The Electronic Journal of Linear Algebra.
A publication of the International Linear Algebra Society.
Volume 6, pp. 31-55, March 2000.
ISSN 1081-3810. http://math.technion.ac.il/iic/ela
ELA
ON TWO{SIDED INTERPOLATIONFOR UPPERTRIANGULAR MATRICES
DANIEL ALPAY
y
AND VLADIMIR BOLOTNIKOV z
Abstrat. ThespaeofuppertriangularmatrieswithHilbert{Shmidtnormanbeviewed as anitedimensionalanalogue ofthe HardyspaeH2 of theunitdisk whenoneintroduesthe adequatenotionof\point"evaluation.Abitangentialinterpolationprobleminthissettingisstudied. ThedesriptionofallsolutionintermsofBeurling{Laxrepresentationisgiven.
Keywords.Interpolation,matries.
AMSsubjet lassiations.47A57,47A48
1. Introdution. Inthispaperwepursueourstudyofbitangentialinterpolation in analogues of the Hardy spae of the unit disk H
2
. To start with we reall the lassialsetting whihhasbeenonsideredin [8℄.
LetH mk 2
denotetheHilbertspaeofmk{matrixvaluedfuntionsoftheform H(z)=
1 X j=0
H j
z j
; 1 X j=0
TrH j
H j
<1;
endowedwiththeL mk 2
innerprodut hH; Gi
L mk 2
=hH; Gi H
mk 2
= 1 2
Z 2 0
TrG(e it
)
H(e it
)dt= 1 X j=0
TrG j
H j
: (1)
TheH mk 2
-funtionsareanalytiin theopenunit diskD andhaveboundaryvalues almost everywhereon theunit irleT. The spae H
mk 2
is aHilbert module (see, e.g,[18℄ andSetion2of [1℄)withrespettotheHermitianmatrix-valuedforms
fH; Gg H
mk 2
= 1 2
Z 2 0
H(e it
)G(e it
)
dt (2)
and
[H; G℄ H
mk 2
= 1 2
Z 2 0
G(e it
)
H(e it
)dt (3)
andhasthereproduingkernelpropertywiththekernels k
^ !
(z)= I
m 1 z!
and k 4 !
(z)= I
k 1 z!
Reeivedbytheeditorson13August1999.Aeptedforpubliationon3Marh2000.Handling Editor: DanielHershkowitz.
y
Department of Mathematis, Ben{GurionUniversity ofthe Negev, Beer{Sheva, 84105, Israel (danymath.bgu..il).
z
ELA
32 DanielAlpayandVladimirBolotnikov inthesense that
H(!)A
=
H; Ak 4 !
H mk 2
and A
H(!)=[H; k ^ !
A℄ H
mk 2 (4)
for every hoie of a point ! 2 D and of amk matrix A. Note that the latter relationsexpressCauhy'sformulaforH
mk 2
-funtions.In[8℄weonsideredageneral bitangentialNudelmantypeproblemforH
mk 2
funtionswithnormonstraints. The multivariable analogueof the problem referred to in theprevious paragraph in the settingofthepolydiskwasonsideredin [4℄.
It iswellknown(see [10℄, [11℄, [12℄, [15℄) that thereare deepanalogies between analytifuntions and upper(or lower)triangular matries. In[2℄welookedat the analogue of the problem referred to in the previous paragraph, for double innite uppertriangular matries. Inthis paperwefous onthe aseof nitematries. In fat, oneouldtry toobtain theaseof nite matries from ourpreviouspaper[2℄ using time varying oeÆient spaes in the spirit of [13, Setion 12℄ or [14℄. This doesnotseemto usnatural,sinetheproblemonsideredhereisnitedimensional; furthermore,theapproahpresentedinthispaperispurelyalgebraiandmuhmore expliit.
Other situations are also possible, suh asthe ase of lowertriangular integral Hilbert Shmidt operators (the ontinuous timevarying ase analogueof [2℄). This wasarriedoutin [3℄.
ConsiderX mk
,thesetofallmkmatrieswhihisaHilbertspaewithrespet totheinnerprodut
hH; Gi=TrG
H (H; G2X mk
); (5)
whihistheanalogueof(1). A matrix H = [h
i j℄
j=1;:::;k i=1;:::;m
is alled diagonal ifh ij
=0 for i 6=j. It is alled upper(lower)triangularifh
ij
=0fori>j(i<j,respetively). ThesymbolsD mk
, U
mk and L
mk
willbeusedforthespaesof diagonal,uppertriangularand lower triangularmkmatries, respetively.
LetZ m
denote themmshiftmatrixdenedby
Z m
= 2 6 6 6 6 6 6 4
0 1 0 0 0 1
. .
. . . . .
. .
. .
. .
. .
0 1 0 0
3 7 7 7 7 7 7 5 (6)
andletZ k
bethekk shiftmatrixdened similarly. We denotebyp +
;p 0
;p the orthogonal projetions of X
mk
onto U mk
Z k
, D mk
, Z m
L mk
, respetively. We alsousetheprojetionsof X ontoitsupperandlowerparts,anddenotethem by
ELA
Interpolationforuppertriangularmatries 33 respetively. ThespaeX
mk
isaHilbertmodulewithrespettoHermitian matrix-valuedforms
fH; Gg X
mk =p
0 (HG
) and [H; G℄ X
mk =p 0
(G
H) whiharetheanaloguesof(2)and(3).
Foraxed H 2U mk
theseforms makesenseforkkandmmmatriesG, respetively,anddenetwodierent\evaluation"mapsforuppertriangularmatries by
F ^
(W)=p 0
(I
m WZ
m
) 1
F
and F 4
(V)=p 0
F(I
k Z
k
V) 1
; (7)
whereF 2U mk
andW andV arediagonalmmandkkmatries,whihusually playtheroleofpointsinthenonstationarysetting. ThetransformationsF
^
(W)and F
4
(V)arenonommutativeanaloguesofthepointevaluation(4). Wealsodene F
℄
(W;V)=p 0
(I
m WZ
m
) 1
F(I k
Z k
V) 1
Z k
: Remark 1.1. It follows from the denition (6) of Z
m
that for everyhoie of W 2D
mm
,thematrixI m
WZ m
islowertriangularwithalldiagonalentriesequal to1. Inpartiular,det(I
m WZ
m
)=1andthematrixisinvertible. Fromthesame reasonthematrixI
k Z
k
V isinvertibleforeveryhoieofV 2D k k
. Notethatifm=k,thenforanyF2U
mk
thereexistuniquelydeneddiagonal matriesF
[j℄ and F
fjg
whihsatisfy F
[j℄ =Z
j m Z
j m
F [j℄
; F fjg
=F fjg
Z j k Z
j k andaresuhthat
F = m 1
X j=0
Z j m
F [j℄
and F = k 1 X j=0 F
fjg Z
j k :
Thelatter\polynomial"representationsallowsustoexpressevaluations(7)as
F ^
(W)= m 1
X j=0
(WZ m
) j
Z j m
F [j℄
and F 4
(V)= k 1 X j=0 F
fjg Z
j k
(Z k
V) j
: (8)
Theseformulasappearin[7℄. Ifm>k(m<k),therst(the seond)formulain (8) is true. In general, for m <k (m > k) therst (the seond) formula in (8) is not valid,andforthisreasonweshalluseformulas(7).
ELA
34 DanielAlpayandVladimirBolotnikov Problem 1.2. Given matries
W j
2D `j`j
; j
2D `jm
; j
2D `jk
(j=1;:::;n) V
i 2D
riri ;
i 2D
rim ;
i 2D
rik
(i=1;:::;`) and
ji 2D
`i`j
, ndall H2U mk
suhthat
j H
^
(W j
)= j
; (H i
) 4
(V i
)= i
and j
H i
℄
(W j
;V i
)= ji (9)
for j=1;:::;n andi=1;:::;`.
Thepaperonsistsof sevensetions. Toset theproblem preiselywerstneed somenotationsanddenitions. Thegeneralproblem(whihinludes Problem1.2as apartiularase)isstatedin theseondsetion. Thedesriptionofallitssolutions (formula (31)) is presentedin thethird setion. Theproofrelies on the analysis of twospeialases,namelyrightsidedandleftsidedinterpolationproblems,whihare onsideredindetailsrespetivelyinSetions4and5. Thegeneraltwosidedproblem isstudiedinSetion6. Thelastsetiondealswiththestrutureoftheminimalnorm solution.
2. Formulationoftheproblemsandpreliminaryremarks. Inthissetion weintroduethebitangentialinterpolationproblemtobestudied,andwhihinludes Problem 1.2 asa partiularase. Given twosetsof positiveintegersf`
i
gand fr j
g, let
n R
=r 1
+:::+r t
; n L
=` 1
+:::+` s
; (10)
letZ rj
andZ `i
betheshiftmatriesdenedvia(6)andlet Z
=diag(Z r
1 ;:::;Z
r t
)2C n
R n
R
and Z
=diag(Z `
1 ;:::;Z
` s
)2C n
L n
L : Therelations
Z
(I Z Z
)=0; Z Z
D=DZ Z
; (11)
(I Z
Z
)Z
=0; Z
Z
F=FZ
Z ; (12)
(I Z Z
)(I Z
D) 1
=(I Z Z
); and (13)
(I Z
Z
)(I Z
F) 1
=(I Z
Z ); (14)
whihholdforeveryhoieofblokdiagonalmatriesD2D nRnR
andF 2D nLnL
, willbeuseful.
In the lass U mk
weonsider the interpolation problem whose data set is an orderedolletion
=fC +
; C ; A
; A
; B +
; B ; g (15)
ofsevenmatries C 2D
mnR
; C 2D k nR
; B 2D mnL
; B 2D k nL
ELA
Interpolationforuppertriangularmatries 35 A
2D
nRnR ; A
2D
nLnL
; 2D nRnL
: Remark 2.1. Byabuse of notation, by A
2 D
nRnR and A
2 D
nLnL we meanthatA
andA
aren
R n
R andn
L n
L
matrieswithdiagonalblokentries (A
)
ij 2D
` i
` j
(i;j=1;:::;s) and (A
) ij
2D r
i r
j
(i;j=1;:::;t): Similarly, byC
+ 2D
mn R
wemeanthat C +
isablokrowandeverymr i
blok is diagonal (in other words, we reserve the symbols n
L and n
R
to indiate blok deomposition of the underlying matries). We shall refer to suh matries as to blok diagonal. The same onvention holds with U and L instead of D for blok upper triangular andfor bloklower triangular matries. Theother notations(suh asD
mnR orD
k nL
)shouldbelear. Wesaythat thedata(15)isadmissible if
Span n
Ran
(A Z
)
j C
; j=0;:::;n R
1 o
=C n
R ; (16)
Span n
Ran
A Z
j B
+
; j=0;:::;n L
1 o
=C n
L ; (17)
= Z
Z ; (18)
andtheSylvesterequalityholds A
Z
Z
A
=B
C B +
C +
: (19)
WedenotebyIPthefollowingtwo{sided interpolationproblem. Problem 2.2. Given anadmissible dataset ,nd allH 2U
mk
suhthat p
0 n
HC ( I Z A
)
1 o
=C +
; (20)
p 0
n H
B
+ ( I Z
A
)
1 o
=B ; (21)
P 0
n I A
Z
1 B
+
HC (I Z A
)
1 Z
o
= ; (22)
whereinthe rstequation,p 0
istheorthogonal projetionofX mnR
ontoD mnR
,in the seondequation p
0
denotes the orthogonal projetion of X k nL
ontoD k nL
,and nally, inthe thirdequation,
P 0
=p 01
p 0s andp
0i
isthe orthogonal projetion of X `i`i
ontoD `i`i
.
Note thatby Remark 1.1and in viewofthe blok strutureof A
, A
, Z
and Z
,thematriesI Z A
andI Z
A
areinvertible.
Remark 2.3. Conditions (20) and (21)generalize the Nevanlinna{Pik ondi-tions(9)andoinidewiththelastonesforthespeialhoieof
ELA
36 DanielAlpayandVladimirBolotnikov
A
Thenext lemma showsthat onditions(20)and (21)ontain moreinformation aboutasolutionH oftheinterpolationproblem.
Lemma 2.4. LetH belong toU mk
andsatisfy (20),(21). Then q
and
p
and therefore,
q
Thersttermontherighthandsideoftheequality (I Z isstritlyblokuppertriangular,andthus, foreveryuppertriangularH,
p Sinetheoperator(Z
isblokdiagonal,itfollowsfrom(20)that p
isblokdiagonal, (Z Makinguseofthethreelastequalitieswededue(23)from(25):
ELA
Interpolationforuppertriangularmatries 37 Takingadvantageoftherelation
p n
H
B +
( I Z
A
) 1
o =
nL 1 X j=0
p
0 n
H
B +
(I Z
A
) 1
Z j
o Z
j ; itanbehekedinmuhthesamewaythat (21)implies(24).
Corollary 2.5. Conditions (20) and (21) are equivalent to (23) and (24) re-spetively.
Indeed,applyingp 0
to bothsidesof(23)and(24)weobtain(20),(21). Therest followsfromLemma 2.4.
Sometimesitwill beonvenientto useonditions(20) and(21)in thefollowing \adjoint"forms
P 0
n (I A
Z
)
1 C
H
o
=C +
and P 0
n I A
Z
1 B
+
H o
=B
: (26)
Remark 2.6. Conditions(20),(21)areequivalenttoonditions(26). Proof. TakingH 2U
mk
in theform H = k 1 X j=0
H j
Z j k
withH j
2D mk
,weget
p 0
n
HC (I Z A
)
1 o
= k 1 X j=0
H j
Z j k
C (Z A
)
j ;
P 0
n (I A
Z
)
1 C
H
o
= k 1 X j=0
(A Z
)
j C
Z
j k
H j
: Comparingrighthand sidesintwolast equalitiesweonludethat
p
0 n
HC (I Z A
)
1 o
=P
0 n
(I A
Z
) 1
C
H
o
andtherefore,(20)isequivalenttotherstonditionin(26). Theequivaleneof(21) andtheseondonditionin(26)ishekedin muhthesameway.
Remark2.7. TheSylvesteridentity(19)followsfrom(20){(22)andistherefore, aneessaryonditionfortheproblemIPtobesolvable.
Proof. Firstwenotetheequality P
0 Z
M(I Z Z
)
=0; (27)
whihholdsforeveryM 2X nRnL
. Indeed,takingM in theform M =
nL 1 X j=0
Z j
M j
+ nR 1
X i=1
M nL 1+i
Z i
with M `
2D n
R n
L ; weget
Z
M(I Z Z
)=Z
nL 1 X
Z j
M j
(I Z Z
)+
nR 2 X
Z j
M j
! Z
(I Z
Z
ELA
38 DanielAlpayandVladimirBolotnikov
andometo (27),sinethersttermontherighthand sideinthelatterequalityis stritlybloklowertriangularand theseond termisequaltozerodueto(12).
Let H belong to U mk
and satisfy (20){(22) (or equivalently, (22) and (26)). Uponapplying(27)to
M= I A taking into aount that the matrix A
is blok diagonal and making use of the equality
Z
nLnL
,weget Furthermore,
A
andC areblokdiagonal,itfollowsfrom(26)that A andthisompletestheproof.
Itwill beshownin Setion6thattheproblem IPalwayshasasolution. Letus denotebyIP
theproblemIPtowhihhasbeenaddedthenormonstraint hH; Hi
X mk
def = Tr(H
H) (28)
ELA
Interpolationforuppertriangularmatries 39 ThevaluekH
min k
HS
(as wellasH min
itself)dependsonlyontheproblemdata(15) and theondition kH
min k
HS
is neessaryandsuÆient fortheproblem IP
to besolvable. The expliit formulafor H
min
and a desriptionof all solutionsof the problemIP
willbegiveninSetion6. WedenotebyIP
R (
R
)theright{sidedproblem (20)(i.e.,whenonditions(21) and(22)arenotin fore)towhih hasbeenaddedthe\matrixnorm"onstraint
fH; Hg def
= p 0
(HH
) R (29)
forsomepreassignednonnegativematrix R
2D mm
. Similarly, wedenote by IP
L (
L
)theleft{sided problem(21)to whih hasbeen addedthematrix\norm"onstraint
[H; H℄ def
= p 0
(H
H) L (30)
forsomepreassignednonnegativematrix L
2D k k
.
Itturnsoutthattheonstraints(29)and(30)donotsuittheleft{sidedondition (21)andtheright{sidedondition(20)respetively. Thatiswhyweonsider atwo{ sidedproblemonlyunder theonstraint(28)whihonaountof
hH; Hi=Tr fH; Hg=Tr [H; H℄ suitstoleftonditionsaswellastorightones.
3. Statement of the main result and rst formulas. The main result of thepaperisnowstated:
Theorem 3.1. The setof allsolutions ofProblem IPisgiven by H=H
min +
L h
R (31)
whereH min
2U mk
isthe minimalHilbert{Shmidtnormsolution, R
2U
(k +nR)k and
L 2U
m(m+nL)
are twopartialisometries with upper triangular blokentries, built fromthe interpolation dataandhisafree parameterfromU
(m+nL)(k +nR) . Inthis setionweonstrutexpliitly
L and
R
(seeformulas(43) and(42)), while the formula for H
min
will be given in Setion 7. Webegin with preliminary lemmas.
Lemma 3.2. The Steinequations IP
R A
Z
IP
R Z
A
=C
C and IP L
A Z
IP L
Z
A
=B +
B + (32)
are uniquely solvable, and that their solutions are the blok diagonal matries given by
IP R
= m 1
X j=0
(A
Z
) j
C
C (Z
A
) j
2D nRnR
IP L
= m 1
X (A
Z
)
j B
+
B +
(Z
A
) j
2D nLnL
ELA
40 DanielAlpayandVladimirBolotnikov
Conditions (16)and(17)areneessaryandsuÆient thatthe operatorsIP R
andIP L areboundedlyinvertible.
Theproofisstraightforwardandwillbeomitted. Lemma3.3. LetIP
R andIP
L
aresolutions oftheSteinequations(32). Thenthe matries
Q
areorthogonalprojetions.
Proof. Thematriesdenedin (33){(36)areevidentlyselfadjoint. Toshowthat Q
R
isaprojetion,westartwiththeequality (I A
whihisanimmediateonsequeneoftherstequationin(32). Itfollowsfrom(33) that
Q
where,onaountof(37), L=(I Z suessively,weget
L=(I Z whih togetherwith (38)leads to Q
2
is a projetion, we rstnotethat
ELA
Interpolationforuppertriangularmatries 41 Indeed,makinguseof(32),wetransformtheleft handsideas
A andobtain(39),sineinviewof(11),
Z
TheequalityT 2 R
=T R
followseasilyfrom (39)andrelations
(Z
whihareonsequenesof(12). EqualitiesQ 2
areveriedquite similarlywiththehelpof(11){(14)and
I A
whihinturn, followsimmediatelyfromtheseondequationin (32). Lemma 3.4. The matries
R
p(m+nL)
denedby
and
respetively,areblok upper triangular andsatisfy
and
whereQ R
arethe orthogonalprojetionsdenedin(33){(36). Proof. Theuppertriangular strutureof
R and
L
followsimmediately from theirdenitions (42)and (43). Theupper triangularstruture hereismeantin the sense of Remark 2.1: aordingto partitions (10),
R and
L
ELA
42 DanielAlpayandVladimirBolotnikov where
M=IP Takingadvantageofthersttworelationsin(12)(withD=IP
1
whihbeingsubstitutedinto(46)leadstotherstequalityin(44). Furthermore,on aountof(42),
Substituting (37)into thethird termon therighthand side of(47)and takinginto aount(34),weget
whihprovestheseondequalityin(44). Theequalities(45)areveriedinmuhthe samewaywithhelpof(43)and(41).
Remark 3.5. The matries R
and L
dened via (42) and (43) admit the representations
ELA
Interpolationforuppertriangularmatries 43
Proof. Inviewof(34)and(48),
andtherefore,
T
whih proves(48). The representation (49) is veried in muh the same way with helpof(36)and(43).
4. Right{sided problem. InthissetionwedesribeallH 2U mk
satisfying ondition (20). We rst exhibit a partiular solution, whih will be shown in the sequelto beoftheminimalHilbert{Shmidtnorm.
Lemma 4.1. LetH R
bethe blokuppertriangular matrixgiven by H
satisestheondition(20)and fH
Proof. Inviewof(37), H
andsine
C
mnR Z
; (53)
ondition(23)(whih isequivalentto (20)byCorollary2.5) followsfrom (52). Fur-thermore,multiplying(52)byIP
1 R
C +
ELA
44 DanielAlpayandVladimirBolotnikov
By(53),theseondtermontherighthandsideof(54)isstritlyblokupper trian-gularwhiletherstoneisuppertriangularandthus,
fH R
; H R
g def
= p 0
(H R
H R
)=p 0
n C
+ ( I Z
A
)
1 IP
1 R
C +
o =C
+ IP
1 R
C + ; whih provesthe rst equalityin (51). Theseond equality is animmediate onse-queneoftherstone.
Notethat everyH 2U mk
satisfyingtheondition(20)is oftheform H =H
R + (55)
where isanelementfromU mk
suhthat p
0 n
C ( I Z A
)
1 o
=0: (56)
Lemma 4.2. The matrix belongsto U mk
and satises (56)if and only if it admitsa representation
= b H
R (57)
where R
isgiven by(42)and b H 2U
m(k +nR) . Proof. Let
b
H beinU m(k +n
R )m
andlet beoftheform(57). Sine R
isblok uppertriangular, 2U
mk
. Furthermore,in viewof(42)and(37),
R
C (I Z A
)
1 =
0 I
k
+ "
Z
IP 1 2 R
( I Z A
) C
# IP
1 R
(I A
Z
) 1
C
!
C (I Z A
)
1
=
0 C
( I Z
A
)
1
+ "
Z
IP 1 2 R
(I Z A
) C
#
( I Z A
)
1 +IP
1 R
(I A
Z
) 1
A Z
IP
R
= "
Z
IP 1 2 R 0
# +
" Z
IP
1 2 R
( I Z A
) C
# IP
1 R
(I A Z
)
1 A
Z
IP R
:
Itiseasilyseenfromthelastequalitythatthematrix R
C (I Z A
)
1
isstritly blokuppertriangularandtherefore
p 0
n
C ( I Z A
)
1 o
=p 0
n b H
R
C (I Z
A
) 1
o =0 foreveryelement
b H2U
(k +nR)m . Conversely,let belongsto U
mk
andsatisfy(56). ByCorollary2.5, q
n
C ( I Z
A
) 1
ELA
Interpolationforuppertriangularmatries 45 whihmeansthat thematrix C (I Z
A
)
1
isstritly blokuppertriangular andtherefore
C (I Z A
)
1 (I Z
Z
)=0: It followsfrom thelast relationand from (33)that Q
R
=0,whihtogether with therstequalityin (44)impliesthat
R
R
= (I Q R
)= :
Thelatterequalitymeansthat admitsarepresentation(57)with b H =
R
. Itremainstonotethat by(42),
b H =
n (0; I
k
)+C (I Z A
)
1 IP
1 R
(I A
Z
)IP
1 2 R
Z
; C
o
whihimplies,inviewof(58),that b
H isblokuppertriangular: b H 2U
m(k +nR) . Using(55)andLemma4.2weobtainthefollowingresult.
Theorem4.3. AllH 2U mk
whihsatisfy(20)areparametrizedbytheformula H =H
R +
b H
R (59)
whereH R
and R
aregiven by(50)and(42),respetivelyand b
H isafreeparameter fromU
m(k +n R
) .
Nowweandesribethesetofallsolutionsoftheproblem IP R
( R
). Lemma 4.4. The representation (59)isorthogonal: for every
b H 2U
m(k +nR) f
b H
R ; H
R
g=0 and h b H
R ; H
R i =0: (60)
Proof. It follows from (36) that T R
ZIP
1 2 R Z
A C
= 0. Making use of this equalitytogetherwith(37),(48)and(50)weobtain
R
H R
=T R
" Z
IP
1 2 R
( I A Z
)
1 C
I
k
#
C (I Z A
)
1 IP
1 R
C +
=T R
" Z
IP
1 2 R
n IP
R Z
A
(I Z
A
)
1
+( I A Z
)
1 IP
R o
IP 1 R
C + C (I Z
A
)
1 IP
1 R
C +
#
=T R
" Z
IP
1 2 R
Z A
C
# (I Z
A
)
1 IP
1 R
C +
+T R
" Z
IP
1 2 R
( I A
Z
) 1
C + 0
#
=T R
" Z
IP
1 2 R
( I A Z
)
1 C
+ 0
# : Itisreadilyseenfrom(34)that theprojetionT
R
isblokdiagonal. Thenitfollows from the last equation that
R H
R
isstritly blok uppertriangular. Thus, for ev-eryuppertriangular
b
H, the matrix b H
R H
ELA
46 DanielAlpayandVladimirBolotnikov therefore,
n b H
R ; H
R o
=p 0
b H
R H
R
=0 whihinturn, implies
h b H
R ; H
R i=Tr
b H
R H
R
=0 andnishestheproofofthelemma.
Theorem 4.5. AllsolutionsH ofthe problemIP R
( R
)areparametrizedbythe formula (59)the parameter
b
H variesin U
m(k +nR)
andissubjetto n
b HT
R ;
b HT
R o
R
C +
IP 1 R
C + ; (61)
whereT R
isthe orthogonal projetion denedin(36). Proof. Inviewof(44),
n b H
R ;
b H
R o
=p 0
b H
R
R
b H
=p
0
b HT
R b H
=
n b HT
R ;
b HT
R o
whihtogetherwith(51),(59)and(60)leadsto fH; Hg=
n H
R +
b H
R ; H
R +
b H
R o
= fH R
; H R
g+ n
b H
R ;
b H
R o = C
+ IP
1 R
C +
+ n
b HT
R ;
b HT
R o
; andthus,thematrixH oftheform(59)satises(29)ifandonlyiftheorresponding parameter
b
H satises(61).
5. Left{sided problem. Inthis setion wedesribeall H 2 U mk
satisfying theondition(21).
Lemma 5.1. LetH L
bethe blokuppertriangular matrixgiven by H
L =B
+ (I Z
A
)
1 IP
1 L
B
: (62)
Then H L
satisesthe ondition(21)and [ H
L ; H
L
℄=B IP 1 L
B
; hH L
; H L
i = TrB IP 1 L
B
: (63)
Proof. Makinguseof(62)and(41)weget H
L B
+ ( I Z
A
)
1
=B ( I Z
A
) 1
+B IP 1 L
I A Z
1 A
Z
IP L (64)
andsinetheseondtermontherighthandsideinthelatterequalityisstritlyblok lowertriangular,theondition(21)followsfrom(64). Furthermore,multiplying(64) byIP
1 L
B
from therightweget H
H
L
=B (I Z
A
) 1
IP 1
B
+B IP 1
I A
Z
1
A
Z
B
ELA
Interpolationforuppertriangularmatries 47 whihimplies
[H L
; H L
℄=p 0
(H L
H L
)=p 0
n
B ( I Z
A
) 1
IP 1 L
B
o
=B IP 1 L
B
; andinpartiular theseond equalityin (63)isin fore.
It willbeshownthat H L
hasminimal Hilbert{Shmidt normamong allU mk
{ solutionsoftheproblemIP
L
. Therestissimilartoonsiderationsfromtheprevious setion: everyH 2U
mk
satisfyingtheondition(21)isoftheform H =H
L + (65)
where isanelementfromU mk
suhthat p
0 n
B
+ (I Z
A
)
1 o
=0: (66)
Lemma 5.2. The matrix belongsto U mk
and satises (66)if and only if it admitsa representation
= L
b H (67)
where L
isgiven by (43)and b H2U
(m+nL)k . Proof. Let
b
H bein U (m+n
L )k
andlet beoftheform (67). Sine L
isblok uppertriangular, 2U
mk
. Furthermore,in viewof(43)and(41),
L
B +
( I Z
A
) 1 =
0 I
k
+ "
Z
IP 1 2 L
( I Z
A
) B
+
# IP
1 L
I A
Z
1 B
+
! B
+ ( I Z
A
)
1
= "
Z
IP 1 2 L 0
# +
" Z
IP 1 2 L
( I Z
A
) B
+
# IP
1 L
I A
Z
1 A
Z
IP L
:
Therefore,thematrix L
B +
(I Z
A
) 1
isstritlybloklowertriangularandthus, p
0 n
B
+ (I Z
A
)
1 o
=p 0
n b H
L
B +
(I Z
A
) 1
o =0 foreveryblokuppertriangular
b H 2U
(m+n L
)k . Conversely,let belongsto U
mk
andsatisfy(66). ByCorollary2.5, p
n
B +
(I Z
A
) 1
o =0 andthus,theoperator
B
+ (I Z
A
)
1
isstritlybloklowertriangular. There-fore
B
+ (I Z
A
)
1 (I Z
Z
)=0 andnowitfollowsfrom(35)that
Q
L
=0. Inviewof(45),
L
=
(I Q L
)=
ELA
48 DanielAlpayandVladimirBolotnikov
Takingadjointsin thelast equalityweonludethat admitsarepresentation(67) with
b H :=
L
,whihbelongstoU
(p+nL)k
, by(43). Using(65)andLemma5.2weobtainthefollowingresult. Theorem5.3. AllH 2U
mk
whihsatisfy(21)areparameterizedbytheformula H=H
L +
L b H (68)
whereH L
and L
aregiven by(62)and(43)respetivelyand b
H isaparameterfrom U
(m+n L
)k .
NowwedesribethesetofallsolutionsoftheproblemIP L
( L
). Lemma 5.4. The representation (68)isorthogonal: for every
b H 2U
(m+n L
)k , [
L b H; H
L
℄=0 and h L
b H; H
L i =0: (69)
Proof. It followsfrom (36)that
A Z
IP 1 2 L Z
; B
+
T L
=0. Takingadvantage ofthelastequalitytogetherwith(41),(49)and(62)weget
H L
L
= B IP 1 L
I A
Z
1 B
+
B
+ ( I Z
A
)
1 IP
1 2 L
Z
; I m
T
L = B IP
1 L
I A
Z
1
A Z
IP 1 2 L Z
; B
+
T L +B (I Z
A
)
1 IP
1 2 L
Z
( I n
L ; 0)T
L = B (I Z
A
)
1 IP
1 2 L
Z
(I n
L ; 0)T
L whihshows(sine theprojetionT
L
isblokdiagonal)that H L
L
isstritlyblok uppertriangular. Therefore,foreveryblokuppertriangular
b
H,thematrixH L
L
b H isalsostritlyblokuppertriangularandhene,
[ L
b H; H
L ℄=p
0 (H
L
L
b H)=0: Inpartiular,
h L
b H; H
L
i=Tr(H L
L
b H)=0; whihendstheproof.
Theorem 5.5. All solutions H of the problem IP L
( L
) are parameterized by the formula(68)the parameter
b
H varies inU (m+n
L )k
andissubjet to h
T L
b H; T
L b H i
L
B IP 1 L
B
: (70)
Proof. Inviewof(44), h
L
b H;
L b H i
=p 0
b H
L
L
b H
=p 0
b H
T L
b H
= h
T L
b H; T
L b H i
whihtogetherwith(63),(68)and(69)leadsto [H; H℄=
h H
L +
L b H; H
L +
L b H i
= [H L
; H L
℄+ h
L
b H;
L b H i = B IP
1 B
+
h T
L b H; T
ELA
Interpolationforuppertriangularmatries 49 andthus,thematrixH oftheform(68)satises(30)ifandonlyiftheorresponding parameter
b
H satises(70).
6. Solutionsofthe two{sidedproblem. Usingtheresultsfromthetwo pre-vioussetions wenowdesribethe setof allsolutionsof problemsIP and IP
. By Theorem 4.3all operatorsH 2U
mk
satisfyingtheondition(20)are given bythe formula (59). It is possible to restritthe set of parameters
b
H in (59)in suh way thattheoperatorH oftheform(59)wouldsatisfyalso(21)and(22). Webeginwith thefollowingauxiliaryresult;fortheproofsee[2,Lemma6.1℄.
Lemma 6.1. For everyhoie of matries 2U nLk
and 2U nLm
, itholds that
P 0
n I A
Z
1 o
=P 0
n I A
Z
1 L
o
whereL2D n
L k
isdenedby L=P 0
I A Z
1
. Lemma 6.2. Amatrix H of the form (59)belongstoU
mk
and satises ondi-tions (21), (22)if and only if the orresponding parameter
b
H belongs toU m(k +n
R ) andsatisesthe ondition
p 0
n T
R b H
B +
(I Z
A
) 1
o =
b B (71)
where b B 2D
nRnL
isdenedby b B =T
R "
Z
IP 1 2 R
Z B
# (72)
Proof. LetH beoftheform(59). Multiplying(59)by R
ontherightandusing (44)weget
b HT
R
=( H H R
) R
: Makinguseof(48),werewritethelastequalityas
b HT
R =H
C (I Z A
)
1 IP
1 2 R
Z
; I k
T
R (73)
C
+ (I Z
A
)
1 IP
1 2 R
Z
; 0
T R
: SineT
R
isblokdiagonal,theseond termontherighthandsidein(73)isstritly bloklowertriangularandtherefore,
P 0
n I A
Z
1 B
+
C
+ ( I Z
A
)
1 IP
1 2 R
Z
; 0
T R
o =0: Thenitfollowsfrom(73)that
P 0
I A Z
1 B
+
b HT
R = =P
0 I A
Z
1 B
+
H
C (I Z A
)
1 IP
1 2 Z
; I k
T
ELA
50 DanielAlpayandVladimirBolotnikov
Suppose thatH satises(22)andtheseondequalityfrom(26)(whih isequivalent to(21)byRemark2.6). Then,asIP
,itfollowsfrom(74)that P
Takingadjointsin thelastequality,usingRemark2.6and(72),weget(71): p
Conversely,let b
and taking into aount (71) we obtain
P
Itfollowsfrom (42),(48)and(72),that b
Furthermore,inviewof(50), B whihbeingaddedto(77)leads,onaountof(19),to
ELA
Interpolationforuppertriangularmatries 51 Substituting (76)into(75)andusing(78)weobtain
P Sine thematrix Z
isstritly blokupper triangular, theseondtermontherighthand sideinthelastequalityequalsto zeroandthus,
P whihisequivalentto(21),byRemark 2.6.
Theveriationof(22)isdonein muh thesameway: inviewof(59), P
(whihisstritly bloklowertriangular)andtakinginto aount(71)weobtain
P
whihbeingsubstitutedto(79),leadsto P Uponsubstituting(78)into thisequalityandusing(37)weget
P Sinethematries
ELA
52 DanielAlpayandVladimirBolotnikov
are respetively, stritly blok lower and blok upper triangular, the rst and the thirdtermsontherighthandsideinthelastequalityareequaltozeroandby(18),
P 0
n I A
Z
1 B
+
HC (I Z
A
) 1
Z o
=P 0
n Z
(I A
Z
)
1 Z
o = Z
Z
= :
ThereforeH satisestheondition(22)whihendstheproofofthelemma. Aording toTheorem 4.3, alloperators
b H 2U
(nR+k )m
whihsatisfy (71)are oftheform
b H =
H
L +
L h (80)
where L
isdenedby(43), H
L =B
+ (I Z
A
)
1 IP
1 L
b B (81)
and hisanarbitraryoperator from U
(k +nR)(m+nL)
. SineT R
istheprojetion,it followsfrom(72)that
H
L T
R =
H
L : (82)
Substituting (80)into(59)weometothefollowingtheorem.
Theorem 6.3. All solutions H of the problem IP are parameterized by the formula
H=H R
+ H
L
R +
L h
R (83)
whereH R
; H
L ;
L and
R
areblokuppertriangularmatriesdenedby(50),(81), (43)and(42)respetively andhisafreeparameter fromU
(k +nR)(m+nL) . Notethat equation(83)isinfat (31)with
H min
=H R
+ H
L
R : (84)
AordingtoLemma 5.1andinviewof(82), h
b H L
T R
; b H L
T R
i =h b H L
; b H L
i = Tr b B IP
1 L
b B
while Lemmas 4.4 and 5.4 ensure that the representation (83) is orthogonal with respettotheinnerprodut(5). Due to(44)and(45),
hH; Hi=hH R
; H R
i+h b H L
T R
; b H L
T R
i+hT L
hT R
; T L
hT R
i =TrC
+ IP
1 R
C +
+Tr b B IP
1 L
b B
+hT L
hT R
; T L
hT R
i This equality leads to the desription of all the solutionsto theproblem IP
(i.e., undertheadditionalnormonstraint(28).
Theorem 6.4. (i) Theproblem IP
issolvable if andonlyif TrC
+ IP
1 C
+Tr
b B IP
1 b B
ELA
Interpolationforuppertriangularmatries 53 (ii) Allsolutions ofthe problem IP
areparameterizedby the formula(83)whenthe parameterh2U
(k +nR)(m+nL)
issuhthat hT
L hT
R ; T
L hT
R
i TrC +
IP 1 R
C +
Tr b B IP
1 L
b B
= kH min
k 2 HS
: Theorem3.1islearlyaonsequeneofthesetwolast theorems.
7. The minimal norm solution. It was shown in the previous setion that the minimal norm solution H
min
of the problem IP is given by the formula (84) and ontainsas anadditiveterm the minimal norm solutionH
R
of the right{sided problem (20). Ofourse, kH
min k
HS
is notlessthankH R
k HS
andtheirdierene is determinedbythesupplementaryleft{sidedproblem(71). Fromsymmetryarguments thepreseneoftheminimalnormsolutionH
L
oftheleft{sidedproblem(21)asaterm intheadditiverepresentationofH
min
should beexpeted. Lemma 7.1. The matrixH
min
denedby (84)anbe representedas H
min =H
L +
L b H R (85)
whereH L
and L
aregiven by (62)and(43)respetively, b
H R
= b C +
IP 1 R
( I A Z
)
1 C
(86)
andwhere
b C +
=T L
" Z
IP 1 2 L
Z C
+ #
: (87)
Proof. Firstwenote that the seond equality in (87) followsimmediately from (36). Asaonsequeneof(87)and(49)weget
L
b C +
=
B +
(I Z
A
) 1
IP 1 2 L
Z
; I m
T
L "
Z
IP 1 2 L
Z C
+ #
=C +
+B +
(I Z
A
) 1
IP 1 L
I A
Z
Z
B +
C +
: Multiplying this equality by IP
1 R
(I A Z
)
1 C
on the right and taking into aount(50),(86),weobtain
L
b H R
=H R
+ B +
( I Z
A
) 1
IP 1 L
n I A
Z
Z
B +
C +
o IP
1 R
(I A Z
)
1 C
: (88)
Furthermore, multiplyingtheequality (77)by B +
(I Z
A
) 1
IP 1 L
from the left andtakingintoaount(62),(81),weget
b H L
R
=H L
+ B +
(I Z
A
) 1
IP 1 L
(Z
A
) B
C Z
IP 1
Z
( I A
Z
) 1
C
ELA
54 DanielAlpayandVladimirBolotnikov Subtratingthis equalityfrom (88)andusing(19),weget
L
b H R
b H L
R
=H R
H L whihmeansthat(84)and(85)denethesameoperatorH
min . Itfollowsfrom (84)and(89)thatH
min
anberepresentedas H
min
(z)=H L
+H C
+H R (90)
whereH L
,H R
aregivenby(62),(50)respetivelyand H
C =B
+ ( I Z
A
)
1 IP
1 L
(Z
A
) B
C IP
1 R
( I A Z
)
1 C
: Therepresentation(90)isnotorthogonal;neverthelessitturnsoutthat
H L
?(H C
+H R
) and (H L
+H C
)?H R
: Indeed,inviewof(88)and(89),
H C
+H R
= b H L
R
; H L
+H C
= L
b H R andthelaimedorthogonalitiesholdbyLemmas4.4and5.4.
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