON (a) (A, B) GL 2k (R). (b) 0 (A, B) GL 2k (R). (c) [α k, α 1 ] [β k, β 1 ] = φ. Proof. (a) (b) is obvious. (b) (c) Suppose [α k, α 1 ] [β k, β 1 ] φ, i.e., either (i) α k β k α 1 or (ii) β k α k β 1. By symmetry we may assume that (i) holds. Because of (2) D 0 (A, B) = D 0 (A D, B D ) we may assume that A = diag (α 1,..., α k ) J, B = diag (β 1,..., δβ k ) J where δ := δ A δ B = ±1. Set h(v ) := Pf (A + V BV 1 ), V SO(2k). Clearly k (α i + β i ) 0 if δ = 1 h(i) = Now let V SO(2k) such that (α k β k ) k 1 (α i + β i ) 0 if δ = 1. V BV 1 = diag ( β k, β 2,..., β k 1, δβ 1 ) J. Then (α 1 β k )(α k β 1 ) k 1 i=2 h(v ) = (α i + β i ) 0 if δ = 1 (α 1 β k )(α k + β 1 ) k 1 i=2 (α i + β i ) 0 if δ = 1. By the path connectedness of SO(2k) the continuity of the Pfaffian, there is W SO(2k) such that h(w ) = Pf (A + W BW 1 ) = 0, so A + W BW 1 is singular. (c) (a) See [6, Lemma 3]. The next two theorems determine D 0 (A, B) D(A, B) respectively. Theorem 2.3. Let A, B R 2k 2k be real skew symmetric with singular values α 1 α 1 α 2 α 2 α k α k β 1 β 1 β 2 β 2 β k β k respectively. Let δ A δ B be the signs of the Pfaffians of A B, respectively. Let D 0 (A, B) = [m 0, M 0 ]. (a) If [α k, α 1 ] [β k, β 1 ] = φ, then k (α i + β k i+1 ) 2 if δ A δ B = 1 M 0 = max (α j β k j+1 ) 2 k (α i + β k i+1 ) 2 if δ A δ B = 1 k (α i β k i+1 ) 2 if δ A δ B = 1, k even or δ A δ B = 1, k odd m 0 = min (α j + β k j+1 ) 2 k (α i β k i+1 ) 2 if δ A δ B = 1, k even or δ A δ B = 1, k odd
DETERMINANT, PFAFFIAN AND SKEW SYMMETRIC MATRICES 5 (b) If [α k, α 1 ] [β k, β 1 ] φ, we have two cases: (I) if k = 1, then 4α1 2 if δ A δ B = 1 D 0 (A, B) = 0 if δ A δ B = 1. (II) if k 2, then D 0 (A, B) = [0, M 0 ]. Proof. We consider two cases. (I) If k = 1, then D 0 (A, B) = s}, where (α 1 + β 1 ) 2 if δ A δ B = 1 s = (α 1 β 1 ) 2 if δ A δ B = 1 (II) Let k 2. (a) Suppose [α k, α 1 ] [β k, β 1 ] = φ. Because of (2) D 0 (A, B) = D 0 (A D, B D ) we may assume that (3) A = diag (α 1,..., α k ) J, B = diag (β 1,..., β k 1, δβ k ) J where δ := δ A δ B. Notice that m 0 M 0 depend continuously on α 1 α k. So we may assume that (4) α 1 > α 2 > > α k > 0. By Lemma 2.2 all matrices in D 0 (A, B) are nonsingular. Assume that D 0 (A, B) = det(a + V BV 1 ) : V SO(2k)} attains its minimum m 0 or maximum M 0 at det(a + B 0 ), where B 0 = V 0 BV 1 0 for some V 0 SO(2k). The optimizing matrix (5) C 0 := A + B 0 is nonsingular C 1 0 is skew symmetric. For each real skew symmetric matrix S R 2k 2k, e ɛs SO(2k). Recall Fiedler s lemma [3]: det(p + ɛq) = (det P )(1 + ɛ tr QP 1 ) + o(ɛ 2 ), where P, Q C n n P is nonsingular. Hence det(a + e ɛs B 0 e ɛs ) = det(a + B 0 + ɛ [S, B 0 ]) + o(ɛ 2 ) = det C 0 (1 + ɛ tr [S, B 0 ]C 1 0 ) + o(ɛ 2 ) = det C 0 (1 + ɛ tr S[B 0, C 1 0 ]) + o(ɛ 2 ) so that tr S[B 0, C0 1 ] = 0 for all real skew symmetric S. Choose S = [B 0, C0 1 ] to have [B 0, C0 1 ] = 0, i.e., B 0 commutes with C0 1 thus
6 T.Y. TAM AND M.C. THOMPSON with C 0, i.e., B 0 (A + B 0 ) = (A + B 0 )B 0. Hence B 0 A = AB 0. Because of (3) (4), for some σ S n, with B 0 = diag (±β σ(1),..., ±β σ(k) ) J (6) the number of subtracted terms is even if δ = 1 odd if δ = 1. This becomes a finite optimization problem. The maximum minimum occur among the numbers (7) (α i ± β σ(i) ) 2, σ S k subject to (6). The maximum is attained at B 0 = diag (β σ(1),..., β σ(k 1), δβ σ(k) ) J for some σ S k we have k maxσ Sk M 0 = (α i + β σ(i) ) 2 if δ = 1 max σ Sk (α j β σ(j) ) 2 k (α i + β σ(i) ) 2 if δ = 1. The expression max σ Sk k (α i +β σ(i) ) 2 can be identified. Notice that [3, p.29] for i < j σ(i) < σ(j), (α i +β σ(i) )(α j +β σ(j) ) (α i +β σ(j) )(α j +β σ(i) ) = (α i α j )(β σ(i) β σ(j) ) 0 for each σ S k. So (8) max σ S k (α i + β σ(i) ) 2 = (α i + β k i+1 ) 2. The expression max σ Sk (α j β σ(j) ) 2 k (α i +β σ(i) ) 2 can also be identified. Notice that for i < j σ(i) < σ(j), (a) if α i β σ(i) 0, then α i β σ(j) 0 (α i β σ(i) )(α j +β σ(j) ) (α i β σ(j) )(α j +β σ(i) ) = (α i +α j )(β σ(i) β σ(j) ) 0. (b) if α i β σ(i) 0, then α j β σ(i) 0 (α i β σ(i) )(α j +β σ(j) ) (α i +β σ(j) )(α j β σ(i) ) = (α i α j )(β σ(i) β σ(j) ) 0. So max σ S k (α j β σ(j) ) 2 (α i +β σ(i) ) 2 = max (α j β k j+1 ) 2 (α i +β k i+1 ) 2
DETERMINANT, PFAFFIAN AND SKEW SYMMETRIC MATRICES 7 Since [α k, α 1 ] [β k, β 1 ] = φ, α j +β k j+1 > 0 for all j, it can be expressed (α as max j β k j+1 ) 2 k (α j +β k j+1 ) 2 (α i + β k i+1 ) 2. The minimum is attained when most subtracted terms are present, i.e., Case 1. k is even: (i) if δ = 1, then for some σ S k B 0 = diag ( β k,..., β 1 ) J yields the minimum k (α i β k i+1 ) 2. It is because (9) min (α i β σ(i) ) 2 = (α i β k i+1 ) 2 σ S k since for definiteness we may assume that α k > β 1 for i < j σ(i) < σ(j), (α i β σ(i) )(α j β σ(j) ) (α i β σ(j) )(α j β σ(i) ) = (α i α j )(β σ(i) β σ(j) ) 0 for each σ S k. Also see [6, Theorem 1]. (ii) if δ = 1, then for some j some σ S k B 0 = diag ( β σ(1),..., β σ(j),..., β σ(k) ) J yields min σ Sk (α j + β σ(j) ) 2 (α i β σ(i) ) 2. Since [α k, α 1 ] [β k, β 1 ] = φ, (α j β k j+1 ) 2 > 0 for all j. For definiteness, assume α k > β 1. Then (α i +β σ(i) )(α j β σ(j) ) (α i +β σ(j) )(α j β σ(i) ) = (α i +α j )(β σ(i) β σ(j) ) 0. (10) min σ S k So (α j +β σ(j) ) 2 (α i β σ(i) ) 2 = min (α j +β k j+1 ) 2 (α i β k i+1 ) 2 which can be expressed as min (α j +β k j+1 ) 2 (α j β k j+1 ) 2 k (α i β k i+1 ) 2. It is attained at one of the B 0 = diag ( β k,..., β k j+1,..., β 1 ) J, Case 2. k is odd: (i) if δ = 1, one of the B 0 = diag ( β k,..., β k j+1,..., β 1 ) J, j = 1,..., k j = 1,..., k yields min (α j + β k j+1 ) 2 k (α i β k i+1 ) 2.
8 T.Y. TAM AND M.C. THOMPSON (ii) if δ = 1, then B 0 = diag ( β k,..., β 1 ) J yields k (α i β k i+1 ) 2. (b) Suppose [α k, α 1 ] [β k, β 1 ] φ. The case k = 1 is trivial. When k 2, by Lemma 2.2, 0 D 0 (A, B) hence m 0 = 0. For the maximum, by continuity argument the fact that the set of singular skew matrices in R 2k 2k is of measure zero, we may assume that (4) holds that A + B is nonsingular. So D 0 (A, B) 0} thus the maximizing matrix C 0 in (5) is nonsingular. Then follow the argument in the proof of (a). Corollary 2.4. (a) When δ A δ B = 1, all matrices in 0 (A, B) are singular if only if rank A + rank B < 2k. (b) When δ A δ B = 1, all matrices in 0 (A, B) are singular if only if rank A + rank B < 2k or α i = β i = c > 0 for all i = 1,..., k. Proof. (a) Suppose δ A δ B = 1. Then D 0 (A, B) = 0 if only if M 0 = k (α i + β k i+1 ) 2 = 0 from Theorem 2.3, i.e., α i = β k i+1 = 0 for some i = 1,..., k. In other words, rank A + rank B < 2k. (b) Suppose δ A δ B = 1. Then D 0 (A, B) = 0 if only if M 0 = max (α j β k j+1 ) 2 k (α i+β k i+1 ) 2 = 0, i.e., either α j β k j+1 = 0 for all j = 1,..., k, or α i = β k i+1 = 0 for some i = 1,..., k. The first amounts to α i = β i = c, a constant, for all i = 1,..., k. Theorem 2.5. Let A, B R 2k 2k be real skew symmetric with singular values α 1 α 1 α 2 α 2 α k α k β 1 β 1 β 2 β 2 β k β k respectively. (I) If k = 1, then D(A, B) = (α 1 β 1 ) 2, (α 1 + β 1 ) 2 }. (II) Let k 2. Then D(A, B) = [m, M], where m = Proof. (I) is trivial. (II) By Lemma 2.1(b) M = (α i + β k i+1 ) 2 k (α i β k i+1 ) 2 if [α k, α 1 ] [β k, β 1 ] = φ 0 if [α k, α 1 ] [β k, β 1 ] φ, D(A, B) = [m 0 (A, B), M 0 (A, B)] [m 0 (A D, B), M 0 (A D, B)],
DETERMINANT, PFAFFIAN AND SKEW SYMMETRIC MATRICES 9 where k M 0 (A, B) := (α i + β k i+1 ) 2 if δ A δ B = 1 max (α j β k j+1 ) 2 k (α i + β k i+1 ) 2 if δ A δ B = 1 k M 0 (A D, B) := (α i + β k i+1 ) 2 if δ A δ B = 1 max (α j β k j+1 ) 2 k (α i + β k i+1 ) 2 if δ A δ B = 1 since δ AD = δ A, according to Theorem 2.3. So M is the maximum of M 0 (A, B) M 0 (A D, B) is k (α i + β k i+1 ) 2. (i) If [α k, α 1 ] [β k, β 1 ] φ, then m 0 (A, B) = m 0 (A D, B) = 0 by Theorem 2.3. So m = 0 D(A, B) is the interval [0, M]. (ii) If [α k, α 1 ] [β k, β 1 ] = φ, we have m = minm 0 (A, B), m 0 (A D, B)} = k (α i β k i+1 ) 2, again by Theorem 2.3. It remains to show that D(A, B) is an interval, i.e., D 0 (A, B) D 0 (A D, B) φ. Let L := (α i + β k i+1 ) 2 l := max (α j β k j+1 ) 2 s := min (α j + β k j+1 ) 2 S := (α i β k i+1 ) 2 Example 2.6. Let k = 2 A = ( ) ( 0 4 0 2 4 0 2 0 Theorem 2.5. Since k = 2, (α i + β k i+1 ) 2 (α i β k i+1 ) 2 Clearly S s l L. To show that D 0 (A, B) D 0 (A D, B) φ it suffices to consider δ A δ B = 1 by symmetry. Suppose δ A δ B = 1. By Theorem 2.3, when k is even D 0 (A D, B) = [s, l] [S, L] = D 0 (A, B); when k is odd D 0 (A, B) = [s, L] D 0 (A D, B) = [S, l]. Hence D(A, B) is an interval. ( ) ( ) 0 3 0 1 B = 3 0 1 0 ). Since [1, 3] [2, 4] φ, S(A, B) = [0, 625] by S(A, B) det(uau 1 + V BV 1 ) : U, V U(4)} = det(u(ia)u 1 + V (ib)v 1 ) : U, V U(4)} = H(iA, ib).
10 T.Y. TAM AND M.C. THOMPSON The eigenvalues of ia are 3, 1, 1, 3 (denoted by α 1, α 2, α 3, α 4), those of ib are 4, 2, 2, 4 (denoted by β 1, β 2, β 3, β 4). None of 4 (α i+ β σ(i) ), σ S 4, is zero. Indeed, if τ = (13) S 4 denotes the transposition, then β τ = ( 2, 2, 4, 4). So 4 (α i + β τ(i) ) = 63 S(A, B). ( ) ( ) 0 4 0 3 Similarly if we consider A = B = 4 0 3 0 ( ) ( ) 0 2 0 1. Since [3, 4] [1, 2] = φ, S(A, B) = [3, 625] 2 0 1 0 by Theorem 2.5. 4 (α i + β τ(i) ) = 24 S(A, B). So the lower bound of Theorem 1.1 is in general not included in S(A, B). We consider 3. Pfaffian sum of orbits P (A, B) := Pf (UAU 1 + V BV 1 ) : U, V O(2k)}. P 0 (A, B) := Pf (UAU 1 + V BV 1 ) : U, V SO(2k)}. Clearly P 0 (A, B) is an interval. Theorem 3.1. Let A, B R 2k 2k be real skew symmetric with singular values α 1 α 1 α k α k 0 β 1 β 1 β k β k 0 respectively. (a) If [α k, α 1 ] [β k, β 1 ] φ, then P (A, B) = [ P, P ], where P = k (α i + β k i+1 ). (b) If [α k, α 1 ] [β k, β 1 ] = φ, then P (A, B) = [ P, p] [p, P ] where p = k α i β k i+1. Proof. Since P (A, B) = P 0 (A, B) P 0 (A D, B) P 0 (A, B D ) P 0 (A D, B D ) (11) P 0 (A D, B D ) = P 0 (A, B), P 0 (A, B D ) = P 0 (A D, B), P (A, B) is symmetric about the origin. Since det A = (Pf A) 2, using Theorem 2.5 we have the desired result. Theorem 3.2. Let A, B R 2k 2k be real skew symmetric with singular values α 1 α 1 α k α k 0 β 1 β 1 β k β k 0 respectively. Let δ A δ B be the signs of the Pfaffians of A B, respectively.
DETERMINANT, PFAFFIAN AND SKEW SYMMETRIC MATRICES 11 (I) If k = 1, then P 0 (A, B) = s}, where α 1 + β 1 if δ A = δ B = 1 (α 1 + β 1 ) if δ A = δ B = 1 s = α 1 β 1 if δ A = 1, δ B = 1 α 1 + β 1 if δ A = 1, δ B = 1 (II) Let k 2. Set P 0 (A, B) = [p 0, P 0 ]. (a) Suppose [α k, α 1 ] [β k, β 1 ] = φ. (1) If δ A = δ B = 1, then P 0 = k (α i + β k i+1 ) k p 0 = α i β k i+1 if k even min (α j + β k j+1 ) k α i β k i+1 if k odd (2) If δ A = δ B = 1, then k P 0 = α i β k i+1 if k even min σ Sk (α j + β σ(j) ) α i β σ(i) if k odd p 0 = k (α i + β k i+1 ). (3) If δ A = 1 δ B = 1, (i) if α k > β 1, then P 0 = max α j β k j+1 (α i + β k i+1 ) min (α j + β k j+1 ) k p 0 = α i β k i+1 k α i β k i+1 if k is even if k is odd (ii) if β k > α 1, then min (α j + β k j+1 ) k P 0 = α i β k i+1 if k is even k α i β k i+1 if k is odd p 0 = max α j β k j+1 (4) If δ A = 1 δ B = 1, (α i + β k i+1 ) < 0
12 T.Y. TAM AND M.C. THOMPSON (i) if α k > β 1, then min (α j + β k j+1 ) k P 0 = α i β k i+1 if k is even k α i β k i+1 if k is odd p 0 = max α j β k j+1 (ii) if β k > α 1, then P 0 = max α j β k j+1 (α i + β k i+1 ) < 0 (α i + β k i+1 ) min (α j + β k j+1 ) k p 0 = α i β k i+1 k α i β k i+1 (b) Suppose [α k, α 1 ] [β k, β 1 ] φ. if k is even if k is odd (1) If δ A = δ B = 1, then P 0 (A, B) = [a, k (α i + β k i+1 )] for some k (α i + β k i+1 ) a 0. (2) If δ A = δ B = 1, then P 0 (A, B) = [ k (α i + β k i+1 ), b] for some 0 b k (α i + β k i+1 ). (3) If δ A = 1, δ B = 1, or δ A = 1, δ B = 1, then P 0 (A, B) is either [c, max α j β k j+1 (α i + β k i+1 )] for some max α j β k j+1 k (α i + β k i+1 ) c 0, or [ max α j β k j+1 (α i + β k i+1 ), d] for some 0 d max α j β k j+1 k (α i + β k i+1 ). Proof. (I) is trivial. (II) Clearly P 0 (A, B) is a closed interval. Since D 0 (A, B) = [m 0, M 0 ], where m 0, M 0 0 are given in Theorem 2.3 det A = (Pf A) 2, P 0 (A, B) is either [m 1/2 0, M 1/2 0 ] or [ M 1/2 0, m 1/2 0 ]. It is also clear that P 0 (A D, B D ) = P 0 (A, B), P 0 (A, B D ) = P 0 (A D, B).
DETERMINANT, PFAFFIAN AND SKEW SYMMETRIC MATRICES 13 (a) Suppose [α k, α 1 ] [β k, β 1 ] = φ. By Lemma 2.2, 0 P 0 (A, B) so that either P 0 (A, B) R + or P 0 (A, B) R. (1) If δ A = δ B = 1, then clearly 0 < k (α i + β k i+1 ) P 0 (A, B). So P 0 (A, B) = [m 1/2 0, M 1/2 0 ]. (2) If δ A = δ B = 1, then clearly 0 > k (α i + β k i+1 ) P 0 (A, B). So P 0 (A, B) = [ M 1/2 0, m 1/2 0 ]. (3) Suppose δ A = 1 δ B = 1. (i) if α k > β 1, then 0 < max (α j β k j+1 ) k (α i + β k i+1 ) P 0 (A, B). So P 0 (A, B) = [m 1/2 0, M 1/2 0 ]. (ii) if β k > α 1, then 0 > max (α j β k j+1 ) k (α i + β k i+1 ) P 0 (A, B). So P 0 (A, B) = [ M 1/2 0, m 1/2 0 ]. (4) Similar. (b) Suppose [α k, α 1 ] [β k, β 1 ] φ. Then 0 P 0 (A, B) by Theorem 2.3. (1) (2) follow from Theorem 2.3. (3) Clearly P 0 (A, B) contains max α j β k j+1 k (α i + β k i+1 ) or its negative by Theorem 2.3. We were not able to obtain a, b, c, d. 4. Complex case Given complex skew symmetric matrices A, B C 2k 2k with singular values α 1 α 1 α k α k 0 β 1 β 1 β k β k 0 respectively. Consider the complex analog C (A, B) := UAU T + V BV T : U, V U(2k)}, D C (A, B) := det M : M C (A, B)} P C (A, B) := Pf M : M C (A, B)}. Theorem 4.1. Let A, B C 2k 2k be complex skew symmetric with singular values α 1 α 1 α 2 α 2 α k α k β 1 β 1 β 2 β 2 β k β k respectively. Then (1) D C (A, B) is an annulus of inner radius r outer radius R, where R = (α i + β k i+1 ) 2
14 T.Y. TAM AND M.C. THOMPSON r = k (α i β k i+1 ) 2 if [α k, α 1 ] [β k, β 1 ] = φ 0 if [α k, α 1 ] [β k, β 1 ] φ (2) P C (A, B) is an annulus of inner radius r outer radius R. Proof. Clearly D C (A, B) admits circular symmetry by the connectedness of U(n), D C (A, B) is an annulus. By Autonne s decomposition [1] (also see [10]) we may assume that A = diag (α 1,..., α k ) J, B = diag (β 1,..., β k ) J. The radii of D C (A, B) can be deduced from Theorem 2.5 [6, Theorem 1] since α s β s are singular values of A B respectively. P C (A, B) also admits circular symmetry since e iθ Pf (UAU T + V BV T ) = det(e iθ/2k I)Pf (UAU T + V BV T ) = Pf (e iθ/2k UA(e iθ/2k U) T + e iθ/2k V B(e iθ/2k V ) T ) using Pf (P AP T ) = (det P ) Pf A. We remark that circular symmetry does not exist if U(2k) is replaced by SU(2k). References [1] L. Autonne, Sur les matrices hypohermitiennes et sur les matrices unitaires, Ann. Univ. Lyon, Nouvelle Série I, 38 (1915), 1 77. [2] N. Bebiano, C.K. Li J. da Providência, Determinant of the sum of a symmetric a skew-symmetric matrix, SIAM J. Matrix Anal. Appl. 18 (1997), 74-82. [3] M. Fiedler, Bounds for the determinant of the sum of hermitian matrices, Proc. Amer. Math. Soc., 30 (1971), 27 31. [4] W. Fulton P. Pragacz, Schubert Varieties Degeneracy Loci, Lecture Notes in Mathematics 1689, Springer-Verlag, Berlin, 1998. [5] R.A. Horn C.R. Johnson, Matrix Analysis, Cambridge University Press, 1985. [6] C. K. Li R. Mathias, Determinant of the sum of two matrices, Bull. Austral. Math. Soc., 52 (1995), 425 429. [7] C.K. Li, Y.T. Poon N.S. Sze, Ranks determinants of the sum of matrices from unitary orbits, Linear Multilinear Algebra, 56 (2008), 105 130. [8] C.D. Godsil, Algebraic Combinatorics, CRC Press, 1993 [9] J.S. Lomont M.S. Cheema, Properties of Pfaffians, Rocky Mountain J. Math. 15 (1985), 493 512. [10] D. C. Youla, A normal form for a matrix under the unitary group, Canad. J. Math., 13 (1961), 694 704.
DETERMINANT, PFAFFIAN AND SKEW SYMMETRIC MATRICES 15 Department of Mathematics Statistics, Auburn University, AL 36849 5310, USA E-mail address: tamtiny@auburn.edu, mct0006@auburn.edu