Internat. J. Hath. & Math. S. VOL. 15 NO. 2 (1992) 261-266 261 ON POLYNOMIAL EPr MATRICES AR. MEENAKSHI and N. ANANOAM Department of Mathematics, Annamalai Universlty, Annamalainagar- 608 002, Tamll Nadu, INDIA. (Received Hay 8, 1989) ABSTRACT. This paper gives a characterization of EP-X-matrices. Necessary and r sufficient conditions are determined for {i} the Moore-Penrose inverse of an EP r matrix to be an EP-X-matrix and {ii} Moore-Penrose inverse of the product of r to be an EP-X-matrix. EP-X-matrlces r Further, a condition for the generalized r inverse of the product of X-matrices to be a X-matrix is determined. KEY WORDS AND PHRASES: EP-X-matrlces, generalized inverse of a matrix. r AMS SUBJECT CLASSIFICATION CODES" 15A57, 15A09. - 1. INTRODUCTION Let be the set of all mxn matrices whose elements are polyrmmtals in X over an arbitrary field F with an involutary automorphism a" a for a e F. l} The elements of are called x-matrlces. For A(X} e X (alj{ Let are X}}of set elements whose all A (X} mxn matrices be the rational functions of the form f( X}/g( X} where f( X}, 8(X # 0 are polynomlals in X. For simplicity, let us denote A(X) by A itself. F F (jt The rank of A eF F, is defined to be the order of its largest minor that is AeF not equal to the zero polynomial {[2]p.259}. is said to be an unlmodular X-matrix {or} invertible in if the determinant of A{X}, that is, det A{X} is a nonzero constant. is said to be a regular X-matrix if and only if it is of rank n {[2]p.259), that is, if and ortly if the kernel of A contains only the r and zero element. is said to be EP over the field .F(X) if rk (A} r $ R(A) R(A where R(A}and rk (A) denote the range space of A and rank of A respectively [4]. We have unlmodular X-matrices regular x-matrices} E P- X-matrices }. Throughout this paper, let AeF]. Let 1 be identity element of F. The + Moore-Penrose inverse of A, denoted by A is the unique solution of the following AF F . AeF set of equations- AXA=A (1.1); XAX=X (1.2)" (AX)=AX [1.31; (XA} =XA (1.41 A + exists and exists, A is (or} {I A+F EPr over if and orfly if rk F(X} + AA (AA*) A+A. rk For (A’A) AeF, rk (A) [7]. When A+ a generalized inverse inverse is defined as a solution of the polynomial matrix equation (1.11 and a reflexive generalized inverse or) {1,2} inverse is defined as a solution of the equations (1.1} and (1.2} and they belong to The purpose of this paper is to give a characterization of an EPX-matrix. Some results on r EP- X-marices having the same range space are obtained, As an application r necessary and sufficient conditions are derived for (AB) + to be an EP-X-matrix r whenever A and B are EP-X-matrices. r F. AR. MEENAKSHI AND N. ANANDAM 262 2. CHARACTERIZATION OF AN EP r-h-MATRIX THEOREM I. AEFn] is EP r over the field F{k) if and only if there exist r an nxn unlmodular k-matrix P and a r x r regular k-matrlx E such that PAP 0" D PROOF. By the Smith’s canonical form, A =1 O where P and O are unimodular-k-matrices of order n and D is a rxr regular diagonal h-matrix. Any p-1 (1) where R2, R3, and R {1} inverse of A is given by A 4 are arbitrary conformable matrices over F(k). A is EP over the field F(k) Q_1 D-11 R3 ----@ R(A)= R(A AA*(1) A* A =- D ==> QP QP R 3 R 4 *{ R* 1.0 0., % 0 (By Theorem 1713]) 2 ,-I O , O 0 0 -1 Partitioning conformably, let, QP T3 o o 4 DT DT2 o o T2 0 1 T4, DO E P DT 1 Conversely, let T1 0 p, oo P T1 0 P* P o is a r x r regular h-matrix. E 0 0 0 PAP where E is a r x r regular h-matrix. Since E is regular, E is EP If 0 ITI 0! QP A where 0 0 0 T3 Hence 0 01 DT2R2D + D (since D is regular). ,-1 Therefore 4 T LT3 =- T =@ R(E) b R(PAP == R(A) = Ar] r over F(X). R(E R(PA P R(A over F(X). A is EP Hence the theorem r EP over the field F{k} then we can find nxn regular and is and K need not be unlmodular h-matrlces. For example, consider A Ik{= .A is 263 POLYNOMIAL EP r MATRICES E P, being a regular K A -1 * The ’,XI Here H A there then inverse then X-matrices. are not EPr-X-matrix H and K for condition nxn unimodular exist AK 1 necessary a gives unimodular X-matrices. If A is an nxn THEOREM 2. {1} ll/k’ and K A" If :0 0 theorem following -I A* A HA then H If A X-matrix. ’-il -1/X and A has a X-matrices H and to be X-matrix K such that AK. HA an nxn PROOF. Let A be an nxn EP r-X-matrix. By Theorem 1, there exists 0 E where E is a rxr regular unimodular X-matrix P such that PAP 0 0 -I is also a X-matrix. inverse, E Since A has a X-matrix {1 A x-matrix. 0 I0 0 p-1 [ E* -I * P P A Therefore E p-1 A Now 0 0 E*E-I p-I 0 I 0 HA where p-I H X-matrix. Similarly we can write A E-1E , K p-1 P 1 - pp-I p-l* E 0 0 0 o P is an nxn AK where is an nxn unimodular X-matrix. 0 AK. HA A REMARK I. The converse of Theorem 2 need not bo true. Therefore ,Ix consider , 0 Since A A, H 0 1} inverse. X-matrix A 10 unimodular K A is an 12 For example, EP1- X-matrix. However A has no 3. MOORE-PENROSE INVERSE OF AN EP -X-MATRIX r The following theorem gives a set of necesssry and sufficient conditions for the existence of the X-matrix Moore-Penrose inverse of a given X-matrix. THEOREM 3. For A F], the following statements are equivalent. rk(A and A A has a X-matrix {1 inverse. i) A is EP r, rk(A) 2) ti) There exists an unimodular X-matrix U with A U U 0 where D is a rxr unimodular X-matrix and U U is a diagonal block matrix. iii) A GLG where L and G G are rxr unimodular X-matrices and G is a X-matrix. + is a X-matrix and EP r v) There exists a symmetric idempotent iv) A , AE EA and R(A) PROOF. (i) rk(A2), rk(A) a X-matrix {1} pp (ii) / A (E 2 , E E such that R(E). Since A is an EP over the field F(X r X-matrix and exists, by Theorem 2.3 of [5]. By Theorem 4 in [6], A A has inverse where 0 X-matrix E, implies P1 is that there exists an unimodular X-matrix P with a symmetric rxr unlmodular X-matrix such that AR. MEENAKSHI AND N. ANANDAM 264 PA where W is a rxn, ol AA is a AA+A A o Pl PAA+P * k-matrix and [0 A(AA+). Hence by Theorem 2 in [6], X-matrix of rank r. Since A is EP 0 !Wo p-1 A Therefore p-1 r’P1AA A+A + ,.o and 0 P o L" oJ 0 thus H is a nxr, X-matrix of rank r. H consists of the first r columns of P p-I Now A rxr regular D 0 0 O p-I U U Since A X-matrix. and D WH is a inverse and P is an -ID inverse. Therefore by Theorem I in [6], D X-matrix a has 0 {I} {I} X-matrix hasa PAA P * unimodular X-matrix, p-I where U 0 is an unimodular X-matrix D Hence A U where D which implies D is an unimodular X-matrix. U 0 0 is a rxr unimodular X-matrix and U U is a diagonal block X-matrix. Thus (ii) holds. O] , =- (i) (iii) U2 u4 U1 Let us partition U as U u3 u3 0 0 u4 U U is Since a dtasonal unimodular X-matrices. (iii) =) (iv) Since A G(G * G) + A + Now AA + A is RP r. GLG 1 Then is a rxr X-matrix. GLG u Lu2 are U where U X-matrices. 3 block X-matrix, UIU1 G G and U3U 3 + L are rxr Thus (iit) holds. L and G G are unimodular X-matrices. -IL-I (G*G)-IG *. One can verify that A+A GLG G (G G) (G G) G G(G G) G implies that Since L and G * G are untmodular, L -1 and (G * G) -1 are X-matrices, and G is a X-matrix. Therefore A + is a X-matrix. Thus (iv) holds. (Iv) - (v) = Proof is analogous to that of (II) {v) = {i) (lil) of Theorem 2.3 [5]. Since E is a symmetric Idempotent X-matrix with R(A) Theorem 2.3 in [5] we have A is EP + E Let e. AE A E R(E) and R(A) and a. denote the =-- X-matrix, Aej aj, since == ej jth r+ AA and rk(A) EE columns + of rk(A E. E R(E) and AE 2) ==> A exists. Since (AA+)A A. Now AE EA and A respectively. Then is a A-matrix, the equation Ax has a X-matrix solution. EA, by + aj where aj is a Hence by Theorem 1 in [6] it follows that + A ha a X-matrix {I} inverse. Further AA E is also a X-matrix. Hence by Theorem 4 in [6] we see that A A has a X-matrix {1} inverse. Thus (i) holds. Hence the theorem. POLYNOMIAL EP r MATRICES 265 The condition (i) in Theorem 3 cannot be weakened which can be REMARK 2. seen by the following examples. EXAMPLE I. is EP and A 1 X rk(A} rk(A 1. A A has X-matrix 1} inverse (since | 2 Consl,dec 2} thel 2XmJatrlx2Xzl’A21 , L2X the tnvariant polynomial of A A is A A, I 4X + 1 1 1 1 X- is not a EXAMPLE 2. Consider the matrix I/ which is not the identity of F). For this X-matrix. Thus the theorem falls A rk{A2), rk(A) A A" 0 0 .0 {I} inverse (since any conformable X-matrix is a this A, A does not exist. REMARKS 3. X-matrix, and A is EP I. Since has a X-matrix over GF(5). X-matrix {I inverse}. For Thus the theorem fails. From Theorem 3, it is clear that if E is a symmetric idempotent A is a X-matrix such that R{E} R{A} then A is EP + z=> AE EA A is a X-matrix and EP. We can show that the set of all EP-X-matrices with common range space as r that of given symmetric tdempotent X-matrix forms a group, analogous to that of the Theorem 2.1 in [5]. COROLLARY 1. Let E E* e Then H(E}={A e R(E}} is s maximal subgroup of A is EP over F(X} and R(A} r containing E as identity. E2 F: F. F PROOF. This can be proved similar to that of Theorem 2.1 of [5] applying Theorem 3. 4. APPLICATION by In general, if A and B are X-matrices, having X-matrix {1} inverses, it is not the X-matrix 1} inverse for both A and B. tnvariant polynomial of AB is 1+2 X2 But AB 0 3 +2 X Since the 0 1, AB has no X-matrix {1 inverse. The followin 8 theorem leads to the existence of X-matrix {1} tnverse of the product AB. 2 THEOREM 4. Let A, B If A A and B has X-matrix {1} inverse and R(A} .C. R(B} then AB has a X-matrix 1 inverse. PROOF. Suppose ABx b, where b ls a X-matrix, is a consistent system. Then b R(AB) R{A) c._ R{B) and therefore Bz b. Since ]9 has a o X-matrix 1 inverse, by Theorem I in [6] we get z is a X-matrix. Since A is o idempotent, so in particular A is a{1 }inverse of A and b R{A}, we have Ab=b. Now Ab b. Thus ABx b has a X-matrix solution. Hence by Theorem1 in [6], AB has a X-matrix 1} inverse. Hence the theorem. The converse of Theorem 4 need not be true which can be seen by the following example. 1 0 1 1 1 EXAMPLE 4. Let A B AB Here 0 0 X X 0 F. ABZo 266 AR. MEENAKSHI AND N. ANANDAM R (A) R(B). we Hence the converse is not true Next shall discuss the necessary and sufficient condition for the Moore-Penrose inverse of the product of EP-k-matrices to be an EP-k-matrix. THEOREM 5. Let A and B be EP-k-matrices. Then A A has a k-matrix rk(A) rk(A and R(A) R(B) if and only if AB is EP and 2) {1} inverse, + + (AB) is a k-matrix PROOF. Since A and B are EP with R(A) R(B) and r rk(A) by a Theorem of Katz [1], AB is EP Since A is a EP-k-matrix rk(A) rk(A and A A has a k-matrix {1 inverse, by Theorem 3 + A is a k-matrix and there exists a symmetric idempotent k-matrix E such that B+A rk(A2), 2) + + Hence AA AA E. Since A and B are EP and R(A) R(B), r + + we have AA BB E Therefore BE EB and R(B) R(E). Again from Theorem 3, for the EP -k-matrix B, we see that B + is a k-matrix r + Since A and B are EP with R(A) we can verify that (AB) + R(B) + + + Since B and A are k-matrices, it follows that (AB) is a k-matrix. + + Conversely, if (AB) is a k-matrix and AB is EP then (AB) is an r EP-k-matrix. Therefore by Theorem 3 there a exists symmetric idempotent r R(A) R(E). A+A B+B. B+A _ + + k-matrix E such that R(AB) R(E) and (AB) (AB) E (AB) (AB). Since rk(AB) rk(A) r and R(AB)iR(A), we get R(A) R(E). Since A is EP r’ + by Remark 3, it follows that A is a EP-k-matrix. Now by Theorem 3, A A has a k-matrix I} inverse and rk(A} Since AB and B are EP R(E) R(AB) R((AB) R(B R(B) and rk(AB) rk(B) implies R(B) R(E). Therefore R(A) R(B). Hence the theorem. rk{A2)r, REMARK 4. The condition that both A and B are EP-k-matrices, is essential r illustrated as follows" in Theorem 5, is Let A and B A and B are 0 A A has a not EP 1 (AB) + ACKNOWLEDGEMENT k-matrix 1 I 1} in verse and R(A) R(B). not EP 1. But AB is 0 is not a k-matrix. Hence the claim 1+4 k 2k 0 The authors wish to thank the referee for suggestions which greatly improved the proofs of many theorems. REFERENCES I. 2. 3. 4. 5. 6. 7. KATZ l.J. Theorems on Products of EP Matrices II Lin Alg. Appl. I0 r (1975), 37-40. LANCASTER, P. and TISMENETSKY, M. Theory of Matrices, 2nd ed., Academic Press, 1985 MARSAGLIA, G. and STYAN, G.P.H Equalities and inequalities for ranks of matrices, Lin. and Multi. A1R. 2 (1974), 269-292. MARTIN PEARL, On Normal and EP Matrices, Mich. Math. 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