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AN ABSTRACT OF THE THESIS OF ELIZABETH LINGFOON YIP for the DOCTOR OF PHILOSOPHY (Degree) (Name) in November 29, 1973 presented on MATHEMATICS (Date) (Major) Title: MATRICES CONJUNCTIVE WITH THEIR ADJOINTS Signature redacted for privacy. Abstract approved: C. S. Ballantine This paper studies necessary and sufficient conditions for a matrix to be conjunctive with its adjoint. The problem is solved completely in the usual complex case, in which it is shown that a matrix is conjunctive to its adjoint iff it is conjunctive to a real matrix. The problem is extended to pairs of fields where [4: 2 and characteristic 2. if a matrix is conjunctive to a matrix over , 6, It is shown that it is then conjunc- tive to its adjoint. To achieve this result, we first show a matrix over any field to its transpose. We is congruent over also show that it is sufficient to consider non-singular pencils by proving the uniqueness up to conjunctivity of the non-singular summand of the pencil H* XH + H and conjunctivity over where K* = K, ) X when and 1.1. are indeterminates over XFI + 1.1.K is decomposed (by into a direct sum of its minimum-indices part and a non-singular part. Matrices Conjunctive with Their Adjoints by Elizabeth Lingfoon Yip A THESIS submitted to Oregon State University in partial fulfillment of the requirements for the degree of Doctor of Philosophy Completed November 1973 Commencement June 1974 APPROVED: Signature redacted for privacy. Professor of Mathematics in charge of major Signature redacted for privacy. Cha. an of Department of Mathematics Signature redacted for privacy. Dean of Graduate School Date thesis is presented Typed by Clover Redfern for November 29, 1973 Elizabeth Lingfoon Yip ACKNOWLEDGMENT I would like to express my gratitude to Professor C. S. Ballantine for accepting me as an apprentice and for his generosity in giving me his time and advice during the preparation of this thesis, which, without his infinite patience, encouragement and understanding would have never been completed. TABLE OF CONTENTS page Chapter I. INTRODUCTION AND PRELIMINARY Introduction Uniqueness of the Non-Singular Core of *-Symmetric Pencil II. GENERAL CASE Congruency of a Matrix to Its Transpose General Results Non-Singular *-Symmetric Pencil with All Eigenvalues in III. USUAL COMPLEX CASE General Results Geometrical Approach to the 2 x 2 Usual Complex Case 1 1 7 17 17 28 34 59 59 65 BIBLIOGRAPHY 70 APPENDICES Appendix I: Pictures of Matrices Appendix II: Miscellaneous Results 71 71 73 MATRICES CONJUNCTIVE WITH THEIR ADJOINTS I. INTRODUCTION AND PRELIMINARIES 1. Introduction Let 3. (i.e., a subfield of index 2 in sion 2 over \6' be be a field of characteristic not equal to 2 and 8 is a vector space of dimen- Then there is a unique automorphism ). S whose fixed field is * of ,....; : (a+b)* = a* + b*, (ab)*= a*b*, and a* = a iff a we also have that conjugate of a. E * If E. Because J is involuntary: A has dimension 2 over E...., (a*)* = a. We call a* is a matrix over -.q.- , we denote by the A* the transpose of the conjugate of A. Then we have the following properties: (A*)* = A, (A+B)* = A* + B*, (AB)* = B*A*, (aA)* = a*A* when A+B S and AB are well defined and a E-7- We shall say is *-symmetric iff S = S*, and say S is *-congruent to a 2 matrix iff there exists a non-singular matrix T C*SC = T. We say a matrix S*; we say is of type 1 iff S is *-congruent to S is of type 2 iff S such that C is *-congruent to a matrix over S I. We also consider our matrix 7f, from an n-dimensional vector space over *-dual of If ( 11) * =iJ Then 5* . 11* into 11. tAr and V:, are being Elements of both Suppose In this context, <x, Sy> = <y,Sx>* for all x E 2f= 4 n x 1 <x, Sy> = <y, S*x>;* for all <x, f> = x*f. Then define E is *-symmetric iff S x,y E if. is *-congruent to S there exists a one-to-one, onto, linear transformation <Cx, SCy> = <x, Ty> for all such that is a basis for if, the matrix of S a subspace of lf if= renrti.. j=1 11,* = if* the is a linear transformation from written as column vectors over x, y E into f(x+y) = f(x) + f(y), f( ax) = a*f(x)1. We call 1)-* = {f: f as a linear transformation S then <e., Se.> x, y 1) e. Se. 1 If E is the C: {el, , we define 7.1..° = {x E If 26 is ,en}. x*11.= 0}. Suppose Define J (1)-1 e "112 e ...e Vj -1 (DV j+1 en} entry of (i, j) with respect to the basis {e1, , iff T 0-)r m-1 v m 3 ,.,, ,..* then V acts as a 4-dual of . LI . n v1r*. n 1-0 i. e. , j;= m03 ti j=1 If and J 3 J _ u, 3 n= and we partition the matrix k., / 1 i=1 Sll . S12 51p S21 s= S . Spl PP ... where is a k.1 x k. S.. 13 J If matrix, we write S = (S..). 13 P , M), and call C is a we write matrix such that C = M1 ED ... El) M , C = diag(M1, C a direct sum of M, ... ,M 1 We have the following facts which will be useful in the work following (we shall not prove those quoted from the literature): Fact 1. If S = (S..), C = diag(Ik, -Ik,Ik, ...), 13 then P ih P = (Pih) = CSC, Fact 2. P = C*SC , If and = (-1)i+hSih C = diag(Ik, jIk, Ik, where . ) j* = -i , then h are odd if i and = jSih if i is odd and = -jSih if i is even and h = -j2Sih if i and Pih = Sih h h is even is odd are even. and 4 The proofs of Fact 1 and Fact 2 are routine computations. Fact 3. where A is non-singular and 0 S=Ae0 If a zero matrix, and if B e 0 where is *-congruent to S then A non-singular and of the same dimension as A, *-congruent to Proof: Suppose C C11AC 1 I = B. C= C11 C12 (C is also non-singular. Thus singular. Suppose K C 21 accordingly. Thus 22) C11AC I I A and Since Suppose H and A B 0 B o 00 -1m-1 H-PI) 0. 0 are both non-singular, is *-congruent to B. are *-symmetric and K is non- (K-1H-PI) is nilpotent for some be a positive integer such that (K-1H-13I)m = 0, (K is is a non-singular matrix such that C*SC = C*(AeO)C = C11 is B B. C*SC = B e 0. Partition Thus is If we define an i nner product from Mal'cev [7], we can choose a basis for If pE 6. Let but (x, y) = x*Ky, then so that K-1H block diagonal with each diagonal block a Jordan canonical block of eigenvalue P, and the Gram matrix of the inner product is con- formably block diagonal with each diagonal block of the form is 5 1. 1 1 where E E Thus we have: s'6° Fact 4. Suppose indeterminants over S = XH + p.K, or , where both is non-singular, and X and p. are and K are H is nilpotent for *-symmetric, K some Then there exists for .11 a basis such that (the matrix of) (K-IH-13I) is block diagonal with each block of the form S X13+p.- where E E . We define En as the n x n matrix with diagonal and zero everywhere else, Fn I 's on the anti- as the matrix with l's on the first super-anti-diagonal and zero everywhere else. Also define Nn = EnFn. Fact 5. (K -1 H-P)(K -1 If H-P*) K and H are *-symmetric and is nilpotent with p p*, and if X, p. are 6 lf such indeterminates over; that XH + p.K is block diagonal with each block of the form: then there exists a basis for , z .X. X 1 XP+p. XP+p. :2. exp*i_p, xp*+11 Fact 6 [14 If symmetric over 6. E, , then the pencil H is symmetric and K is skew- and and X. p. are indeterminates over is congruent over XH + p.K ing form 0 A L E. 0 1 1 where z 0 A L 0 e (XH 0+p.K 0 ), to the follow- 7 is called the minimum-indices part of the pencil L = E and L IL XH + p.K, and = i 1 det(XH0 +1.1.K0 is a non-zero polynomial. ) Fact 7. Witt's Theorem [9]. A, B1, B2 If *-symmetric non-singular matrices over s-4 implies *-congruent to A e B2 is B1 are A e Bl' *-congruent to B2. then Uniqueness of the Non-Singular Core of a *-Symmetric Pencil 2. Theorem I. 2. 1. pencil XH + pa, Let H where is *-congruent to L M, and and X K be *-symmetric. Then the are indeterminates over p. where 0 L=0 Lei 1:13 i=1 LE L is the conjugate-transpose of is as defined in Fact 6, LEi i and M = LEi, 0 XH0 + ill<0 is *-congruent also to is a non-singular pencil. If XH + 1.1.K L Ef) N where N is a non-singular pencil, 8 then M is *-congruent to N. Turnbull proved that a *-symmetric pencil can always be reduced to the form L where is the minimum-indices L part and M is a non-singular pencil [11]. In this section we shall only prove the uniqueness (up to *-congruency) of the non-singular summand M. Before we proceed with the proof of the theorem, we shall discuss some of the properties of the pencil Suppose 14) is a subspace of 7.f transformations from , L IFD M. S, T are linear and V*. For our discussion in this sec- to tion, we need to define, for 24 c 7J (s-1T)2L= {x lf:SX alt}) E Ell:SX and, inductively, define for = 2,3,... i E T(S-1T)i-1210}, E (s-1T)°U . Also note that if we let A H + I.LK be a pencil of *-symmetric matrices, considered as the matrix of a pencil of bilinear forms, then, for each co-ordinate subspace, the matrix of this pencil of forms restricted to this subspace is the corresponding principal submatrix of A, Lemma 1. If(j = 1, 2, and let Let and conversely. be the *-dual of be subspaces of k) If , let (11 such that if= tOk =be the corresponding *-dual direct sum =1 j defined in Section 1 of this chapter. If S and T are linear 9 transformations such that S, T: TIT C 3 /I* i _<h-i. J [lirm(S-1Thf.]. 3 J The proof proceeds by induction on i = h. It is enough to show Consider ,r -1 r T)h.V.) T)h-LiCe3(U.r(S J k k i=1 x- and s V. C V3*, the lemma is obviously true. Suppose it is true for i = 0, -1 ® j=1 3 Proof of Lemma 1. If k (S-1T)i/l= then , . V*, x., Suppose (s with EV.. Then x (S-1T)11, and k 3 3 xE Sx = Sx. T. E (S-1T)h-111 3 j=1 j=1 By our induction hypothesis, T(63V. (--(S-1T1/ 111-1.) T(S-1T)1-1-11). 3 T(V. rm(S-1T)h-hr) = 3=1 Thus Sx )Sx. = yi Therefore (.174( ( J J J k Thus xE 43) if i=1 Lemma 2. K = K* (Th Ty. and Sx. = Ty. E T(S-1T)h-1 3 jVel j=1 j=1 where Ty. ?I . 71f. But Sx. J 3 SO X.VE J . rm ::'. 3 (S-1-T)h7f. J 3 3 (S - 1 T )h 1 ..* q. e. d. 3 i Let (as usual) characteristic be linear maps of if into If . Let i 2, H = H* and X, p. be 10 indeterminates over to co M00f$ M 0 L Suppose . where M1, has no elementary divisors, L has elementary divisors only of the form elementary divisors only of the form divisors neither of the form M = Moo EH M Let U. and y of *-bilinear forms A and x*(XH 0 e Mo ED M1, to the and A to the pencil of is *-congruent over 0 ED M Without loss of generality, we may Proof of Lemma 2: =L 2/ 1110y restricted to 26 is *-congruent over *-bilinear forms whose matrix is as sume XH + p.q has elementary 745.= (H-1K)ng. (Here n = dim pencil of *-bilinear forms whose matrix is restricted to M1 has M0 is determined as follows: as usual.) Then the pencil x and Xcl, nor of the form Xq (K-1H)nif = and is a non-singular pencil. Then the *-congruency M1 0 type of M over g. with is *-congruent over X.1-1 + iK Mec Mo 0 i=1 L 0 k' L= and M1 63 L'E. )1/4 0 1 Let 11= Irooe Here denote the co-ordinate subspaces corresponding to respectively. Denote by = span {x1, L, Moo, Mo, M1, 'xzrnfl }, m > 0, one of 11 the co-ordinate subspaces corresponding to one of the direct summands in L, Let i(R of where x = span{x 1 , thus XH + 1.1.K ,5Z2m+1} be the corresponding subspace of acts as a *-dual of 6 restricted to (;'i co-ordinate vector. jth denotes the Note that the matrix . is m+1 IL X 11, X P. P. m+1 For simplicity in notation we drop the subscript Hx1 = xm+2 Kx1 Hx 2.= x m+3 Kx2 Hx x m Hxm+1 = Hx m+2 Kx 2m+1 0 , =x l , P. Hx2m = xm-1 Hx2 m+1 = x m in =0 = . xm+2 x m+3 Kxm+1 = X2m+1 Hxm+3 7 x2 A 3 i Kx Kx m+2 -x2 A m+3 x3 Kx2m =xm Kx 2m+1 X m+1 Thus 12 Thus _ = span {x1, x2, . (K1 H) span {xi x2, grm (K-1H)2R orm -1H)rn+i . (This oo mapping on V-00. ; K-1H 7f (- so f" oo E - 1H)V0 Then x E (K Thus li 0 -1 = H)k if 0. l(../r Suppose iff 0 (Th (K 1 Mil Li-0 = Kx E is 0- H is a nilpotent K1 and (K-1H)n iloo = 0. Thus (K-1H)1/C K-1Y-*= {x:Kx if*1 0 0 0 (K for all k > m+1 hence K LLoo = 15*co and is non-singular on 1/0. Thus, Thus x2m-1} here.) m+1 is non-singular on 1j."co C /f* ' restricted to 0'? rm(K-1H)110';) A matrix is of order 0 x2m} - span {x1, x2, ... ,xm+1} 0:?rm(K-1H)ko-,z Thus the matrix of , = H 7).0" = ,* () 0. 170. Ifo, since K( /f)C 0 (K-1H)1 270 = Vo Elm-le-1)Vr0 for i <k-1 .... - 0Vr0 - -)Lir0 . . 1- TO. and H restricted to -1X1 are non-singular. Thus (K-lipnLi)r Therefore 26= (K-1H)11Y, x*Ay x*(XH+p.K)y, with x, y restricted to is *-congruent to a pencil of *-bilinear forms whose matrix is 0 ® M0 M1- Similarly, we can show that x*Ay, with x, y restricted to 13 741-= (F1-1K)nV is *-congruent to a pencil of *-bilinear forms whose matrix is 0 ® M 00ED M1. 1 Write Proof of Theorem 1.2. 1. =A, L e M = XH + L ED N = XH1 + F.LK1 = Al . Without loss of generality, we may assume M = Moo]. El) mo1 e mu N = M002 0 MO2 ED M12, where Moo 1 and Mon2 have elementary divisors only of the form flq, divisors only of the form Xq, and M01 and M11 have elementary MO2 and M12 nor of the form elementary divisors of the form recall that M and N are non-singular pencils. singular matrix over map of then 21 such that into and C*FIC = H1 By Lemma 2, (K-1H)n U. and u*(0 ED (we also consider , x*Ay, and u,v E (K-1H)nir and C and is a non- as a non-singular C C *AC = C*(L and x*Aiy, M)C = L (14 N, with x, y restricted to y respectively, is equal to u*(0 ®M )n?r, (K-1H1 1 ED M12)v, respectively, with respectively. We want to show by induction on i -1 - 1H) ic 11. [(C *KC) -1(C *HC)]7f= X.q, C*KC = Kl. (K1-1H1)n MO1 ED M11)v If have no that 14 i = 1: For xE [(C*KC)-1C*HC)]?,7 iff C*KCX E C*14C iff KCx E iff Cx K 114C 2.7 iff xE C-1(K 1H)C HCV Suppose [(c*Kc)-1(c*Hc)]k-12i. Then 1(C*FIC)PV X E [(C*I<C) iff (C*KC)x E C*1-1C[(C*KC)-1(C*FIC)P-12/ iff KCx E FIC[(C*KC)-1(C*HC)11C-1Y. (by our induction hypothesis) iff = HCC-1(K-1H)k-101- = H(K-1H)1(-1Clr Cx E K iff XE -1 H(K-1 H)k-1 C v C-1(K-1H)kC Letbe a'xm} basis for (K-1H)91 Then Lx1' x2' {C x1,C x ,C xm} is a basis for C(K-1H)ny' C-1(K-1 -1 = [(C*KC) n (C*HC1f1 15 Therefore 'x.) x.)*A (C (C _1 A C- x. x.) = x. (C*) 1 1 - * = x. (C*)1 C*ACC 7 lx = x. Ax.3 . 1 Thus x*Ay with ° IT' 3, E is *-congruent to Hi) Thus is *-congruent to 0 El) MO2 e M12. Thus by Fact /) Ml l MO1 -1 with u 1(OA) MO2 ® M U x,y restricted to M01* M11 is *-congruent to M.. = XH.. +i = 0,1, MO2 j = 1,2, 13 e M12. Let where and H.. *-symmetric. Then C1(H01 ED H11)C1 = H02 Cl(K01 A) H12 K11)C1 = K02 ED K12 Thus -1 -1 -1 C1 (H01 K01 H11K11)C1 = [C 1(H01 ® H11)C11 -1 H02 K02 Partition - C1 B.. = H..1K... 13 13 1j * [C1 (K01 K11 )C1] -1 e H12 K12 . accordingly: Then -1 C1 = C11 C12 C21 C22 ( ) . Let K..13 are 16 = C1(B02 ® B12) (B01 e B11)C - _ B01C12 B01C11 IMli C11B02 C12B12 C21B02 C22B22 = B11C22 B11C21 _ - B11 C21 = C21 B02 . Since [12] B11 and B02 Similarly, have no eigenvalues in common, C11M01C11 - MO2' * C22M11C22 = M12 . thus C11 and and M01, MO2, M11 C22 0 * Thus C21 = 0. C12 = are non-singular. are non-singular, M12 So M01 is *-congruent MO2 andM11 is *-congruent to M12. Likewise, L ® M restricted to (H - le?)-- is *-congruent to 0 0 M001 0 Mll to L ED N )n-lf 1 (FI-1K1 restricted to o e Moo2 e M12, where Moo 1 is *-congruent to and M002 have elementary divisors of the form ilq. With an approach similar to that of our consideration on Mc02. ?/../ , we can show that M oo 1 Since MM = O1 ® A4 .01 e m 11 N = MO2 ED M.02 ED M12 we have that M is *-congruent to N. q.e.d. is *-congruent to 17 II. GENERAL CASE 1. Congruency of a Matrix to Its Transpose If p. H and K are over H = H', K = K', are indeterminates over E then , XH + 1.1.1c , and X and is congruent to the following: L where L M only of the form Xq 00 M1 is described in Fact 6 of Chapter I, divisors only of the form form M 0 nor , M1 and p.q, and Xq, Moo M0 has elementary has elementary divisors has neither elementary divisors of the M0 eM00 M1 is non-singular. In this section, we shall provet Theorem II. 1.1. Every matrix over E is congruent over to its transpose. Lemma 3. Let ?./ be finite dimensional over always) characteristic g K: 'if' -.f be skew, and 2. Let H: B = KH: with (as 71' be symmetric, be non-singular. Let q(t) be irreducible in [t] polynomial for VT(Here R1 is the (ordinary) dual of Y.) B on and let q(t2)ni be the minimum Then there is a maximum B-cyclic subspace 21 of 21 such that 18 lLm (H ni = 0. U)0 Proof of Lemma: Let thus there exists e H1 = Hq(B2)m-1. e'Hle 1 0. Let such that E f(t) [t], E ,-,(H it (f(B)e)tH q(t2)4(t). Let r(t) it is sufficient to show that for = 0, f(t) = [q(t2)]ip(t) Write = 0. g. c. d (p(t), p(-t)). and some k > 0, some s(t) E g [t], q(t2) tn(t2). r(t) for some / >0 m>0 and some h(t) E E [t] m(t) = n(t2). Thus SO which in turn implies . n(t) = 1. Thus r(-B) Thus so 1 or B E t*s(t2). We 6[t], such that q(t2)= t2mh(t2) for some Thus til,'h(t2 ). is non-singular which implies q(t2)1 p(t), k = 0, n(t) q(t), n(t) = q(t), n(t) = q(t). If then which is a contradiction. is non-singular, so r(-B)2t =2t , for m(t) = 1. Since g. c. d (r(-t), q(t2)) = m( -t) = 1. r(t) = a(t)p(t) + b(t)p(t), 0 = (f(B)e)TI-fit n(t) such that n(t2) g(t2), n(t) n(t2) = m(t)Ir(t), q(t2) and since q(t) = tmh(t). But q(0) 1 0, tks(t2), m(t) = g. c. d (r(t), q(t2 )) It follows that Celq(t2). Thus where is even or odd, then is even and m(t) = tn(t2) r(t) Then such that claim that g. c. d (r(t), q(t2)) = 1. Let Since be the = 0 => q(t2)11.1 f(t). (f(Ble) TH Suppose )0 ?./..." 0, 21 is maximum B-cyclic subspace generated by e. To show that cyclic and that HI1 = H1 Then for some a(t), b(t) which implies for all E i > 0, . Let [t]. Now, that 19 0 = (f(B)e)11-1Bie [II. 1. 2] Since = [q(B2)ep-(B)ePHB1e K = HB and H' = H, B'H = (HB) ' = K' = -K = -HB. Induc- tively, we can show HBi = ( Thus, -Bi) 'H. [II. 1.2] = ((-B)ie)'H[q(B2re p( -B)e] = [((-B)ie)1H[q(B2)113(-13)e111 = [q(B2) ep(-B)ePH(-B)ie . Thus [q(B2)ep(-B)ePH 11 = 0 = [q(B2)1p(B)ePH 2L, SO 0 = [q(B2)ep(-B)e1'H ../Z D [q(B2)1p(-B)e]'Hb(-B)7t = [q(B2)ep(-B)b(B)ePH21.- and o= [q(B2)p(B)epa. D [q(B2)e p(B)e]'Ha(-B)a = [q(132rep(B)a(B)ePH ??.., Thus 0 = [q(B2)e[p(B)a(B)+p(-B)b(B)]ePHit = [q(B2)1r(B)ePH/L = [q(B2)1 e] 1Hr( -B)1 = [q(B2)1e]'H2t 20 [q(B2)1e]' H q(B2)ie e'Hq(B)m-le = for all e'Hq(B2)1fie = 0 >m. Thus q(t2)mlf(t). 0, q. e. d. Without loss of generality we may Proof of Theorem 11.1.1. consider the pencil [II. 1. 1] i > m. Since L Mo KB M described in the M1 first paragraph of this section; and we replace the indeterminates X., with 1. 11 We shall show separately L, Mo, Moo, and M1 are congruent over Con L, 8. to their respective transposes. L. For each non-zero block (of order 2c.+ multiply on the right and left by IEi+1 6 E E. . of ) lsider Thus in L, the E. first column becomes the last column, the second column becomes A the second to the last column and etc. ; in L the first row , Ei becomes the last row, and the second row becomes the second to the last row and etc. Thus, we have for each non zero block of L z o [II. 1.3] 2 E. 1 where ii 1 'el 1 Z 4. E. = 1 1 1 1 61 Multiply [II. 1.3] on the right and left by diag(1, -1,1, -1, , 1); 21 we get the transpose of [11.1.3]. Thus Next consider corresponding to the direct summand M0 S=H+K=L ED Mx *Fa 0 is congruent to L'. be the co-ordinate subspace Let Mo. L in H is non-singular and B = (H-1K) e)N41. is nilpotent on Vo. Let m be a positive integer such that B rn Br11-1 = 0, and We shall consider the cases m even and 0. m odd separately. We shall prove the following: is odd, there exists a maximum B-cyclic subspace m if k(Th (HiL)° = 0; ?I. such that if m is even, there exist two maximum B-cyclic sub- U and 24j such that spaces 71/ 0= )° = riff (H2L)° =(1 = Suppose m is odd. eE ir such that (HBm-1)1 e'HBrn-le 0. space generated by e. Suppose p(t) E nomial of p(-B)W. Thus such that E,,[t] B = on M.. Thus t.i(p(t). p(-B) (BW))° 74jc (H26) °. and 11. C (HU )°, (H (. ) = HBm-1 There exists Let W./ be the B-cyclic sub(131p(B)e)lHa..= 0, Since tm where is the minimum poly- is non-singular and hence 0 = (131p(B)e)'HU, = (Ble)'Hp(-B)2t= (Ble)1H2L > m. We therefore have completed the proof of (i). The following discussion is not needed in the proofs of (i) and 22 (ii), but useful for later purposes. We note that B UC _ U. because is B-cyclic. 21.... B(H M )o C (Hit)o because if which implies x'Plit. D x'FIBU, = -(B,o'HU., m-1 a e ; we can choose al ' a2 ' Let e1 =e I 0 x Bx , ,-- i.B * i=1 k <m-Z. Let for e'1HBke1 = 0 2.1.,1 * . (H76)°, then E E (H46)0. so that ' a m-1 be the B-cyclic subspace generated by el. 21.1 has all the properties proved for 21.. restricted to 21.1 is the Jordan canonical block of eigenvalue H restricted to is a matrix ./4,1 H0 K0 0, with non-zero elements on the anti-diagonal and zero everywhere else. Thus restricted to 261 is a matrix B K = HB with non-zero entries on the first super-anti-diagonal and zero everywhere else. Multiply on the right and left by diag(1, -1, ... , H0 + K0 1). Since m is odd, by Fact 1, we have multiplied the first super-anti-diagonal by and have not changed the anti-diagonal. Thus we obtain the -1, transpose of Ho + Ko. In the next paragraph, we shall prove (ii). Suppose m is even; since HBm-1 i 0 there exist e,f Y" such that E e1l-IBm-1f e'HBITI-lf = 1. Let U. f and If xEUt--N Zti E E [t] , e, f so that Zti be the B-cyclic subspaces generated by respectively. First, we want to show e p(t),q(t) , we can choose 0; then x = Bkp(B)e such that 0(0), = B1q(B)f, Orq(t). 7/4. (ThW = 0. for some Without loss of generality, 23 assume k If / > m, then x = O. Therefore assume I < m. Note, since to(t), tirq(t), p(B) and q(B) are non-singular, r(B) = p(B)q(B)-1 is non-singular and we can even take r(t) E 6[t] here, and r(-B) is also non-singular. x = Bkp(B)e = Blq(B)f and k >/. Therefore r 0 = Bi Lq(B)f-Bk-/p(B)e] = q(B)Bi (f -Bkr(B)e) 0 = B (f-Bk-/r(B)e) since is non-singular. 131(f-Bk-Ir(B)e) q(B) 0= = If implies =0 &FIBm-1(f-Bk-/r(B)e) etHBm-1f - e'HBm-1+k4r(B)e . k >1+1, &FIBr(B)e = e'FIBmBk-4-1 r(B)e = 0; 0 = &FIBm-1f e'HBIn-lr(B)e because m-1 which is a contradiction. If k = e'HBril-lr(0)e = = = -eHB e'HB111-lf , is odd (and hence = 0 = e,HBm-lf eIm-le HB which impliex thus HBm-1 is skew). Thus which is a contradi ction. Thus x = 0. Thus r1,6 , 2ii m- ' >m, = O. Secondly, we want to show ?A-(--) (H741)0 = O. Suppose (Bkp(B)e)11-1111= 0 p(B) and p(-B) for some p(t) E 6 [t] such that are non-singular, so p(-B)t, = 0(13(t). Thus 24 0 = (Bkp(B)e)II-124).= (Bke')Hp(-B)241 = eiH(-B) e'H(-B)k+ilif for all i > k >m Thus 7A. Thus 7)7, (H)0 (H Zii)° = 0. Likewise we can show that = 0. Thirdly, we want to show If xE ( where .1103, W) rTh assume (Hlle) W))°, [t] p(t) , q(t) E > k. Suppose 0= 11.1) and t then r (FT( U et ?Ai))° = o. x = Bkp(B)e + Blq(B)f p(t),q(t). Without loss of generality, k <m-i. Then [Bkp(B)e+Blq(B)f]'H(Zi e141) 03W) [Bkp(B)e+Blq(B)f]tH(-B)m-1-k( = [Bm-1p(B)e+Bl+m-1-kq(B)f]'H(72. fdAj) [Bm-1 p(B)e+B = [Bni-lp(B)el'H7f1+ [131+m-l-kq(B)f]TH241 p(0)[Bm- 1 ePHf + [B If 1>k, then B +m -1 -k m- 1 - k = 0. If , ,B)f clkJ'Hf. = k, [Bm-lq(B)fillif [B/ +m- 1 -1(q(B)fi = q(0)[Bm-lf]'Hf. 25 (-1)111-1VHBm-if. Since = is skew. Thus Both fIFIBm-1f = 0. is odd, then HBm-1 m-1 p(0) >m-1. Thus nonzero. Thus we have a contradiction. Therefore we have shown (U el()) ,Th are (Bm-1e)Ilif and [H(11 e 145)]0 = o. Fourthly, let p(t) = and let a.t2j j>0 f = f + Bp(B)e = f+ .B )e j> 0 be the B-cyclic subspace generated by f. We can choose Let a ,a 0 1 such that C H fa°- -A 3 3 j>0 j>0 = Thus we can solve for VI-1Bm-2f a 0 . a.B2j+le) a.B2j+le),HBm-2(f + 0 = fIFIBm 2f = (f+ - 2a0e'HBm-1f, If we consider f'HB m-4A we will f = 0, get an expression in terms of f'HBm-4f, a0, f'HBm-3e, e'HBm-2f and and solve for e'HBm-lf, by considering f'HB m-2(k+1)A f. al. Likewise, we can solve for ak A In like manner, let e = e + Bq(B)f A where q(t) = p j>0 2.tj. We can pick rf01 . . so that (2/6 , the 26 satisfies the property H B-cyclic subspace generated by A Zt, and kr obviously satisfy the first 3 properties we proved for Thus we have completed the proof of (ii). Further, 21- and A let f = f + fJ a a2,a3, ... we can choose j.B -If so that A A for e HBkA f = 0 j-2 A, k < m. Let 7/tY be the cyclic subspace generated A by f; then the pair 7.6 also satisfies the properties we have , A It proved for the pair B restricted to 'IL 63, w is the direct sum of 2 m-dimensional Jordan blocks of eigenvalues 0. A H, restricted to Ho = 1.) ® 0 -S [S 0 , is where S= as. A and A A K, restricted to 'ILEB"lif , K0, is NM, Ko = r-T where T= o 0 0 .0 Multiply by r I -H0 -K0 . diag(1, -1, Since , -1) I the right and left of Ho + Ko we get 0 H0 is of even dimension, if we multiply by on the right and left of the anti-diagonal has been multiplied by -1, -H0 -K0' then by Fact 1, and the first 27 super-anti-diagonal is not changed. Thus we get In case m is odd, V--0 ZIED (Hit )°, = H(HU.)° C /C), KU, = HBll C 0 ),0 ; 1, 0 ), 0 K(Htt..) = HB(H GC) C H(Hit) C 1-10-K0. ITO = H (Es and process with 0 thus we can repeat the above -)/ Likewise, for m even, (H)°. replaced by we can repeat the above process with replaced by 0 j)))°. Thus, after a finite number of steps, we show that (H( U is congruent to its transpose. 0 The process of showing M to in this case, instead of HK, we consider Now we consider N/Icc K-1H. (and thus we assume H and MI non-singular here). Suppose 1 I K.i = 2(N.-N!), +NI), that if H.i = (Ni i i i 2 1 of Note that K(K1 H)K-1 = -HK nomial of K-1H - xolynomial = are ) f.(x) and i i=1 such N is the minimum is prime to f.(x) for j. so the minimum poly- -(K-1H)', is even; thus we place another restriction on the f.(x); f.( p f.( then p K is a non-singular matrix such that C2 C'2 M1C 2 = M. Without loss of generality, write M = K.-1H., is similar; = xolynomials [q.(x2dm where q(x) is irreducible x)rq(x)). Without loss of generality, we may assume (and M1 = HI + K1. Lemma 3 (proved in this section) assures us that we can find a basis for 1 (Mil+M'il) _Jr , such that and K il 1 M1 = P el i=1 M.1, il' if il -M' ) then Kil-1 Hu. is similar to a companion matrix. Thus it suffices to prove the following: H H. = 2 = 2 (M 28 Lemma 4. and K as defined above are non-singular, H Let m be the degree of the minimum polynomial of Proof. Thus, for some A = K-1H. U. basis for is congruent to S'. is non-derogatory, then S = H + K K-1H and If Write ,Am le} is a e E V--, {e, Ae, A2e, e. = Ai-le. Recall is skew when KAi i 3 is even, is symmetric when i e!Ke. = (-1)je1KAi+j-Ze1 = 0 for i+j even, for i+ odd. Let 3 is odd. Thus 1 e!He. = (-1)je1HAi+j-2e1 = 0 3 1 Ji C: 11-- 2r such that Ce. = (-1)i-1e.. Then e (1.C'FICe. Ce.)'H(Ce.) = 1 because if i+j 1 3 is odd, (-1)1+3e!He.1 3 = 3 1 e!He.1 = 0. Also 3 = -e.'Ke. el.C1KCe. = (Ce.)1K(Ce.)1 = (-1)i+je!Ke. 3 because if 2. 1 1 3 i+j is even, e!Ke. = 0. Thus 3 3 C 'SC = S 1. q. e. d. 1 General Results In this section, we shall prove some simple results, some of which lay the ground work for the following section. Proposition 11.2.1. Let S be a non-singular matrix; then: 29 if is of type 1, then det 5* = ±det S. S if if is of type 2, S S E. det then is of type 2, then is *-congruent to 5, where S is the conjugate matrix of S. if Proof. S (i) If (C*)2 S*(C2 ) = S*. If is of type 2, then it is of type 1. Therefore C*SC = T det S = (det T)(det C If then C*SC = S*, -1 S is over )(det C*SC = T det(CC*) )*, C1 so we have C*S*C = S, Thus det S* = ±1. det S. then , which is in is over , C*S*C = T*= T' = C'S'C. Therefore then (CI) C*S*C(C) -1 = 51. (C1)-1C4SC(E)-1 = (S')*= If C*SC = T 6, is over and from (iii) of Proposition 11.2.1, S is *-congruent to Proposition II. 2. 2. where j* = -j, and let is similar to A If If K H T, from Theorem (hence *-congruent over I. 2.1, is congruent over then then S to ) is *-congruent to 5; (5)' = 5*. Let A= (A H= 1 (5-S*), and let K 1 2j 51S*. Then -1 )* is non-singular, is non-singular, B = K-1H B1 = H-IK is similar to B*. is similar to B. TI, 30 If is of type I (hence also if S is of type 2), S is A similar to A*. If is of type 1, i.e., S singular, and is non-singular, then C-1K K Proof. (i) SAS' KBK -1 - S(S-1 S*)S1 = K(K-1H)K-1 = = If C -1 is also similar to C*HC = H = = -K -1 (A*) 1, -1 -1 1-1HC H. = B.* 1 (C*S*C) = (S*)-1S thus A = then A-1. Since A is similar to A*. is non-singular such that C*SC = S*, C - (A*)'. is non-singular such that C*SC = S*, C AC = C -1(S -1S*)C = (C*SC) If - S*S1 is non- C HK-1 = B*. -1-1 = H(HK)H= KH = -1 HB1H where C*SC = S* then and C*KC = -K. Thus ,-1 C-1(K-1H)C = [C*KCJ (C*HC) = -K-1H Proposition II. 2. 3. S=S0 OS co A oo = S'S *co oo (1) S 1 where (i) -1 * A0 = S0 S 0 has eigenvalues only at eigenvalues neither at I nor at -1. matrix such that C*SC = S*, o C is such that CSC 000 = S, CI Suppose is such that CISIC, = Si S q. e. d. is non-singular, and has eigenvalues only at -1, If and C - * 1, has Al = S11 S1 is a non-singular where Co 1, C=C0 f9C co is such that C00S0000 Coo = S oo and then , 31 (ii) Suppose H = singular. Let 1 K-1H B 1 where the B., (f) j=1 1 some qi(t) E e[t], = and g. c. d (f.(t), f.(t)) j are such that if B.1 f.(t) = qi(t2) then is the minimum polynomial of B., f.(t) and non- and K = -2- (S-s*) (S+S*) iij Suppose when 1 for 1 S = ED S. . 1=1 where if H. 1 =12 (S.+S.) 1 1 1 K. = --(S.-S. ) 2j then B. = K.1H. 1 1 If . 1 C*SC = S*, then C is a non-singular matrix such that C = e C. such that C.1 S.C.1 = S. . 1 1 Proof. (i) Since C' AC AC = C -1(S = [C*SC]i[C*SC] = (A*)-1 AC = C(A*) Partition C = (Cih) - 1S *)C conformably to A. -1 . i = 0,00,1 and h = 0, 00, 1. Then AC = C(A*)-1 A. and (A*) - I implies *- Ai Cih = Cih (A, )1 If i h, have no eigenvalues in common and thus C ih = 0. 32 We have shown that the off-diagonal blocks of C are Thus (i) 0. is proved. (ii) Since C-1BC = (C *KC)-1(C*HC) = -B Thus BC = -CB. Partition B.C.. = -C..B., i,j 1.3 13 C 1, 2, ... ,p. conformably to B. Thus j, B. and If i then -B. J = 0. We have shown that have no eigenvalues in common, thus the off-diagonal blocks of C are 0, thus we have proved (ii). The next result is based on Taussky-Zassenhaus' result [10]. Proposition 11.1.4. If S is non-singular and S-15* is non-derogatory, and C*SC = S* for C is *-congruent to a symmetric matrix over non-singular, then .......4 C*S . Proof: First assume A = S-15* is a companion matrix. is similar to A* hence similar to a is of type 1 therefore A matrix over since matrix over [2]; ."-8 A is a companion matrix, A is a 33 Since C*S(A)S-1(C*)-1 C*S(S-1S*)S-1(C*)-1 = -1 C *(S ) (C *) = (C*S*C)(C*SC)-1 = S(S*)-1 = A* = A' . Since A is a companion matrix, by the Taussky-Zassenhaus result, is symmetric. C*S Now suppose A = SS* is not a companion matrix, but is non-derogatory. There exists A0 D such that D-1AD = Ao, where is a companion matrix. Let T = D*SD. Then S = (D*) - 1 TD-1, 5* = (D*) -1 T *D- 1 . C*SC = C*(D*)-1TD-1C = S* = Thus (D*)-1T*D-1 (D1 CD)*TD-1 CD = T*. Note T-1T*= D-15-1(D*)-1D*S*D = D-1AD = A0 is a companion matrix. From what we have shown for the case S-15* being a companion matrix, [D-1CD]*T A is a symmetric 34 matrix over [D-1CD]*T (D*)C*(D*)-1D*SD = D*C*SD Thus 3. is *-congruent to a symmetric matrix. C*S Non-Singular *-Symmetric Pencils with All Eigenvalues in Let and K be *-symmetric. Then Theorem I. 2. 1 H asserts that the pencil XH + pa, where is *-congruent to terminates over and X L 1.1. are inde- M where is the L minimum-indices part as defined in Theorem I. 2. 1 and M is the non-singular core, unique up to *-congruency. If T = H1+ jKl, where matrix over the matrix pencil and H1 K1 then H1- jK1 T*=T1 = HI1 +jK 1 Thus proving that S = H + jK is of type 2 is equivalent to proving that the XH + H.K is *-congruent over * Hi = Hi = Hi , jKi = -(jKi) are *-symmetric and j* = -j , and jKi is a E *-congruent to -K1 = .t K' = K,i 1 e. to X.H1 +1 where rxwl. and H1= H1 We say a pencil is of type 2 if XH + pa X.H1 +1 where H1 above. We say a pencil is of type 1 if XH - pa. First we want to show that: and XH + K1 is are as described is *-congruent to 35 where H and K are Proposition II. 3. 1. If S = H + jK *-symmetric and j*= -j, core of the pencil then is of type 2. XH + p.K M where XH + i.K = L Proof: Write is of type 2 iff the non-singular S is the minimum- L indices part and M is the non-singular core. ("only if") *-congruent to If XH + p.K XH1 + p-K1 . 1K (51K1)* = -3- thus Ko = j -1K = * -KI1 = K1 = k to K 1 Let is a matrix over 3-1K is XH + p.K H1 = H1 = H1 where is congruent over XH1 + p.K0 . is of type 2, then L0 41) M1 where L 0 1 L0 = 0 13) A i= I L 0 1. A are as defined in Fact 6 and M1 = XH1 + p.K0 is a E. Lei A A A A and HI1 = H1, non-singular pencil and HI, KO are over 1 and L 1 A A (Ko)' = -K0. A Thus A L 1e X.H1 + iljKo, kl-11 + 1.1K1 = XHi + jp.K0 where kl 1 where WEi = 0 ED L i=1 kV* 0 is *-congruent to 36 E. 1 W 0 i E Because of the uniqueness of the minimum indices, W 0 Li *-congruent to * - Since 0 kH1 + pK1 and 1 * is 0 E. 1 kH + .1.1K have LEi Ll is *-congruent to L [11]. (Recall in the beginning of the proof, we assume kH + 1.1K = L Eli M, the same minimum indices, where L is the minimum-indices part and M is the non-singular core.) Thus kH + i.LK is *-congruent to *-congruent to L A = jKl; kH1 + (jKI)* = H1 = HI = H1, A L1 A (j11)* = A )* = H?1 = H1. (H1 A* A L EDkH1 + FIKl A where p.K1 in turn is which in turn is *-congruent to A A singular core of pencils before + Ki = jKo Thus kH + iK= L + pjKo. Let *-congruent to .A kH1 + 1.1.3K0, .A kH1 A 5K1 = A.2A, -jKi = --J(JK0)=,' = J K o - is By the uniqueness of the non.2 A Ko A A Hal. We said and is a matrix over is *-congruent to M M kH1 + A -J.2AKo - -3K1 Thus M is of type 2. (UHT?) We only need to prove the minimum-indices part L of 37 the pencil is of type 2. Therefore without loss of generality, assume A = X.H + 1.1.1C IMO where L E. is defined in Fact 6 of Section 1 of Chapter I. Multiply on the right and left of A by By a calculation similar to I EEi. a calculation appearing in the proof of Theorem II. 1. 1, we can see that A becomes [iii. 1. 1 A2 where Multiply on the right of [III. 1. 1] by C1 = diag(1,j, ,j, 1) and the left by Cl. Since [III. 1. 1] is of odd order, by Fact 2 of Section 1, Chapter I, we have multiplied the entries on the anti-diagonal 38 alternately by j alternately by that E {1 , 1 and -j and -3 .2 Thus . 6 is a matrix over H2 entries are E {j, -j} is such C 1A2C1 = XH2 + H.K2 because its non-zero entries are E because is a matrix over and jK2 -j2} and the entries on the anti-diagonal K2 Is non-zero q. e. d. . In this section, we shall consider 5 = H + jK where H, K are *-symmetric and j*= -j, such that the pencil XH + p.K is non-singular, and such that all the eigenvalues of the pencil are in (Note that 5-1S* = is non-singular, (H+jK)-1(H-jK) If H then [H(I+jF1-1K)I1H(I-jH-1K) S-1S* having eigenvalue at Therefore being nilpotent. Likewise, equivalent to Suppose and and - HI1 K1 K. I -1(I-jFI-1K) = is equivalent to H-1K 1 having eigenvalues at 5-15* -1 is K-1H being nilpotent. ) S= P ED i= 1 S. and Hi = * 1 * 1 and Ki = Ti (Si-Si) (Si+Si) 1 is nilpotent, - is nilpotent, and for i > 3, K21 H2 are non-singular, and K.-1H. 1 1 is similar to -K-1.IH.1 H. 1 and ,, has eigenvalues only at Pi, the minimum polynomial of 13i -13. , 1 K. 1H 1 1 and for R "i ' i > 3, an if f.( i td ) is then g. c. d. S (f.(t),f.(t)) j = 1 for i i j. Then Proposition II. 2.3 asserts that is Of type 1 iff each Of the S. is of type 1. Therefore, without loss 1 1 39 of generality, we consider the following cases separately: K-1H has eigenvalue only at is similar to K-1H and p ±3, K-1H and is similar to is similar to H-IK where p, 1, -R and P = P -I3 -K-1H. K-1H has eigenvalues only at and K-1H -p -K-1H. K-1H has eigenvalues only at p and p, -13 where p -K-1H. is nilpotent, H is nilpotent. K1 Theorem II. 3. 2. If has eigenvalues only at K-1H then S P' p -p , -p -K- 1H, where and P is of type 2. Proof: chapter, is similar to K- 1H S First consider p = -p By Fact 5, in the introductory is *-congruent to a block diagonal matrix whose diagonal blocks are of the form: 40 1 13+i 1 -a+i -1j 1 1 -13+J Icrote: j= (R+j)* = (-13+j)* = R - j = Therefore j(p+j) and j(-i3+j) are elements in 6. We shall Therefore show that each block is *-congruent to a matrix over 6 without loss of generality, we may assume S = [ H.3.1]. Multiply . by C = diag(11 j, 1,j, ...) to the right and by C* on the left of then by Fact 2, the entries on the first super-anti-diagonal of .2 or -3 or (in case of the middle entry) become either 1 the entries on the anti-diagonal are multiplied alternately by -j. Thus S S; S 0, j and and is of type 2. Next consider 13* ±p. By Fact 4, matrix with diagonal blocks of the form: S is 4-congruent to a 1+d- 1-F9- . f+d- 1+44d- .. - f+*d_ 'NI e JAI = 1 1 f+dI . I 1 f+*d . ' f+,:cd . 1+,4 I 1 I f+d . , f+d 1 , I f+d 1 42 where M is the upper left 2m x 2m block and N is the lower diag(1,j, 1,..., right 2m x 2m block. Let C = diag(1, 2m Consider N)C = P e Q. p.. = q.. = 0 C*(M and also 2m, p m, 171 q 2 Pi,2m-i 2m-i -j =1 i-1 2m+1 - q.( = (-1) i-1j(P+i) 2m+1- i 2m+1 when i is even when i is odd. = (-1)i-l[j13*+2j2] i<m = (-1)i-i[jP-Fj2 ] i > m. -1)i - 1( -j P+j 2) =(1)i-1(-j*+j2) = i+j and =0 M, M for i<m i > m. Consider [I I PlI 1:-) P+1-5 P2+P2 P*P+PIP P*P2 +PI2P too. which is over 6 Note that P* PP 15 P and 43 = det(P -P) det ±i2m(p+p*)2m V because Therefore M 0. ±j(13±P*) N 0 is of type 2. Thus is of type 2. From this point let and B. denote non-singular diagonal D. matrices over Theorem 11.3.3. basis for such that 11 is nilpotent, then there is a H-1K If 0 0 where ED T. i=1 1 S T. = (E 1 +jF n.1 Bi ) , ni and n. i n. for i i j. 3 1 that If h S= Ett i=1 co p, and 0, p 1=1 . 1 1 , E 1 U.- such n. i n. for i r/j. 1 3 - -K11H and has eigenvalues only E, then there exists a basis for where 1- = F n. CS) ( B AD D.) + En. 1 1 i is similar to K-1H -p, = (jE+F) ,0 B. n. n. T. 1 such that S = @ T. P. cc where T. If at is nilpotent, then there is a basis for K-1H [(13+j)B. ED 1 i 1 (T -P+j)D.1 1 (Pictures of blocks of Tr, z = 0, .0 , p where n. 1 13 /0, n., for pE Appendix I). Proof: By Fact 4, (i) is *-congruent to a matrix in S diagonal blocks with each block of the form: WO E, n = E(E +F ) E /n, EE _E i j. 3 are in 44 Thus rk. S= Wo E., n. j=1 1=1 3 1 ) k1 where.n. / n.i i / j. Suppose S = 0 W:3, n if and B = H-1K, {el ' Be 1' Write n = n1, n-1 is a 1 k = kr Then ... ,Bn- le elifor basis for i i=1 J 1 , e2, Be2, . some . ,B ek , le2, el, ... ,ek . . ,B ek} If' Rearrange the basis E elements in the following manner: {e1,e21.. ,ek,Bei, Be2, .. ,Bek, . .,Bn-1er respect to this new basis S becomes To .,Bn-1ek}. With 8 = 1 (En1+jFnl) B1 B1 = diag(er . . ,Ek). Repeat the above process for where j = 2, 3 , 0 h . . . Thus if , h. S= h ki Q3 3-1i1 is *-congruent to S W Ei' nj ED T j=1 nj If K-1H is nilpotent, by Fact 4 is *-congruent to a S matrix in diagonal blocks with each block of the form: W E$ E / 0, n = E(jEn+Fn) H-IK being nilpotent, Going through a similar process as in the case we can show S is *-congruent to Eft T. j=1 6. EE . 3 (ii) When K-1H has eigenvalues only at ±p by Fact 4, we have S o, *-congruent to a direct sum of blocks with these blocks paired off in the form U 13 E ,n et U 13 , E $n where 45 n = E(Fn+(3+j)En), /0, E 5 7' 0, 6, n = 6(Fn+(-13+j)En), Without loss of generality, assume k. s= e [( e u [II. 3 . 2] j=1 1=1 e( ED V E'i n j 1-1 6,nj )] 1 First consider ;1( [H.3.31 S= k 3 U ED E i=1 n = n1, Write n-1 manner - . , i .. - 1 - H-3I), A1 = (K1 H+PI). A An1I dk} . e2,...,A1n-1 ek,d1,Aidi,...,A1n-1di, Rearrange this basis in the following {ei,e2,...,ek,di,d2, ...,dk,A0ei, n- 1 n- 1 Ao new basis ,A0 , Ai Then in this dkl. is of the form S (-13+j)D1] where n1 n1 B1 = diag(E ,A0ek,Aidi, n-1 n-1 ek, Ai di, (303 D1) + E__ 0 [(P-1-j)B1 F S '=\\'i is of the form [II. 3. 3] is of the form: S {ei,A0e1,...,A0n-1 2 ,Aidk,Aidk, . ,(13 e v Let A0 = k1 = k. If for which basis of i n, -\\ 3 ,E Ek), , is of [II. 3. 2] S D1 = dial( 5 is *-congruent to 6k) . d k ED T'. . E Thus if q. e. d. j=1 In the following theorem we let z Tz = T1, where z = 00, 0, p 46 and where 13 / 0 13E , B = B, E = E,n l (i) (a) For Theorem II. 3. 4. hence of type 1. For (ii) For (i.e., iff even, n odd, n n T0 (b) jB odd, is of type 2 and T0 is of type 1 iff is B is of is of type 1), and (c) T0 is of type 2. type 2 iff jB For even, n (i.e., iff -B n = nl, k = k1, F=F and *-congruent to we also write ; T on T is of type 2 and hence of type 1. is of type 1 iff B is *-congruent to is of type 1), jB oc and -B of type 2 iff T is of jB type 2. Proof: Suppose Recall is odd. Let n T0 = (E+jF) 0 B. . , j,1). Then )*T°(C(xj Ik) = (C (C C = diag(1, j, 1, Ik)*((E+jF) = C*(E+jF)C B)(C T.k) B By Fact 2, the entries on the anti-diagonal of C*(E+jF)C are alternately or -j2 alternately ±j(j) = ±j2, zeroes. Thus -B. Suppose P the rest of the entries of C*(E+jF)C* are is of type 2, and hence is of type 1. To (b) Suppose those on the first super anti-diagonal are 1, n is even, and suppose B is *-congruent to is non-singular such that P*BP = -B. Let C = diag( 1 , -1,1.. . ); then 47 P)*[(E+jF) (C P*BP P) = C*(E+jF)C 13](C = (-E+jF) 0 -B = (E-jF) 0 B. Conversely suppose Let H= 1 is non-singular such that CI * C1H(H1 K) H(H -1 * 1 and K = 7(T0 -(T0 (T0 +(T0 ) ), K) n- 1 = [E C1 = =0 (T 0)* C1:',E Then ). -) -H(H1 K)n1 131[E F 45) n-1 Ik] = B Eft 0 Thus This computation will be used again in the proof of part (c). 0. Thus by Fact 3, 0)C1 = -B CI(B (c) Now suppose exists a non-singular C2 = diag( I, j, , j). is even and jB n is *-congruent to B is of type 2, i.e. , there * such that jC013C0 C0 -B. is over Let Then (C2 0 CO) T°(C2 0 CO) = [C2 0 C0][(E+jF) 0 B](C2 0 CO) [II. 3.4] = C2(E+jF)C2 C0BC0 The anti-diagonal entries of C2(E+JF)C2 are alternately +j and the first super-anti-diagonal entries are alternately -j3 and j. Thus * j1- C2(E+jF)C2 factor by j-I we get that is over 6. Thus (if we multiply the first and the second factor by [II. 3. 4] = j -1 j in the tensor product) . C2(E+3F)C2 0 jC013C0 is over E Thus 48 is of type 2. To * Conversely suppose 1 1 (w+w*), K1 = C3T0 C3 = (W-W*) so jici is over C3 Since matrix matrix, -- of D 1 1 C3R -1 jR*H1(H1 K) n1-1 R is a matrix and the first principal sub- is non-singular. Hence jD*BD is the upper left block of j(C3R)*(B&O)C3R=j(C3R*(H(H-1K)n-1)C3R which is over Thus . (ii) jB is of type 2. The proof of (ii) is similar to that of (i). For the following discussion, write p 0, and 1 = jR*H1(H1-1K11n-1R . Recall H(H-1K)n-1 = B e 0 matrix 6. Likewise H = H * = HI has its first k x k principal sub- is a permutation matrix, over Thus non-singular. Thus D, R K1 = -K1 . 1 is over * jR*C3(H(H1 K)n1 )C Since Let is non-singular, there exists a permutation such that C3R R is over then W = H1 + jK1, W* = H1 - jK1 = H'1 + jKl1 . Thus (jKI)* = -jKi = jKi, W pE , T q.e.d. as T. Recall that ni= n. (Refer to picture in Appendix I.) Theorem II. 3. 5. T is of type 1. T is of type 2. The following are equivalent. 49 (3) B is *-congruent to (-1)nD. M= ( Lemma 5. If C C* then C*) , C such that cE c* 1 ±c, C = cI, E E (c*)2 A+Ec2 A [A M= M* 0 c2A+E(c*)2A* cc*(A+EA*) [cc*(A+EA*) EA* The proof of Lemma 5 is routine computation. Lemma 6. Suppose C, B, D are non-singular. C*(B (-1)nD)C = -(B El) (1)D) C):c(B ED (l)n1 -1D)C = B If and ( - 1)n -1D then is *-congruent to B (-1)n-1D. Proof of Lemma 6. Partition C= C11 C12 conformably: then C22); (C21 [B * 0 C* C= 0 where C E ED 1 C 11BC 11+EC 21DC21 * * * C 11BC 12+EC 12DC22 * C12BC 11+EC22DC21 * C 12BC 12+EC 22DC22 . Referring to the first condition, we have C11BC11 + (-1)n C21DC21 = -B. 50 Referring to the second condition we have -1 + (-1)n Cl 1BC1 1 * C2 1DC21 = B. If we add the above two equations together, we get 2C1 1BC1 n- * B Thus B is *-congruent to Thus (-1)1 = 1 C2 1DC2 1 Lemma 7. 1 K =23.-(r-T*), If T 1 is as in Theorem 11.3.5, and H = 2(T+T*), then (0)n-1(B K[(K-111)24,2i]f-1 H[(K- (0)n- lp(B 1H)2..p2I]n- 1 Proof of Lemma 7. (-1)n-1D) ® 1 )nD) Recall (-P+3)D] T = F 0 ( B sEs D) + E 0 [(P+j)B (Refer to and picture.) F (B D) +E® (PB Q3 -PD) K=E (B D) H K Let -1 = E 0 (8 -1 , Q3 D -1 ) . k = dim B. K-1H=EFOI 2k +I nGD(PI N=EF is the nxn matrix with 1 's - ). on the first sub-diagonal 51 and zero everywhere else. 1 1 H- I=N I2k + In 0 (0k =N H+ + I2k - 2 61 ) 0 (2PIk $k) SO = - H- (K 13I)(K H+PI) I2k + N N2 (26 1k ED -23I ' k ) = (N (8) I2k)(N 0 I2k + I 0 (2pIk 0 -23I )) . Thus , -1 1.(K H)2 -p2ij,n-1 M = (N is a blocks = (N I2k n-1 - 2PIk))n ® I2k)(N® I2k+10 (213Ik +I ( -1 -26Ik)) 261k 2kn x 2kn block lower triangular matrix with the diagonal = (2)n '1k(Er (-23)n-1I ) I ,n 1 . Thus [E (8) (B ED D)][Nn-1 0 I2k]1\.4 = ((213)n-1B (13 (-23)n-1D) (213)n-1(8 ED (-1)n-1D) ED 0 eo Similarly, H[K-1H)2-p2]n-1 = (2P)n- 1PB e (2p)n-1(-3)D = (0)n-113(B (-1)nD) e 0 0 q. e. d. 52 Proof of Theorem 11.3.5. We shall prove (3) => (2) => (1) => (3). (3) => (2). Suppose Q is non-singular k x k such that Let C = In 0 (Ik e Q). Let p = C*TC. Then P = [Ina) (Ik Ef31Q)]*[F (B e D)+E [(P+j)Bek (-P+j)Dfi[InCi0 (Ik Q*DQ = (-1)2IB. Q)] = F 6 (B 6 (:)*DQ) + E 0 [(p+i)B e (--P-I-j)Q*DQ] (B 9 (-1)nB) + E =F Let such that cE [(P+j)B e (--P+j)(-1)nB] c* ±c, M= . and cIk c*Ik By Lemma 5 R0 = M*(B Let 1.1* = p_j p. = p+i, R (c*)2B+(-1)nc2B c2B+(-1)n(c*)2B cc*(B+(-1)B Thus = M*(p.B e -P+j = -p.*. Then again by Lemma 5, np.*B)m (-1)1 n-1 LB) n-1 2 [cc*(p.B+(-1) c2p.B+(-1) (c*) 1.1*B Note i cc*(B+(-1)nB) (-1)nB)M = (c*)2p.B+(-1) n-1c2p.*B cc*(B+(-1)n-1B) 53 (In 0 M*)P(In 0 = (In® M*)(F =F M) (-1)nB) + E 0 (p.B+(-1)11-14*Bl(In 0 M) (B (M*(B e (-1)nB)M) + E 0 M*(p.13+(-1)*B)M =FOR0 +FOR1. Note that if if n n is odd, is even, jR0 = diag(1, j, 1, j, and ). and jR, R0 are matrices over are matrices over R1 i. Thus let Then (UEfiI2k)*(FOR0 +EOR1)(U$I2k)=U*FU 61)Ro+ U*EU OR1, by Fact 2, is a matrix over 6 This T is of type 2. Proposition 11. 2. 1 (iv) asserts that (2) => (1). C*TC = T*. Recall H = , ,n-1 1 c = -KL(K H) -p K = -23(T-T*), C*KL(K1 H)2 -(32 I] (1) => (3). Suppose C*H[(K-1H)2-132I]-1C C*(H+jK)[(K -1 H[(K-1H)2-132P-1. 1 (T+T*), , Thus H)2 -p 21]n-1 c = (H-jK)[(K -1 H)2 -p 2 n-1 Lemma 7 asserts that H[(K-1H)2I]n-1 (2)n-l(B e (_i)no) K[(K-1H)2-p2Ir-1 (mn-1(B (-1)n-1D) 0 . 54 C*[(2P)11-113(B ED (-1)nD C*(H+jK)[(K-1H)2-I32I]n-1C i(o)n-1(B [(2p)n-l(B _ 1)11-1D)4) 01C Ela (..unD) j(2p)n-1(Be (_un-1D] By Fact 3, there exists C11 C11(B (134 non-singular, such that (-1)nD ( -1)nD)C 1 1 = B C11(Be (-1)n-1D)C11 = B e (-1)n-1D Thus by Lemma 6, Theorem II. 3.6. i j, where is *-congruent to (-1)'D. B z = co (i) If or 0, S Tz. = z i=1 then S , with q. e. d. for n. n. 1 is of type 1 iff each Tz. is of type 1. (ii) Let H = 1 2 (Sz +Sz ), K j (S-S) . 1 2 z z Then if z = 0, Sz is of type 1 iff H(H-1K)2m+1 is *-congruent to -H(H-1K)2m+1 for m = 0, 1, is of type 1 iff K(K-1H)2m is ; *-congruent to Proof: if z = co, Sz -K(K-1H)2m. The "if" parts of both (i) and (ii) are trivial. To prove the "only if" part of (i) for z it is enough to prove 0 type 1 for n.%even 0 (see Theorem 11.3.4(i) ). Assume n. > of for 55 i < h. Write If as S0 is even, then n1 by Fact 3 is *-congruent to 0 1 K.1 = 2j (T. -(T.0 )* ) -1 H(H is of type 1. 1 is H(H so -B1. Hence by Theorem 11.3. 4(i) is of type I also ifn1 is even. Let 1 0 -1K) n1-1-1K)n1-1 = B1 EP 0 , Also H(H -H(H TO T is of type 1 implies S *-congruent to B1 is odd then If n1 S. H. = 0 0 1 (T. +(T. )*) 2 I then ; K) n2-1 n2-1 -1 n2-1 -1 H2(H2 K2) = H1(H1 K1) ED 0. Also n2-1 -1 H2(H2 K2) and (if n2 is even then) H(H -1n2-1 -H(H K) *congruent to *-congruent to Thus (if n2 1K) n2-1 -H1(H1 K1) -1 n2-1 is *-congruent to n2-1 _1 Thus by Fact 3 _1 = B2 ® 0, H1(H1 K1) then -H1(H1 K1) is H1(H1 K1) n -1 is is *-congruent B2 is even and hence in any case) B2 2-1 e -B2* Since n2-1 e T2 -B2. is of type 1. If we repeat the same process by considering in like manner ni- 1 _1 H(H K) for i = 3, 4, ... , h, type 1. Thus we have shown for type 1 if So then we can show each i 1, ,h, that is of type 1. Thus we have proved (ii) The proof for z = co T.0 for 0 T. is of is of z = 0. for both (i) and (ii) are similar to the above discussion. In this case we would only replace "even" by "odd" n -1 by T° and H(H K) (and vice versa) and T0. by T. i-1 56 n.- 1 K(K-1H) Theorem II. 3. 7. If S= (13 , i=1 00/p/o, 13E6, then the following are equivalent. S is of type 1; S is of type 2; is of type 1 for each T1 i; is of type 2 for each T1 Proof: Theorem 11.3. 5 asserts (3) <=> (4) and Proposition 11. 2. 1 asserts (2) 3(1). Thus the following implications are obvious: => (1) (2) A A II I 4 <=> 3 Thus to complete the proof we have only to show (1) => (3). Let K. = Let H= 2j 1 (s+s*), (TP)-(T)*) , P. = KL(K = -P., n.i >nh for -1 K= . H) and 2 1(S-S*). Suppose 2 , n.- 1 C*SC = S* , C-*Q.0 = Q.. (-1) n1 Q1 = and Q1 - jP1 = C*(Q1+jP1)C : such that 13(B1 ED 13 13 (T i+(T. )*), is non-singular. where C 2 2 ni-1 ; then Without loss of generality, assume i < h. From Lemma 7 n1-1 1 2 Q. = Fl[(K H) -p 1] and (213) H. = P 1 = (23) n1 ni-1 -1 (B e(-1) 1 D1)450' 0. Apply Fact 3 to Q1 + jP1 there exists a non-singular C1 01) EP 57 n1-1 CI(BI ED (n1-1-1)D1)C1 = -(B1 ® (-1) D1) and n1 ( -1)1DI)C1 = B1 ED (-1) D1 CI(B1 Then by Lemma 6, is *-congruent to B1 (-1)nD1. Thus TR 1 is of type 1 (see Theorem 11.3.5). Now, 2 1 P2 = K1[(K1 H1) -2P I] 2 1 2 Q 2 = H1 [(K1 H1 ) -(3 I] n2-1 n2-1 n2-1 D2) e o, e (-1) (B2 n2-1 P(B2 e (-1) 2D2) e o, El) (2P) n2-1 ED (20) and C*(1P2+Q2)C = -017)2 + Q2' Applying Fact 3 to this last *-congruency, we get that -1 2 PK1[(K1 H1) 2 2 I 2 n2-1 + HI[(K1 HI) -p I] n2-1 e (2P) n2 (B2 ED 0) is *-congruent to -1 -PICI[(K-11HI)-pI]n2 Also, since T1 - 2 + H1[(K11 H1)2 -0 I] is of type 1, we have that n2-1 (0) n2 (0 013 n2 (-1) D2). 58 is *-congruent to -1 2 2 -PK1[(K1 H1) -p I] Thus by Witt's Theorem, B2 n2-1 eil 0 - n -1 + H1[(K11 H1)2 -p2 1]2 n2 is *-congruent to 0 $ (-1) n2 (-1) D2. Hence T2P is of type 1. is *-congruent to B2 r If we repeat the above process by considering P. and Q. so i = 3, 4, . then we can show T.P 1 is of type 1 for each i. D2 59 III. 1. THE USUAL COMPLEX CASE j General Results In this section, we take to be the complex field and Ee to be the real field and we consider the problem from the view point of a criterion Dina Ng proved in her thesis [3]. Proposition III. 1. 1. X1, X2,.. real roots ,n and j2 r. (2) {[h(1)(y)?1 Lh (y)] , H= (S -S *) 1 Suppose (s+s*). -12m+1 is of signature K) if have that = 0, 1, 2} such that grg2'..gn is non-singular. S is an n x n matrix and Then H1 K is nilpotent, we have that if H(H [h -"(y)] Refer to Dina Ng's thesis [3]. Proposition IlL. 1.2. 1 , matrix P(i,j) = (sgn gj(X.i)) nxn Proof: K= .. then there exist and n = deg h(y), the is a real polynomial of simple h(y) If S m = 0, 1, 2, ... H 0 S is of type 1 iff for m = 0, 1, 2, ...; and K are as in (i) and K-1H is nilpotent, we is of type 1 iff K(K-1H)2m has signature 0 for 60 (i) From Theorem II.3.6(ii), Proof: is *-congruent to H(H-1K)2m+1 is of type 1 iff S (for m = 0,1,2,...). -H(H-1K)2m+1 In the usual complex case, this is equivalent to saying that has signature H(H-1K)2m+1 0. q. e. d. (ii) The proof follows in a similar manner as that of (i). Theorem III. 1.3. Suppose and K are non-singular, and 1 1 H = -2- (S+S*), H K = Ti- cs -s *) , K-1H is similar to -K -1 H. H and f(x) = nomial of K1 is an n x n matrix, S If p(x) is the characteristic poly- p(x)then g. c. d. (p(x), p T(x)) ' f(x) = h(x2) for some real polynomial h(y). Let {[h(1)(ydji[h(2) (y)] r = deg h(y). where has signature of and all 5) Proof: for all even polynomials 0, 1, 2, ... ,n-1 Suppose If we can assume S = , n is the order S1 ED ' j = 1, 2' 52, sig Kq(K1 H) = 0 [x]. q(x) E has signature zero K[f(K- 1H)mg((K-1H)2)] * -Si ) Kj = T (S.-S.) (where is of type 1 then S for m = 0,1, 2, ... , n-1 and all 1 0, 1, 2} j. gE ("only if") ("if ") ir is of type 1 iff K(f(K-1H))m g((K-1H)2) S for m 0 (r) [h(r) j2 g Without loss of generality, * 3233 where if H. = 1 (S,+5.) and -1 such that all eigenvalues of K2 H2 are 61 non-real, and all eigenvalues of are real. We have shown in Ki-1H1 Chapter II that 52 is of type 1. Therefore without loss of generality, as sume S = SI. There exists a basis for Irsuch that S of the form r ®T S 131 k=1 1=1 where P1 Tk and B As sume k =F nk and n. >nh (Bike D/k) + (-p +OD/ [(p+i)Bik Enk are real diagonal matrices, with entries Dik i < h . Note if 2 2 (x -Po), f(x) = 1=1 and h(y) = (y-i32 m=1 'L ) m From our hypothesis, 1. 1] 0 = sig(Kf(K = sig[K - 1 11 H) n1-1 g[(K -1H)2 ]) [(K -1 H) m=1 Let K 1 = R R (T1 -(T1 * ) ) , 2 2 -,111-1 -BmIl g[(K-1H)2] , ±1. 62 [III. 1. 1] equals to sig Et) ly 11 = sig mil 1=1 X ((K -1 Z n1-1 2 -pmin -1 2 g(aCji ) -1 -1 n1-1 2 2 ]gUK1 Ki -1 Hi) -f3mI) ) 2 2I (p.f) 2 ) n -1 2 2 H ) -p m1) / 2 (K./ H1) -1 m=1 / =1 Note ((K 1 + M where M is strictly lower triangular. Thus - 2 g((K/1)2 ) = g(P1)I + M1, mil (because H ((K-1H1 ./ (132-I32 m mil ) = )-62m I) =)I + M2 where hl(P2)), M1 and M2 lower triangular. -1 (K,e 2 - p 2 = Nn1 H./) M.(N n1-1 nl Ci) I2k) I2k =0 for j = 1, 2. Thus , [III. 1. 1] = sig 2 -,111-1 1(1[11'(1)n 1=1 n1-1 , guytN_ it) 12k) are strictly 63 Recall K [III. 1 . =E 1 n1 0 (B1 1 et (-1) 2 1]= sig 0 h ) 1=1 n1-1 n1-1 DI 1) . n1-1 2 (B1 ED (- 1 ) g(P1 )[En N n1-1 DI 1)] "" 2 sig[hI(131)n1- 1 2 g(1:3,e)(131 - e (-1) n1 1Di 1)] 1 1=1 7 n -1 n1-1 sgn[h'(r) 1 g(rdsig(B1 e (-1) DI 1) = 1 1=1 p(j,i) = (sgn such that ,g g (successively) = g 1, g2, Choose matrix. (See Propo- is a non-singular r x r gj(1312)) 2 sition III. 1. 1.) Then Q(j,i) = P(j,fl(sgn(111(I31)) singular r x r matrix. Thus sig(Bi e (-1) each Thus 1. sig 1= -sig(-1) n1-1 = sig(-1) Thus B1 1 is *-congruent to (-1) Di 1 n1- 1 is also a non- Di 1) = 0 1 for each 1. Thus is of type 1 for each 1. Write r S0 = 1=1 r p R T11, for Dil DI 1 n1-1 and let S1 = ED ( 131 ED T. j=2 1=1 ) Pi T1 64 1 1 H0 = ), 2 (S0 +S 0 K0 = 2i (S0 -5 0 ) 1 1 H1 = K1 = 2i (S1 -S 1) (51+51), . Then - 0 = sig K(f(K1 H) n2-1 -1 = sig Ko(f(Ko H0) g((K-1H)2)) n2-1 -1 + sig Ki(f(Ki H1) Because S 0 is of type 1 and f(x) -1 g((K0 H0) n2-1 n2-1 2 )) g((lC11HI)2)), 2 g(x) is an even polynomial, repeating the same process as above, by replacing withn2, we can show K with K1, is *-congruent H with to (-1) /. The proof of the nif" part is completed by similarly considering n2 HI, n1 Di 2 , for each /; PI thus T2 B./ 2 is of type 1 for each the fact that n -1 0 = sig K(f(K - 1 H)k for -1 H)2 ]) k = 3, ... , p. Proposition III. 1.4. 1 iff r gL(K S S is of type is of type 2. Proof: implies In the usual complex case, S It is only necessary to prove that is of type 2. S is of type 1 65 Suppose is of type 1. Here all roots of K-1H are in S so, in view of Chapter II, where we have shown that when K and H are non-singular, type 1 is equivalent to type 2, it is only neces- sary to consider the cases K-1H nilpotent and H-1K nilpotent. or Without loss of generality, assume S = Tz1 where z = 0 In view of Theorem II.3.4, it follows that showing implies S matrix B Let c E of type 1 S of type 2 is equivalent to showing that, if a diagonal with entries type 2. Suppose signature 00. then is a diagonal matrix with entries such that c*I cI cI c* 4 *c . Let M = ( iB is of and of ±1 B = I ® -I. then without loss of generality, write 0, C B is of signature 0, ±1 ) then by Lemma 5 of Chapter II 0 iM*BM = i 2 [c - (c *) Thus 2. iB is of type 2. 2 III 0 q. e. d. Geometrical Approach to 2 x 2 Usual Complex Case In the 2x2 case, using geometric ideas from Ballantine's work on Positive Definite Matrices {l}, we can characterize the prob- lem geometrically. Define F(S) r(s) = {x*sx Ix is a line; S E C2x1}. We say is contradefinite iff r(s) S is bidefinite iff is the whole 66 complex plane. If S is non-singular; let sgn det S = e2iP S [a bdb 2 - 1 det s If such that y>0 define Lc det S -y ac71 + = - - a- 21 det S 1 If S is non-singular, S is *-congruent to one of the follow- ing forms: 0] [i S SI is bidefinite; If S = 1 eiP 0 -i S2 = eiP is contradefinite if y > I. S2 is singular and non-zero, S is *-congruent to one of the following forms 0] [0 0] S3 = 20 S4 00 where here eiP = Proposition 111. 2. I. S S is type 2, is type 1, trace S = sgn tr S 'trace S1 The following are equivalent: 67 (iii) if either is positive or det S if is non-singular then S S is bidefinite or contradefinite; is singular, then either S has real determinant, and S is contradefinite or S trace S is real. In view of Proposition 111.1.4, it is enough to prove Proof: (ii) => (iii) and conversely. We consider the non-singular cases first. (ii) => (iii) Suppose such that then C,',(SC = S*, is of type 1, then these exists det(CC*)det S = det S*. Also det S*= det S 1 det(CC*) det CC* = +1 Since det CC* = (det C)(detC)* > 0, det S >0; det S = det S*. Suppose Without loss of generality, assume to 0 1 SZ = ( Zy 1) , det S < 0, eiP = +i If S S = det CC* 1 . det CC* = 1. Thus det S = ±1. then e iP 1 det S I j e iP = 1. If S , is *-congruent is of type 2, hence is of type 1. Now suppose ip Without loss of generality, assume e = i. is *-congruent to *-congruent to C det(CC*)det 5* = det S. Thus thus C*S*C = S S S 2 = i( 1 0 Zy 1 ) , S is of type 1 iS* = - (iS)*. Let L = iS + (iS)* = iS - iS* = -2y -2 if iS is 68 implies is of type 1 S 4-y2 4(1-y2) < 0 = contradefinite. If iff 1 < y2 iff is *-congruent to S det L < 0; which in turn implies iS* - is = -(iS)* - iS = -L, det L = 4 - is *-congruent to L iff y>1 and is S then in all cases, S1 S is bidefinite. (iii) => (ii) e i_/ det S If det S < 0, then C= det S < 0, ( 01 0 1 and is real. If det S > 0, 0 , k ) i 0 , S 0 is bidefinite, if det S > 0; then Thus C*SC = 5*. is *-congruent to i( is of type 2 hence, is of type 1. Suppose det 5 > 0, is *-congruent to S ( hence is of type 1. If det S < 0, By Lemma 4 of [1], ( provided . (Note: y > I sin a I y>1 = sin i6 101) 2N i 0 0. _i). 0. 0 ) =( -1 0 Thus . ) 0 1 is contradefinite. If 5 1 0 2-y 1 S is *-congruent to Thus ) i is of type 1. If S is *-congruent to Choose I. ) Thus 1. = is 5 Without loss of generality, let S = ( ) then e Without loss of generality, assume eiP = ±i. First suppose *-congruent to Let det S - ±1. Without loss of generality, assume I det SI eiP = +i. Suppose S is of type 2, i( 1 0 2y 1 0 ( i sin a e a = Tr/2. S is of type 2, hence is of type 1. q. e. d. Now suppose S is singular. If which is contradefinite and real then S S is *-congruent to S3, is of type 2, hence is of type 1. Therefore we assume S = 54 Here S is *-congruent to S 69 S* iff there exists e = cc* iff CE ,i3 ci(ce = e C such that e13 = ±,./7"Tv iff E S is over iff q. e. d. . The above interpretation cannot be applied to higher dimensional matrices. Consider T1 = diag(i, i) 0 j1 T3 = is the y-axis. Thus F(T1) F(S) C r(T) F(S) 10 T1 63- is bidefinite. If S = 2y is the complex plane. Thus definite. Det T3 = 1. But T., j = 1,2,3 3 K. = J 2 (T.-T. ) J J are not of signature zero. o ( F(T2) 1 ), then is contra- are not of type 1, because 70 BIBLIOGRAPHY Ballantine, C.S. Products of Positive Definite Matrices III, Linear Algebra and Its Applications 3 (1970) 79-114. Carlson, D.H. A Note on Matrices Over Extention Fields, Duke Mathematics Journal, Vol. 33 pp. 503-505, 1966. Gantmacher, F. R. The Theory of Matrices, Vol. II, Chelsea Publishing Company, New York. 1964. Greub, W.H. Linear Algebra, 3rd Edition. New York, Spring-Verlag, 1967. S. Haynsworth, E. V. and Ostrowski, A. M. On the Inertia of Some Classes of Partitioned Matrices. Linear Algebra and Its Application. Vol. I. pp. 299-316, 1968. Jacobson, N. Lectures in Abstract Algebra, Vol. II, New York. Van-Nostrand, 1953. Malicev, A. I. Foundations of Linear Algebra, San Francisco, W.H. Freeman, 1963. Ng, D.N. An Effective Criterion for Congruence of Real Symmetric Pairs, PhD. Thesis, Oregon State University, 1973. Smiley, M. F. Algebra of Matrices, Boston, Allyn and Bacon, 1965. Taussky, 0. and Zassenhaus , H. On the Similarity Transformation Between a Matrix and Its Transpose. Pacific Journal of Mathematics. Vol. 9, pp. 893-896, 1959. Turnbull, H.W. On the Equivalence of Pencils of Hermitian Forms. London Math Soc. (2) 1935. Vol. 39, pp. 232-48. Turnbull, H. W. and Aiken, A.C. An Introduction to the Theory of Canonical Matrices, 4th Edition, New York, Dover, 1961. APPENDICES 71 APPENDIX I Pictures of Matrices _ jB jB B B 0 T1 = jB jB _ 4 B B .1=1. n1 blocks B B jB jB T= oo B B jB jB _ n1 blocks 72 TR B 0 o D B 0 'FIB :0 D I 0 = B 0 1 p.B 0 1 D 1.1.B 0 1 I . I a 0 --I 0 - p.*D 1 i 1 1 0 -ii*Di n1 Let p. = (3+j , 11* = p.*= blocks 0 -1.1.*D I 0 - 1.1.*D 73 APPENDIX II Miscellaneous Results Here we observe that any non-singular 8, has determinant over -1 2x2 type 1 matrix We also make observations on fields with . as a norm. Proposition IV. 1.1. Lemma 8. If of type 1, then and some 5 S non-singular matrix of is a 2 x 2 is over det S type 1, then If is a S non-singular matrix and 2x2 S S* -s+as s [ is *-congruent to ] for some is S SE aE It has been proved earlier that Proof of Lemma 8. is *-congruent to A* and A-1. Therefore the char-1, acteristic polynomial of A is x2 - ax + 1, i.e., A = [01 a J A = S-15* - Suppose * s 12 -s11+as12 S22 -s21+as22 -1S* = SA = * * s12, s22 = s 12, * s21 = -s11 + as12. Let s = s11 Therefore If T T* s 11 11 s11 - s22 and s* s We have S = (_s*.as t a so = is similar to s . A = (01 1 a ), i. e., 1 74 if there exists a non-singular then S = D*TD such that D1 (T1 T*)D = A, D has the canonical form we have just shown. Thus it is enough to consider A being a companion matrix. Proof of Proposition IV. 1. 1. From Lemma 8, when 2x2 non-singular type 8* = SO a - s ,;(-1-as ). matrix -1 2 (det C)(det C*)(det S) = det S0, Therefore is *-congruent to S 2 det S0 = s + (s*) + ass* non-singular matrix, such that C*SC = is S S0' If C is a then - det S = (det So)[(det C)(det C*)]1 so det S E Proposition IV. 1.2. 1 H= Suppose (5+5*), K = 231(S-S*). S If is an n x n matrix over is a norm, and one of the -1 following holds: H-1K is nilpotent, (s) K-1H is nilpotent, (1) then S is of type 1. Proof: From Theorem 11.3.4, it is enough to show every is *-congruent to its negative. Since diagonal matrix over is a norm, over e , -1 = act*, for some (aI)*B(aI) = -B. Proposition IV. 1. 3. aE ; if B is any matrix q. e. d. Here we assume E and to be -1 75 finite fields, and S, H, and K are as defined in Proposition IV. 1.2. If the following holds: are not nilpotent. H-1K and K-1H is similar to are in é3., then K-1H and all eigenvalues of K-1H -K-1H is of type 2, hence is of type 1. S In view of Theorems II. 3. 2, II. 3. 3, II. 3. 4, II. 3. 5, and Proof: is 11.3.7, it is enough to show that every diagonal matrix over *-congruent to any diagonal matrix of the same order over Let diag(E , E2, . B . . is a norm. Let finite fields, every element in that a.a. = E. 1 1 1 Let -1 -1 C = diag(al , az , Thus every diagonal matrix over matrix. and Since Ei E , Ek), E -1 , ak ) are such a. E then C*BC = Ik. is *-congruent to the identity q. e. d. Proposition IV. 1.4. Here es. S, , H and K are as defined in Proposition IV. 1.3. Suppose one of the following holds H-1K is nilpotent, K-1H is nilpotent. In view of Theorems II. 3.3 and 11.3.6, we may, without loss of gen- erality, assume S = Tz , z = 00, 0. (Refer to pictures in Appendix I.) Then the following holds: (i) if z = 0, (recall diagonal matrix over ) S= then 0 = (En+jFn) B, where B is a 76 when n is odd, when n is even, (ii) if z = 00 oo n is even, when n is odd, (b) is even; where is B then ), when (i) is of type 2 iff dim B 5 (recall S = Tn = (Fn+jEn) 0 B, a diagonal matrix over Proof: is of type 2, S is of type 2 S S is of type 2 iff dim B is even. (a) Refer to Theorem II. 3. 4. In view of Theorem 11.3.4, it is enough to show that if is a diagonal matrix over is even. Suppose dim B e , then jB is of type 2 iff dim B is even. In the proof of Proposition IV. 1.3, we have shown that every diagonal matrix over is *-congruent to any other diagonal matrix over Ex of the same order without loss of generality, assume and rank. If dim B=2m, m > 1, B = (1,-1,1, -1,...,1,-1); let then C*BC = diag(Do, Do, C = diag(C0), , Do) where (c*) 2-c 0 DO = Thus jC*BC = C*jBC c2-(c*) 2 is over versely, suppose dim B E . 0 Thus is odd and jB jB is of type 2. Con- is of type 2. I.e. , there 77 exists over non-singular matrix a 8, such that C C4<jBC = D, and D is Thus . (det C*) det(jB)(det C) = jn(det B)(det C)*(det C) = det W L which is over But . n . is odd, we have a contradiction; so jB SO n J Corollary 1. If E . Thus is not of type Z. q. e. d. The proof is similar to that of (i). (ii) is not over S, H, K are finite fields, and are as Proposition IV. 1. 4, and if one of the following holds: over H-1K is nilpotent K-1H is similar to -K-1H and has all its eigenvalues then , Proof: is of type 1. S Refer to Propositions IV. 1. 2 and IV. 1.4. Corollary 2. If 2 x 2 matrix over the pencil XH + I.LK is similar to -K-1H, Z( and such that 4" are finite fields, if1 H= (s+s*), K = is S 1 (S=S*), is non-singular. K is non-singular, then S K-1H is of type 2. Proof: Since the characteristic polynomial of K-1H would be of the form x2 - c, where cE . Since is finite, .4 is 78 the only quadratic field over e , thus it contains all the roots of all second degree polynomials. So we can apply Propositions IV. 1. 3 and IV. 1.4.