Math. Nachr. 184 (1987) ZS7-271

A Schur Type Matrix Extension Problem

By Bmm l b m z s c ~ ~ and BEEND -STEIN of Leipzig

(Received J~nnarp 9, 1986)

Let p , q and n be positive integers. Denote APxq the set of complex p x q matriw and .#," the set of non-negative Hemitian q x q matrices. Further, let I,, be the unit matrix of order p . If 4, A,, . . ., A,,-, E Apxp then we put

We shall consider the fol3owing extenmn * p o b l m (8): Let 4, A,, . . -, &-I E A p x q

& E A p x q 4 that Dn+, = Du+,(Ao, A,, * a - 9 As-1,An) scstiefies I ( * i ~ p - Du+$X+i d thd D,, = D,,(A,,, Al, . . ., Awl) fulfih I,,, - 09,' E At. Describe th? set of aU

E J e + l , P . We shall show that this set c8~. be represented 89 a matrix ball. This is a generalization

of a resdt due to GERONJMDS [13] who considered the m e p = g = 1. The matrices D,, of the shape (0) are intimately connected to the SCWB c h

p x q - Y of all p x q matrix fanctions cb which are holomorphic in D := (w E C: lzol < 1) and fulfil the relation

I p - @(w) @*(w) E A!:, w E D .

The following basic theorem shows the connection between p x q - Y and the matrices of the shape (0). It was proved by SCHW [l?] in the case p = q = 1, by GAISTJAX 1121 in the case p = g and, finally, by DIJBOVOJ [7] in the general case.

m

k-0 Theorem. (a) Let Q, E p x q - Y, and let @(w) = 2 A&, w E D. Ukng (0) then we

h V e

(1) Imp - 03: E A&,, m E N. 17 W. 19iohr.. Bd. 134

258 Math. Nachr. 184 (1987)

(b) Conversely, let (A&, be a sequence of complex p x q nzatrices such that (1) i s fulfill- ed. Then the series

OD

@(to) := ZA@L+ k=O

ie convergent for w E D, and it h l d s @ E p X p - Y. This theorem shows the connection between the one step extension problem (S) and

the well-known SCWR problem (8') : Let A,,, A,, . . ., A,+l E Apxq such that r a p - DnD: E A:,,. D d b e the set of aU Q, E p x q - Y with first a TAYLOR coefficients A,, A,, . . .,

The problem (S') was treated by GERONIMUS [13] in the case p = q = 1. Using methods of POTAPOV'S J-theory (see [15]) the matrix version of the problem (S') was investigated in the quadratic case p = p by GALSTJAN [12] and in the general case independently by hov/ICBm [3] and DUBOVOJ [7]. h o v and Kmm concluded the solution of the problem (S') from their results about the NEH~RX problem (see also NEHARI [14], ADAMJAS/AROV/KREIB [l], [2]). The investigations of GALSTJAN and DUBOVOJ are concrete realizations of POTAPOV'S general conception of application of the theory of J-expansive matrix functions to the solution of classical problems of analysis. From their results one can see that the pmametrization of the set of solutions of the problem (S') is realizable with the aid of matricial versions of the classical SCWR algorithm [see also S c m [17], GERONIMUS [13], CKllMm [6], BOYD [5], ABSZNE/ CE~USESCU/FOLAS [4]). In [lo], [ll] the authors discussed the extension problem (P) for non-negative Hermi-

tian block TOEPLPTZ matrices. Using the CAYLEY transformation in the special case p = Q the problem (S) can be transformed in an equivalent problem (P). This method will not be explicated here. In a forthcoming paper we shall show how the problem (S) can be transformed also in the non-quadratic case in an extension problem (P) for a special type of non-negative Hermitian block TOH~PUTZ matrices.

Using methods and results of ARSENE/GEBONDEA [lS] and ARSENE/CXUWSESCU/ Foras 141 in his paper [20] CONSTAXTXNESCU considered the CABATEI~ODORY-FEJI~ problem which is intimately connected to the problem (S). His Theorem 2.3 yields a parametrization which is related to Corollary4 of this paper in such a way how the TOEPLITZ case parametrizations of [20] and [lo] are connected. Later we shell discuss this connection more explicitely. Further, we s h d also show the interrelations to the theory of stationary sequences, and we shall give a Sm~o type entropy formula.

4 1 *

1. Preliminaries

Let p and Q be positive integers. The MOOEE-PEN~OSE inverse of A E APxq i ig denoted by A+. We shall often' use the following' identities for S E ,nc,' (see e.g. RAO/MITRA (16, Ch. 1-31);

@ =p=s+T/;F=ps+, p = SfZ = s

s s + = s + s = ~ ~ = ) / s + ~ = . ~ s ) T s + = ~ s + ~ . and

Fritzsche, A Schur Type Matrix Extension Problem 259

The following lemms (see e.g. EB?~~OV/POTAPOV [9, p. 881) is basic for our considera- tions. We shdl formulate it using M O O R E - ~ O S E inverses.

Lemms 1. Let

(a) It hold3 S E (i) A E A:, (ii) AA+B* = B*, (iii) E - BA+B* E A:.

(iv) E E At, (v) E E + B = B ,

if and d y if the following three conditions are satisfied:

[b) It h o h S E A:+, if and only if the folbwinq three ccnzditim are satisfied:

(Vi) A - B*E+B E &:-

Further, using the singnlar value decomposition of matrices one obtains the follow- ing result :

Lemma 2. (a) Let A E ,?. Then it holds

A(I, - A*A)+ = (1, - a*)+ A and

A*(I, - M*)+ = (Iq - A*A)+ A*.

(b) Let A be a c o m p h p x q matrix such that I , - AA* E A!:. TAen it irdds Iq - A*A E 4: and

A f m = f w A ,

A* 1- = fm A*,

A ~ - = V ~ A

A*I- = [&TiiW A*. and

Finally, denote A; the set of all positive Hermitiirn p x p matrices.

2.

Throughout this section, p , q and 7a denote fixed positive integers. Asenme A,, A,, . . ., A,, E

The solution of the problem (S)

Using (0) we define for j E (1,2, . . ., n}

Dj := Di(A0, All . . ., Aj-J,

Pj := Ij, - D#,

Qj := Ijq - DfDj.

I?*

260 Math. Nachr. 134 (1987)

It is easyto see that fork E {1,2, ..., R - 1)

and

This implies

and

Obviously, we have Pl = 1, and Q1 = rl. The set of all np x nq matrices B of the shape

B = DU(A0, 4, * * -, 4 - 1 )

where 4, .. ., Awl E APxq is denoted by Ap,q;n.

Efibtsohe, A Schur Met& Extension Problem 261

Lemma 3. Let D,, = DM1{AB, Al, . .., 4,) E dB,Q;r+t. Then the f o f i n g s t d m are eq.uivdent: (i) PNCl E J-E+1Ip.

(W Qn+1 f d $ + l j g *

(iii) The folbwirag three rn- are sah'sfk3:

(6) P n E A$, (7) PJ';D,,X$ = Dp: , (8) In+, E *

(iv) The following three conditions are satisfied:

(9) Qn €A$,

(11) Tn+, f C q .

(10) QnQn+D:yn = W n

Proof. Obviously, we have (i) if and only if (ii) holds. h m Lemma 1 (a) and (4) we obtain that (i) is fulfilled if and only if (iii) is satisfied. Finally, Lemma 1 (b) and (5) yield that (ii) and (iv) are equivalent. I

Remark 1. (a) Let Dn = Dn(Ao, A,, ..., An-J E A p , q ; n . Then

det P,, = det Q.

rank P, = rank Q,, + %(p - 4). and

(b) As~nae Shut Owl = Dwl{&, All ..., A,) E C A ~ , ~ ; , , + ~ fulfils (7) . T7m in dew of Lemma 2 in [lo] it h l d a

det Pel = det P, det

rank Pn+l = rank P, + rank I,, . and

(c) ASSUW tW Ds+1 = Dn+1(Ao, A,, - t &) E dp,q;n+l fdfa (10). Then cxecordi.ng to Lernm42 in [lo] it holds

and det Qn+, = det Q,, . det rNC,

rank Qn+* = rank Q,, + rank ~ i + ~ .

Remark 2. 8uppme tMt D,, = D,,(Ao, A,, . . ., A,,-,) E AP,$;* satiafh (6). Then n

+I f-1 detP, = det Qn = n d e t 1, = D d e t q

and m I

rankP,,=~'rankZf, rankJZ,=zCrj. j-1 j-1

Now we shall give necessary and sufficient conditions for each of the conditions (7), (8), (10) and (11). Let Bp.g;el be the set of all matrices DMl E cRp,q;r+l such that (6) holds.

262 Math. Nachr. I34 (1987)

L 6 m 4. Dn+l = Dn+,(A,, A,, - * - 9 An) E Bp,q;n+l- Thm (7) holds if 0d3 if (12) rnv;(A: - mr) = A: - rnr . Proof. First assume that n 2 2. From Lemma 2 we obtain

(13) P;Dn = ( Inp - DnDZ)+ Dn = Dn(Inq - D:Dn)+ = DnQ:.

Obviously, me have

(14) PnDn = D,, - DnD:Dn = DnQn-

Now suppose that (7) holds. Then the formulas (13) and (14) yield

(15) DZD,Z: = D:P,P;D,,ZZ = D:P,,D,,&;Z: = D:D,Q,,Q~x~.

According to Lemma 3 and (6) we get (9). Hence,

Qn = QnQnQ; = (Inq DfDn) QnQ;.

This implies

(16) (Inq - D:Dn) 2: = Qnz: = ( I n q - DP.1 QZn?2':4 *

The addition of the equations (15) and (16) yields

(17) 8nQ:g = 4- Conversely, we now suppose that (17) is satisfied. We show that (7) follows. In fact, &pplying (13) and (14) we get

P n P i ~ n z ~ = PnD,,Q:~Z = DnQnQ:~Z = Dnzf.

Therefore, (7) and (17) are equivalent. Now we shall see that (17) holds if and only if

is fulfilled. Setting

Fritzmhe, A Sahur Type Matrix Extension Problem 263

.From (6) and Lemma 3 we get

(20) P*lPLDn-le--l= Dn-14-1.

Using (20) and

we obtain from (19)

This implies

From (22) we see that (17) and (18) are equivalent. Finally, we show that (18) holds if and only if (12) is satisfied. First suppose that (18) is fulfiIlecl. Then there exist a q x q matrix X and a (n - 1) q x q matrix Y such that

from (5) and (24) we get

Thus in view of (12) we obtain

Henee, (18) is stitisfled. Therefore, (12) and (18) are equivdent. In the ~ 8 8 % n 2 2 the proof is cbmpleb. analogons~y, we can treat the Simple c898 n = 1. we omit the details. I

264 Math. Nachr. 134 (1987)

Let us present a further result which can be proved similady.

Lemma 6.8uppse D,, = Dw1(A0, A,, . . ., A,) E Bp,q;*l. men (10) ir0ld.s if and only if

(26)

and (10) hold if and only if there exists a p x q matrix U such that

lnZi(An - m,) = A, - m,.

Lemma 6 . Assume Dn+, = D,,+,(A,, Al, . . ., A,) E .3p,q;n+l. Then the two c m ? i t i m ( 7 )

(27) An = m n + u G - Moreover, in this w e

(28) An =mn + GPUE lwlds where

(29) p u : = Z,Z~Ur~r,.

If for V E APxq the e q w t i m

i s satisfied then it follows pv = pu. Proof. From (6) and Lemma 3 we see that I , E A: and r, E A:. First assume

that (7) and (10) are fulfilled. Applying the results in Lemma 4 and Lemma 5 we ob- tain

A, - m, = l,$(A, - m,) = l,Zi[r,r,'(AX - m:)]* =mJK

where u:= (A, - m,)

If 'CT E & x p fulfils (30) then

An = m n + fi fC+ fi 'V 16 fi = mn + it P ~ G and

P, = fi vfKfL+ = ilnf (An -mn) 1/;;;;' = yrfi u fi = p v .

Conversely, we now assume that a p x q matrix U exists such that the representation (27) is satisfied. Then we get

z,z:(A, - m,) = Z,Z; fi ~ f i = u~J;;;; = A,, - m,.

Analogously, we obtain (12). From Lemma 4 and Lemma 5 we see that (7) and (10) hold. This completes the proof.

Remark 3. Let D,, = D,,+,(Ao, Al, . . ., A,,) E d p , q ; n + l . 8uppose P, E A!:p. .According to the Remarks 1 and 2 it i s exwy to see that det 1, =I= 0 and det rn + 0 hold. Therefore, pu = U and, m q w n t l y , there i s a uniqw p x q d r i x U such that (27) i s fulfilled.

I

Fritzache, A Sohttr Tspe &Wrix Extension Problem 265 Fritzache, A Sohttr Tspe &Wrix Extension Problem 265

The following lemma plays a key role. The formdm for the development of the semi-radii seem to be charaderistic for certaiu dwea d one step extension procedures (compare D r n / G m i ~ m [S, Lemma 5.51 and F a r n s m / I C r r ~ [lo, Lemma 61).

= Dn+l(Ao, Al, . . ., A,) E Bp.q;n+l. Assum th# there eziskr a p x Q matriz U SUcJr tha.t (27) is satisfied. Then

Lemma 7 . Suppose

Moreover, tk folkwing conditions are equivalent:

Pro of. First we consider the case n = 1. We have

(33)

Applying Lemma 2 we obtain

This implies

In view of (33) it follows

(34)

To check ZI Z 4 note that

(35)

Now assume n 2 2. Applying (21) and (25) we get

(36)

266 Moth. Nachr. 184 (1987)

&om (13), (14) and QnD: = D:Pn we have

(37) QJQ'iDnQn = Q J W n Q n + Q n = D z P n D n Q , + Q n = D Z D n Q n

= (Inq - Qn) Qn.

Using (37) and (25) we obtain

s : Q n D X D n Q n S n = s*,(Inq - Qn) QnS,

= [SZ - (Tn, 0)]

From (37), (21), (25) and (13) it follows

Mhche, A Schur Type Metrix Extension Problem 267

Because of (%), (34, (38) and (39) the fornula (31) is proved. In a similar way one proves that (32) holds. In order to complete the proof we need to show the equivalence of the conditions (if, (ii) and (iii). First assume that (i) is satisfied. Then (31) yields

0 =z 1, - zn+1 = ( f i ~ u ) (1 /T~o)*

pu = f < p u = 0 = 0 .

o = ~ n - r m + l = ( ~ U f i ) * ( ~ u f C ) .

pLf=pJ;;;)/F7;f = O K =o .

A, = mn +

Hence,

Berefore, (ii) holds. Now suppose that (ii) is fulfilled. Then from (32) we get

This implies

Then from (27) we obtain

U:ur:rn fi =mn + K.~t,fi = mn.

Therefore, (iii) holds. Finally, assume that (iii) is W e d . Then

)I?;Uf i=An-mn=O. Hence

p u = W f i u 6 Q = o . Then (i) follows from (31). This completes the proof. I

hm-k 4. lkt Dn = Dn(A0, Al, . . - 9 A,+J E 0 4 ~ ~ ~ ; ~ . S U ~ S ~ that (6) hMe. According to Lemma 3 and h m 7 we have

O 5 Z, #,,-I S - - * 5 11 = Ip - A& a d

OSr,,Sr*1 S.* . I r ,= I , -A&.

Proposition 1. LCt D,,+, = D,,(&, A,, ..., A,,, 4) E 3p.g;,,+l a d & E APxr Sup pose that there ex&p x q 4lardriccs U a d 8 suci5 tirat the repeseniatiow (27) and

268 Math. Neck. lw (1987)

#

Proof. We begin by verifying that [a) is fulfilled. l?irst assume that 1,+1 I; 1,,+1

holds. Applying Lemma 7 we have

1, -fCnpupg

fC ( P U P : - PEP;)

pup% - p s p z = fa,' fC (pup$ - pi&) YCn $2 E Jq 3

= tn+, s L+l= in - f l p d f i n

This implies

€ Jq - It folIows

1-e.,

(40) PZP% s PUP: holds. Conversely, now suppose that (40) is satisfied. Then from Lemma 7 we obtain

in+, - ln+1 = fi ( ~ p - P C P ~ ~ ) I t - fi [ ~ p - PUP:) dln = di;; ( P U P % -Pap;) fi € Jq.

Thus (a) is proved. The proof of [b) is similar. We omit the details.

d = Dwl(A0, Al, . . ., An-1, An) E .Bp,,,;n+l. Suppose that (6), (7) and (10) ho7d. Denote DHl := Dncl{Ao, Al, . . ., A,,+ m,). Then l,, I in+, and T.+~

Pro of. Choose U = 0 in Proposition 1. I

Proposition 2. Let DH1 = D,+,(Ao, Al, .\., A,) E .Bp,q;,+l. Assume that there ezists a p x q matrix U sueh that A,, can be represented in the fm (27). Then the f01bwin.g state- ments are equivalent:

I

Corollary 1. Let

5 +n+l-

(i) p*1 E AE+,,,. (iii) L1 €A?". (iv) T,,+~ E .A&,'- (v) POP% S 4&,+. (6) P%Pu 5 TAr,'.

(3 Qn+1 E uflz+l)q-

Proof. Prom Lemma3 and Lemma6 we obtain the equivalence of the conditions [i), (ii), (iii) and (iv). Applying Lemma 7 we have

SC - pup; = fi fC+ (I, - PUP:) lJln fC+ = iC+ Zn+l /C+ and

fi [u: - PUP:) fi = l n + l -

Thus we can gee that [iii) holds if and only if [v) is satisfied. Analogously, one can show that (iv) and (vi) am equivalent. I

Now we are able to formulate our main result of this paper.

Fntzsche, A 8ahw Matrix Extension Pmblem 269

wkre U ~ s e o m c ~ ~ q ~ ~ ~ ~ t ~ U U * ~ I , ~ l o l d e . Proof. First suppose that (41) is satisfied, According to Lemma 3 and Lemma 6

there exista a p x q matrix E' such that (30) is fulfilled. Put U := py . Then (42) holds. We have

pu = ln l , tPpm+~n = PV.

Note that Z,Z: is an idempotent and Hermitian matrix. From (41) and Proposition2 we obtain

uu* = p"p; r 1.r; 4 Ip.

Conversely, now suppose that A, admits a representation (42) where UU* S Ip. Clearly, r,r: I,,. From

Inl; - prrtp; = 1.1; (Ip - UT,Tf U*) E,C1, = 2,ZRf{(Ip - UU*) + U(l , - r&) Lpc] (l,l;)*

then we get pup& S 1.1.'. Proposition 2 yields (41). This completes the proof. GWONIMUS El31 had been considered the problem (S) in the special case p = q = I.

He s u p p e d that det PWl =!= 0 snd P, = @j&)j,k-o,l....,,,-l E &c,Z and showed that P,, E

I

if and onIy if A, belongs to the disc

det P, det P,l

2 E c: (2 - Eml 5

where A1 POO POI PO.^-*

A2 PlO PI1 ...P1.a-a

As-1 P~(-x.o ~ n - 2 ~ 1 * * . ~ n - 2 . ~

0 Pa-1.0 Pn-1.1 **.lla-1.n-a

8, := det (-'In P,+ det (; ; i i ) From Remark 1, a well-known S m determinant formula for block matrices and straightforward calculation it is essy to see that 6, = m, and

det P, det PW1 ' G 6 =

Corollary 2. L& A,, A,, . . ., A,, be p x q mdrices. Pd DbCl := D,,+,(&, Al, .. ., A,,). E .Al$ and &re exisba a sequence (Uf)in_l Then it hoMs (41) if a d a l y if Ip -

from Apxp mch that

and Ip - U j q E A ~ , j E {1,2, ..., n},

(43) A, = mj + fi tlj fi, j E ti, 2, ..., n).

270 Math. Nachr. 134 (1987)

Proposition 3. Let D,, = D,,(Ao, A,, . .., A,,) E Bp.p;lH1. Suppse that there exists a p x q matrix U such t k t tAe representation (42) hokEs. TAen

rank P,+l = rank Pn + rank (ZnEz - pup;)

rank Q,,+, = rank Q n + rank (TnTz - p g p v ) . and

Proof. Using Lemma 6 end Remark 1 Proposition 3 can proved analogously to 1 Proposition 3 in [lo]. We omit the details.

Corollary 3. If det P,, =I= 0 then

r e d P,,+l = np + rank (Ip - UU*)

rank Qn+, = np + rank (Iq - U*U) . and

Theorem 2. Let Df i l = D,,+l(Ao, Al, . . ., A,) E Bp,,,;n+l. Then the following two c o d - tiom are necessary and sufficient for

(i) det (Inp - D,$:) + 0. (ii) There exists a p x q mat& U auch that (42) and Ip - UU* E .Alp' hold.

Proof. First we show the necessity. From PWl E Jl(>n+l)p and (4) it follows P, E && immediately. Then (i) holds. Because of Theorem 1 there exists a p x p matrix U such that (42) and Ip - UU* E ApL hold. Prom (i) and Remark 3 we have det 2, =+ 0. Hence,

- D,,+lPar~+l E At;. l ,p:

(44) &a; -pup: = Ip - uu* Applying Proposition 3 we have

rank (Ip - UU*) = rank P,,+I - rank P,, = (n + 1) p - nip = p

Therefore, we obtain the statement (ii), too.

Moreover, from (i) and Remark 3 we get (44). Then Conversely, suppose that (i) and (ii) hold. Then from Theorem 1 we get P,+l t: &$ l)p.

rank (ZnZn+ - pop$) = rank (Ip - UU*) = p .

Using Proposition 3 we see that det P,, 4 0. Thus P,, E &$+,,,. This completes the

D,,(Ao, A,, ... )An). proof. I

CoraltsrJr 4. Let Ao, A,, ...,A, be p x p matrim. Put Then it Jbolds

I(n+l)p - Dn+*D:+l E d < + i ) p

if and only if Ip - a' E such that

and there &ts a sequence (Uj)&pl of p x q matrices

Ip - U f q E , j E {1,2, . . ., nl , and the relatioras (43) are fulfiuea. I n th& case the se~iuence ( U,)&l is unique.

Fritzsche, A Schwr Type Mstrix Extension Problem 271

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