vol6_pp31-55. 268KB Jun 04 2011 12:06:15 AM

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 INTERPOLATION FOR
UPPER TRIANGULAR MATRICES



DANIEL ALPAYy AND VLADIMIR BOLOTNIKOVz

Abstra
t. The spa
e of upper triangular matri
es with Hilbert{S
hmidt norm
an be viewed
as a nite dimensional analogue of the Hardy spa

e H2 of the unit disk when one introdu
es the
adequate notion of \point" evaluation. A bitangential interpolation problem in this setting is studied.
The des
ription of all solution in terms of Beurling{Lax representation is given.
Key words. Interpolation, matri
es.
AMS subje
t
lassi
ations. 47A57, 47A48
1. Introdu
tion. In this paper we pursue our study of bitangential interpolation

in analogues of the Hardy spa
e of the unit disk H2 . To start with we re
all the
lassi
al setting whi
h has been

onsidered in [8℄.

mk denote the Hilbert spa
e of

Let H2

H (z ) =
endowed with the
(1)

k
Lm
2

X
1

j =0

m  k {matrix valued fun
tions of the form

X1

Hj z j ;

inner produ
t

hH; GiLm2 k = hH; GiHm2 k = 21

j =0

Z

2
0



Tr Hj Hj

it

< 1;

 H (eit )dt =

Tr G(e )

mk

X1
j =0



Tr Gj Hj :

The H2
-fun
tions are analyti
in the open unit disk D and have boundary values
m k is a Hilbert module (see,
almost everywhere on the unit
ir
le T. The spa
e H2



Z

e.g, [18℄ and Se
tion 2 of [1℄) with respe
t to the Hermitian matrix-valued forms

(2)

fH; GgHm2 k = 21

and
(3)

[H;

G℄Hm
k
2

=

1
2

Z

2

0
2

0

H (eit )G(eit ) dt

G(eit ) H (eit )dt

and has the reprodu
ing kernel property with the kernels

k!^ (z ) =

1

Im
z! 

and

k!4 (z ) =

1

Ik
z! 

 Re
eived by the editors on 13 August 1999. A

epted for publi
ation on 3 Mar
h 2000. Handling
Editor: Daniel Hershkowitz.
y Department of Mathemati
s, Ben{Gurion University of the Negev, Beer{Sheva, 84105, Israel
(danymath.bgu.
.il).
z Mathemati

s Department, The College of William and Mary, Williamsburg, Virginia, USA
(vladimath.wm.edu).

31

ELA
Daniel Alpay and Vladimir Bolotnikov

32
in the sense that
(4)



H (!)A = H; Ak!4 Hmk
2

and A H (! ) = [H; k!^ A℄Hmk
2

for every
hoi
e of a point ! 2 D and of a m  k matrix A. Note that the latter
k
relations express Cau
hy's formula for m
In [8℄ we
onsidered a general
2 m-fun
tions.
bitangential Nudelman type problem for 2 k fun
tions with norm
onstraints. The
multivariable analogue of the problem referred to in the previous paragraph in the
setting of the polydisk was
onsidered in [4℄.
It is well known (see [10℄, [11℄, [12℄, [15℄) that there are deep analogies between
analyti
fun
tions and upper (or lower) triangular matri

es. In [2℄ we looked at the
analogue of the problem referred to in the previous paragraph, for double innite
upper triangular matri
es. In this paper we fo
us on the
ase of nite matri
es. In
fa
t, one
ould try to obtain the
ase of nite matri
es from our previous paper [2℄
using time varying
oeÆ
ient spa
es in the spirit of [13, Se
tion 12℄ or [14℄. This
does not seem to us natural, sin
e the problem
onsidered here is nite dimensional;
furthermore, the approa
h presented in this paper is purely algebrai
and mu
h more
expli
it.
Other situations are also possible, su
h as the
ase of lower triangular integral
Hilbert S
hmidt operators (the
ontinuous time varying
ase analogue of [2℄). This
was
arried out in [3℄.
Consider X mk , the set of all m  k matri
es whi
h is a Hilbert spa
e with respe
t
to the inner produ
t

H

(5)

H; Gi = Tr G H

h

H

(H; G 2 X mk );

whi
h is the analogue of (1).
=1;:::;k is
alled diagonal if h = 0 for i 6= j . It is
alled
A matrix H = [hi j ℄ji=1
ij
;:::;m
upper (lower) triangular if hij = 0 for i > j (i < j , respe
tively). The symbols Dmk ,
mk and Lmk will be used for the spa
es of diagonal, upper triangular and lower
U
triangular m  k matri
es, respe
tively.
Let Zm denote the m  m shift matrix dened by
2
3
0 1 0


0
6
. . .. 7
6
. . 7
0 1
6
7
6
7
.. ..
Zm = 6 ..
(6)
.
.
0 7
6 .
7
4
1 5
0




0
and let Zk be the k  k shift matrix dened similarly. We denote by p+ ; p0 ; p the
 Lmk , respe
tively. We
orthogonal proje
tions of X mk onto U mk Zk , Dmk , Zm
also use the proje
tions of X onto its upper and lower parts, and denote them by

p = p0 + p+ and q = p0 + p ;

ELA
33

Interpolation for upper triangular matri
es

respe
tively. The spa
e X mk is a Hilbert module with respe
t to Hermitian matrixvalued forms

f

gX mk =

H; G

(

p0 H G

 ) and [H;

℄X mk = p0 (G H )

G

whi
h are the analogues of (2) and (3).
For a xed H 2 U mk these forms make sense for k  k and m  m matri
es G,
respe
tively, and dene two dierent \evaluation" maps for upper triangular matri
es
by




) 1F
(7) F ^ (W ) = p0 (Im W Zm
and F 4 (V ) = p0 F (Ik Zk V ) 1 ;
where F 2 U mk and W and V are diagonal m  m and k  k matri
es, whi
h usually
play the role of points in the nonstationary setting. The transformations F ^ (W ) and
4(V ) are non
ommutative analogues of the point evaluation (4). We also dene
F


℄
 1 F (Ik Z  V ) 1 Z  :
F (W; V ) = p0 (Im
W Zm )
k
k
It follows from the denition (6) of Zm that for every
hoi
e of

Im
W Zm is lower triangular with all diagonal entries equal

to 1. In parti
ular, det (Im W Zm ) = 1 and the matrix is invertible. From the same
reason the matrix Ik Zk V is invertible for every
hoi
e of V 2 Dkk .
Note that if m = k , then for any F 2 U mk there exist uniquely dened diagonal
matri
es F[j ℄ and Ffj g whi
h satisfy
Remark 1.1.

W

2 Dmm , the matrix

F[j ℄

j j
f g = Ffjg Zk Zk

j Z j F[j℄ ;
= Zm
m

F j

and are su
h that
F

=

X1

m

j =0

j

Zm F[j ℄

and

F

=

X1

k

j =0

fg

j

F j Z :
k

The latter \polynomial" representations allows us to express evaluations (7) as
(8)

F

m 1
^(W ) = X (W Z  )j Z j F[j℄ and
m
m
j =0

4(V ) = X Ffjg Z j (Z  V )j :
k

F

1

j =0

k

k

These formulas appear in [7℄. If m > k (m < k ), the rst (the se
ond) formula in (8)
is true. In general, for m < k (m > k ) the rst (the se
ond) formula in (8) is not
valid, and for this reason we shall use formulas (7).
The maps (8) have been introdu
ed in [5℄ and [6℄ for bounded upper triangular
operators in innite dimensional Hilbert spa
es. For more on this setting and on the
related interpolation problems for uper triangular
ontra
tions, see [9℄, [13℄, [16℄, [17℄.
Now we present the analogue of the
lassi
al Lagrange interpolation problem,
whi
h is posed below in terms of the \point" evaluations (7).

ELA
Daniel Alpay and Vladimir Bolotnikov

34
Problem 1.2.

and

(9)

Given matri
es

2 D`j m ; j 2 D`j k
2 Dri m ; i 2 Dri k
`i `j
, nd all H 2 U mk su
h that
ji 2 D
j H

Wj
Vi

2 D`j `j ;
2 Dri ri ;

j
i


^

(Wj ) = j ;

(Hi )4 (Vi ) = i and

(j = 1; : : : ; n)
(i = 1; : : : ; `)

j Hi


℄

(Wj ; Vi ) =

ji

for j = 1; : : : ; n and i = 1; : : : ; `.

The paper
onsists of seven se
tions. To set the problem pre
isely we rst need
some notations and denitions. The general problem (whi
h in
ludes Problem 1.2 as
a parti
ular
ase) is stated in the se
ond se
tion. The des
ription of all its solutions
(formula (31)) is presented in the third se
tion. The proof relies on the analysis of
two spe
ial
ases, namely right sided and left sided interpolation problems, whi
h are
onsidered in details respe
tively in Se
tions 4 and 5. The general two sided problem
is studied in Se
tion 6. The last se
tion deals with the stru
ture of the minimal norm
solution.
2. Formulation of the problems and preliminary remarks. In this se
tion
we introdu
e the bitangential interpolation problem to be studied, and whi
h in
ludes
Problem 1.2 as a parti
ular
ase. Given two sets of positive integers f`ig and frj g,
let
(10)

nR

= r1 + : : : + rt ;

nL

= `1 + : : : + `s ;

let Zrj and Z`i be the shift matri
es dened via (6) and let
n n
Z = diag (Zr ; : : : ; Zr ) 2 C R R and Z = diag (Z`

1 ; : : : ; Z`s

t

1

) 2 C nL nL :

The relations
Z (I

(11)
(12)
(13)
(14)

(I
(I
(I

Z Z )

= 0; Z Z D = DZ Z ;

Z Z )Z = 0; Z Z F = F Z Z ;
Z Z )(I
Z D) 1 = (I
Z Z ); and
Z Z  )(I
Z F ) 1 = (I Z Z  );




whi
h hold for every
hoi
e of blo
k diagonal matri
es D 2 DnR nR and F 2 DnL nL ,
will be useful.
In the
lass U mk we
onsider the interpolation problem whose data set is an
ordered
olle
tion

= fC+ ;

(15)

C ; A ; A ; B+ ; B ;

g

of seven matri
es
C+

2 DmnR ;

C

2 DknR ;

B+

2 DmnL ;

B

2 DknL ;

ELA
35

Interpolation for upper triangular matri
es

A

2 DnR nR ;

A

2 DnR nL :

2 DnL nL ;

By abuse of notation, by A 2 DnR nR and A 2 DnL nL we
mean that A and A are nR  nR and nL  nL matri
es with diagonal blo
k entries
(A ) 2 D`i `j (i; j = 1; : : : ; s) and (A ) 2 Dri rj (i; j = 1; : : : ; t):
Remark 2.1.

ij

ij

Similarly, by C+ 2 DmnR we mean that C+ is a blo
k row and every m  ri blo
k
is diagonal (in other words, we reserve the symbols nL and nR to indi
ate blo
k
de
omposition of the underlying matri
es). We shall refer to su
h matri
es as to
blo
k diagonal. The same
onvention holds with U and L instead of D for blo
k
upper triangular and for blo
k lower triangular matri
es. The other notations (su
h
as DmnR or DknL ) should be
lear.
We say that the data (15) is admissible if
n



n



(16)

Span Ran (A Z )j C 

(17)

Span Ran

A Z


j

(18)
and the Sylvester equality holds
(19)
A Z  Z






o

; j

= 0; : : : ; nR

1 = C nR ;

 ;j
B+

= 0; : : : ; nL

1 = C nL ;



=

Z Z ;

A

= B C

o

 C+ :
B+

We denote by IP the following two{sided interpolation problem.
mk su
h that
Problem 2.2. Given an admissible data set
, nd all H 2 U
o
n
1 =C ;
(20)
Z A )
p0 HC (I
+
o
n
1 =B ;
(21)
p0 H  B+ (I Z A )
(22)

P0

n

A Z

I




1 B  HC (I
+

Z A )

1 Zo =


;

where in the rst equation, p0 is the orthogonal proje
tion of X mnR onto DmnR , in
the se
ond equation p0 denotes the orthogonal proje
tion of X knL onto DknL , and
nally, in the third equation,

= p01 


 p0s
and p0i is the orthogonal proje
tion of X `i `i onto D`i `i .
Note that by Remark 1.1 and in view of the blo
k stru
ture of A , A , Z and
Z , the matri
es I
Z A and I
Z A are invertible.
Remark 2.3. Conditions (20) and (21) generalize the Nevanlinna{Pi
k
onditions (9) and
oin
ide with the last ones for the spe
ial
hoi
e of
P0

B+

= (1 ; : : : ; n ) ;

B

= (1 ; : : : ; n ) ;

C+

= (1 ; : : : ; ` ) ;

C

= (1 ; : : : ; ` ) ;

ELA
Daniel Alpay and Vladimir Bolotnikov

36
"
A

=

W1

0

..

0 #
.

"
A

;

Wn

=

V1

..

0

0 #
.

"

11

..
.
n1

=

;

V`

#
1`
..
:
.
n`










The next lemma shows that
onditions (20) and (21)
ontain more information
about a solution H of the interpolation problem.
mk
Lemma 2.4. Let H belong to U
and satisfy (20), (21). Then
n

(23)

q

and

n

(24)

p



H B

. For every

Proof



(I

HC

+ (I

Z A

)

1

Z A

)

1

2 X mnR ,

o

o



n
q

HC



(I

Z A

)

1

o

=

p

j =0

0

)

1

= B (I

Z A

it holds that

nX
R 1

1

Z A

q

n

0 Z
j

p

j =0



(I

HC

:

nX
R 1

=

therefore,
(25)

)

= C+ ( I

Z A

)

1

j

Z

o




j and

Z

j

Z :

The rst term on the right hand side of the equality
(I



Z A

)

1

j

Z

=

j 1
X
`=0

(Z A )` Zj + (I



Z A

) 1 (Z A )j Zj

is stri
tly blo
k upper triangular, and thus, for every upper triangular H ,
n
o
n
o
HC
(I Z  A ) 1 Z j = p0 H C (I Z  A ) 1 (Z  A )j Z j :
p0










Sin
e the operator (Z A )j Zj is blo
k diagonal, it follows from (20) that
o
n
o
 n
HC
(I Z A ) 1 (Z A )j Zj = p0 H C (I Z A ) 1 
p0
 (Z A )j Zj
= C+ (Z A )j Zj :
Sin
e A is blo
k diagonal,
(Z A )j Zj Zj = (Z A )j ;
j = 0; 1; : : :
Making use of the three last equalities we dedu
e (23) from (25):
n
q

HC

(I



Z A

)

1

o

=
=

nX
R 1



j



j

+ (Z A )

C

j =0

nX
R 1

+ (Z A )

C

j =0

= C+ (I



Z A

)

1

j

j

Z Z

:

ELA
37

Interpolation for upper triangular matri
es

Taking advantage of the relation
n
p

 + (I

Z A

H B

L 1 n
o nX

H B+ ( I
p0
) 1 =
j

Z A

=0

it
an be
he
ked in mu
h the same way that (21) implies (24).
Corollary 2.5. Conditions (20) and (21) are equivalent

) 1 Zj

to

o

j


Z ;

(23)

and

(24)

re-

spe
tively.

Indeed, applying p0 to both sides of (23) and (24) we obtain (20), (21). The rest
follows from Lemma 2.4.
Sometimes it will be
onvenient to use
onditions (20) and (21) in the following
\adjoint" forms
n

o



(26) P0 (I

A Z

n

0

o
 
1 B H = B :
+

I

P

A Z

Conditions (20), (21) are equivalent to
onditions (26).

Remark 2.6.

Proof.

) 1 C  H  = C+ and

2 U mk in the form

Taking H

0

n

p

Z A

=

=0
k
X1

) 1 =

j

n

o



0 (I

A Z

P

1

k
X

Hj Z

j

o



(I

HC

H

) 1 C H =

=0
1

k
X
j

=0

j
k

with

Hj

2 Dmk , we get

(Z A )j ;

j
k

Hj Z C

(A Z )j C  Zkj Hj :

Comparing right hand sides in two last equalities we
on
lude that


p

0

n



(I

HC

Z A

) 1

o

n



= P0 (I

A Z

) 1 C H

o

and therefore, (20) is equivalent to the rst
ondition in (26). The equivalen
e of (21)
and the se
ond
ondition in (26) is
he
ked in mu
h the same way.
Remark 2.7. The Sylvester identity (19) follows from (20){(22) and is therefore,
a ne
essary
ondition for the problem IP to be solvable.
Proof. First we note the equality
(27)

P

whi
h holds for every M
M

=

1

nX
L
j

=0



0

Z M



(I

Z Z




) = 0;

2 X nR nL . Indeed, taking

j
 Mj +

1

nX
R

Z

i

=1

Mn

L

M

in the form

1+i Zi with

M`

2 DnR nL

we get


Z M (I




Z Z ) = Z




1

nX
L
j

=0

j Mj (I
Z



Z Z ) +


2

nX
R
i

=0

j M j
Z


;

!
Z

(I



Z Z

)

ELA
Daniel Alpay and Vladimir Bolotnikov

38

and
ome to (27), sin
e the rst term on the right hand side in the latter equality is
stri
tly blo
k lower triangular and the se
ond term is equal to zero due to (12).
Let H belong to U mk and satisfy (20){(22) (or equivalently, (22) and (26)).
Upon applying (27) to

1 
B+ HC (I Z A ) 1 ;
M = I A Z
taking into a

ount that the matrix A is blo
k diagonal and making use of the
equality
Z  (P0 X )Z = P0 Z  XZ ;




whi
h holds for every X 2 X nL nL , we get
A Z Z


1 
Z
B+ HC (I Z A ) 1 Z


 1

= P0 A Z I A Z B+ HC (I Z A ) 1 Z Z


 1

= P0 A Z I A Z B+ HC (I Z A ) 1


 1

 HC (I Z A ) 1 (I Z Z )
B+
A P0 Z I A Z



 1
1




:
I A Z
I B+ HC (I Z A )
= P0

Furthermore,
A



= A Z P0

= P0
= P0





I

I

A Z

I

A Z






A Z




1 
B+ HC (I Z A ) 1 Z A


1 
B+ HC (I Z A ) 1 I

and sin
e B+ and C are blo
k diagonal, it follows from (26) that
o
n
o
n

1 
 HC (I Z  A ) 1
P0 B+
B+ HC
A Z Z
A = P0 I A Z

o
n
n

o


1
 P0 HC (I Z  A ) 1
= P0 I A Z B+ H C
B+

 C+ ;
= B C
B+
and this
ompletes the proof.
It will be shown in Se
tion 6 that the problem IP always has a solution. Let us
denote by IP
the problem IP to whi
h has been added the norm
onstraint

hH; H iX mk def
= Tr (H  H ) 
for some preassigned number
 0. Be
ause of the Hilbert spa
e stru
ture, we will
(28)

see that there exists a unique solution Hmin of the problem IP with minimal norm.

ELA
39

Interpolation for upper triangular matri
es

The value kH kHS (as well as H itself) depends only on the problem data (15)
and the
ondition
 kH kHS is ne
essary and suÆ
ient for the problem IP
to
be solvable. The expli
it formula for H and a des
ription of all solutions of the
problem IP
will be given in Se
tion 6.
We denote by IPR (R ) the right{sided problem (20) (i.e., when
onditions (21)
and (22) are not in for
e) to whi
h has been added the \matrix norm"
onstraint
min

min

min

min

(29)
fH; H g = p (HH  )  R
for some preassigned nonnegative matrix R 2 Dmm .
Similarly, we denote by IPL (L) the left{sided problem (21) to whi
h has been
added the matrix \norm"
onstraint
(30)
[H; H ℄ = p (H  H )  L
for some preassigned nonnegative matrix L 2 Dkk .
It turns out that the
onstraints (29) and (30) do not suit the left{sided
ondition
(21) and the right{sided
ondition (20) respe
tively. That is why we
onsider a two{
sided problem only under the
onstraint (28) whi
h on a

ount of
hH; H i = Tr fH; H g = Tr [H; H ℄
suits to left
onditions as well as to right ones.
3. Statement of the main result and rst formulas. The main result of
the paper is now stated:
def

def

0

0

Theorem 3.1. The set of all solutions of Problem

(31)

H

where
and

Hmin

L 2 U

2 U m k

=H

min

IP is given by

+ L hR

is the minimal Hilbert{S
hmidt norm solution,

m(m+nL )

R 2 U k

k

( +nR )

are two partial isometries with upper triangular blo
k entries,
(m+n ) (k+n )
is a free parameter from
.

U L R
In this se
tion we
onstru
t expli
itly L and R (see formulas (43) and (42)),
while the formula for H will be given in Se
tion 7. We begin with preliminary
lemmas.
built from the interpolation data and

h

min

Lemma 3.2. The Stein equations

(32)

IPR

A Z IPR Z A

= C C

and

IPL

A Z IPL Z A

= B B
+

+

are uniquely solvable, and that their solutions are the blo
k diagonal matri
es given
by

X(  )
X
= (  )

IPR =
IPL

m

1

j =0
m 1
j =0

A Z j C  C

(Z A )j 2 DnR nR

 B+ (Z A )j
A Z j B+

2 DnL nL :

ELA
Daniel Alpay and Vladimir Bolotnikov

40

Conditions (16) and (17) are ne
essary and suÆ
ient that the operators IPR and IPL
are boundedly invertible.
The proof is straightforward and will be omitted.
Let IPR and IPL are solutions of the Stein equations (32). Then the
matri
es
Lemma 3.3.

(33)

QR = C (I

(34)

TR =

(35)

QL = B+ (I

(36)

TL =







Z A

0





Z Z

"



0

Z Z

) 1 (I

0

0

#

C

Z A



IPR2 Z A
1

Z

Ik

Z Z

)IPR1 (I

) 1 (I

IPR1

 )IP 1 (I
L

Z Z

"





Z Z

 IP 21 Z A
L
B+

#

Z

Im





)



A Z

IPR2 Z ;

) 1 C ;

1

A Z

Z Z

IPL 1



) (I

C




;

 
1 B ;
+

I

A Z

  IP 21 Z ; B 
+
L



A Z

are orthogonal proje
tions.
Proof. The matri
es dened in (33){(36) are evidently selfadjoint. To show that
QR is a proje
tion, we start with the equality


(I
(37)

A Z

= IPR (I

) 1 C  C (I



Z A

) 1 + (I



Z A



) 1=

A Z

) 1 A Z IPR ;

whi
h is an immediate
onsequen
e of the rst equation in (32). It follows from (33)
that



Q2R = C (I

(38)

Z A

) 1 L (I



A Z

) 1C

;

where, on a

ount of (37),



L = (I

(

Z Z



Z Z

I

n



)IPR1 (I

)IPR1 (I

Z Z



n



Z Z

= (I

Z Z

= (I



)IPR1 IPR (I
n



) (I

1

Z Z )IPR (I

(39)

) 1 + (I



A Z

) 1 A Z IPR

o

)

Z A

D



= IPR1 ) and of (11) and (13) (with

Z A

) 1 + (I

=

) 1 A Z IPR IPR1 (I



A Z


Z Z );

D

o



A Z

) 1 IPR1 + IPR1 (I

o

) 1 A Z (I



Z Z

A

)

Z Z

)



)



whi
h together with (38) leads to
rst note that
IPR1



Z A

Z Z :

Taking advantage of (12) (with
su

essively, we get

L = (I

) IPR (I



Q2R = QR . To show that TR is a proje
tion, we

1


2 
A Z IP Z ; C



R





"

IPR2 Z A
1

Z

C

#

IPR1 = IPR1 :

ELA
41

Interpolation for upper triangular matri
es

Indeed, making use of (32), we transform the left hand side as


IPR2   IPR2   +  = IPR
and obtain (39), sin
e in view of (11),
1

A Z



1

Z Z

Z A



1

A Z

C

C

IPR2 (



1

Z Z

I

)IPR2

Z A



2
2
(40)
 IPR (
 ) = (
  )IPR = 0
The equality TR = TR follows easily from (39) and relations
1

1

Z

I

Z Z

Z

I

Z Z

:

2

(



 )2 = Z Z  ;

Z Z



0



"

0

Z Z

IPR2

Ik

#



1

Z

Z A

C

=

"

IPR2

#



1

Z

Z A

C

whi
h are
onsequen
es of (12). Equalities QL = QL and TL = TL are veried quite
similarly with the help of (11){(14) and
2

2

 




) = IPL (
+
 )


 
  IPL
+
(41)
 
 
whi
h in turn, follows immediately from the se
ond equation in (32).
k nR k
Lemma 3.4. The matri
es R 2 C
and L 2 C p m nL dened by
I

A Z

1

B+ B+

(

1

Z A

I

I

I

( +





R = 0k +

(42)

"

I

IPR2 (

Z A

I
C

(

#

) IPR (
1

and


(43) L = 0

;

Im



+

B+

(

Z A

I

1

A Z

)



1

Z

1

Z A

) IPL
1

1



+

A Z

I



A Z

)

)

1

C

 
IP 12 Z ;

I

A Z







B+

L

respe
tively, are blo
k upper triangular and satisfy

(44)

R R =

Ik

LL =

Im

Q

and

(45)

Q T ,Q

R;

Q

L;

R R = TR
L L = TL

;

;

R
L and TL are the orthogonal proje
tions dened in (33){(36).
The upper triangular stru
ture of R and L follows immediately from
their denitions (42) and (43). The upper triangular stru
ture here is meant in the
sense of Remark 2.1: a

ording to partitions (10), R and L are ( + 1)  and
 ( + 1) blo
k matri
es, respe
tively, and ea
h blo
k entry is an upper triangular
matrix in a usual sense. The veri
ation of (44) and (45) is quite straightforward: by
(42),
 )


(46)
R R = k
(
IPR MIPR (
 )


where R ,
Proof.

t

s

t

s

I

C

I

Z A

1

1

1

I

A

Z

1

C

;

ELA
42
where

Daniel Alpay and Vladimir Bolotnikov

M = IP

R



(

+(
In view of (32),

)+
(
 ) IPR

Z A

I



I

A Z

M=(





1

A Z

I

) IPR2



1

Z Z

IPR2 (

Z A

I

)

C

C

:



2
) IPR2 (
  )IPR (
 )
1
Taking advantage of the rst two relations in (12) (with = IPR2 ), we
ome to
M = (   )IPR(  )
whi
h being substituted into (46) leads to the rst equality in (44). Furthermore, on
a

ount of (42),

 

i
0
0
0
1 IP 1 h(


  ) IP 21 

(
R R = 0 k +
 )

R
R 


1

1

A Z

I

I

Z Z

I

Z A

D

I

+
+

I

C

I

"

1

Z

Z Z

IPR2 (
1

Z

IPR2 (

) IPR1 (



) IPR1 (

h

C

A Z



)

1



)

1 C C

I

A Z

I

A Z

#

Z A

I

;

I



C

"

Z A

#

Z A

I

Z Z

I

0

; C

Z ;

C




(



Z A

I

1

)

i

  ) IP 2 

IPR1 (
(47)

R 
Substituting (37) into the third term on the right hand side of (47) and taking into
a

ount (34), we get
#
 "

1
1

2
0
0
(
)
IP

 
IPR 2 (  0)
R R = 0 k +  R
1

I

A

Z

Z ;

Z

I

I



C

:

Z A

Z ;

C



h

  ) IP 2 
+ 0 IPR 2 (

R 
"
#
1
h

2
 IPR (
 )
IPR1 (
Z

1

1

I

Z

I

=



Z Z

Z ;

Z A

C



A Z

"



0

I

1

Z

IPR2

#



Z A

C

i





IPR1

) IPR2
1

A Z







Z ;

C



i

1

IPR2





0
k
whi
h proves the se
ond equality in (44). The equalities (45) are veried in mu
h the
same way with help of (43) and (41).
Remark 3.5. The matri
es R and L dened via (42) and (43) admit the
representations
#
"
1
  ) 1 
2 (
IP

R
(48)
R = TR
I

C

I

Z

(49)

L =

A Z

C

Ik



+ (I

B

) 1 IPL 2

;



1

Z A

A Z

Z ; Im

T

L:

Z ; C

;

ELA
43

Interpolation for upper triangular matri
es

Proof.

"

In view of (34) and (48),

IP 2 (

TR



1

 R

Z

I

A Z

k

I



"

#

1 C

)



Z

IPR1

 R Z A

Z

C



1

By (40),  IPR2
Z

"



Z

TR

=

"

I

= 0k +

k

I

A

k

"



A Z

 IPR (







k



)

#

1 C







IP 2

1

  R Z Z

A Z



IPR 2 (
1

 

I

A Z

) 1+


I

C

#

1 C

)

#

 Z ) 1 C 


 21
Z IP (I

Z A

C

2
 IPR

#




1

#

 

R

"
Z

I

I

A Z

1

I
Z

I

Z

1

1
2



1

=  IPR2 and therefore,

 

Z Z

IPR 2 (

IPR 2 (

I

#



1
2

IP

=

"



Z A

) IPR1 (



I

A Z



IPR1 (

I

C

)

1 C





A Z

= R

)

1 C

;

whi
h proves (48). The representation (49) is veried in mu
h the same way with
help of (36) and (43).
4. Right{sided problem. In this se
tion we des
ribe all
2 U mk satisfying
ondition (20). We rst exhibit a parti
ular solution, whi
h will be shown in the
sequel to be of the minimal Hilbert{S
hmidt norm.
H

Let

Lemma 4.1.

be the blo
k upper triangular matrix given by

R

H

(50)

R

H

Then

H

R

(51)

satises the
ondition

f

H

Proof.
H

RC

=

R;

H

R

g=

C

In view of (37),
1

(
 ) =
=
I

Z A

C

+ IPR1 (I



)

1 C :

R

i = Tr

 

A Z

(20) and

+ IPR1 C+ ;

h

H

R;

H

1
+ IPR1 (I A Z ) C  C (I
1 + C IP 1 (I

C+ ( I
Z A )
+ R
C

C

+ IPR1 C+ :

 ) 1
1 A Z IP ;

A Z )
 R
 

Z A

(52)


and sin
e
 1
  ) 1   IPR 2 U mnR 
(53)
+ IPR (


ondition (23) (whi
h is equivalent to (20) by Corollary 2.5) follows from (52). Furthermore, multiplying (52) by IPR1 + on the right we get
1 1 
1

 ) 1   
(54) R R = + (
  ) IPR + + + IPR (


+
C

I

A Z

A

Z ;

Z

C

H

H

C

I

Z A

C

C

I

A Z

A Z C

:

:

ELA
44

Daniel Alpay and Vladimir Bolotnikov

By (53), the se
ond term on the right hand side of (54) is stri
tly blo
k upper triangular while the rst one is upper triangular and thus,
f

H

R;

H

R

g= (
def

p0 H

n

 ) = p0

R HR

C+

(



 

I

Z A

) IPR
1

1



o

C+

=

C+

IPR

1



C+ ;

whi
h proves the rst equality in (51). The se
ond equality is an immediate
onsequen
e of the rst one.
Note that every 2 U mk satisfying the
ondition (20) is of the form
(55)
= R+
where is an element from U mk su
h that
o
n

(
(56)
=0
 )
H

H

C

p0

H

I

1

Z A

belongs to U mk

The matrix
admits a representation
Lemma 4.2.

(57)

:

(56) if and only if it

and satises

= b R
where R is given by (42) and b 2 U m k nR .
Proof. Let b be in U m k nR m and let be of the form (57). Sin
e R is blo
k
upper triangular, 2 U mk . Furthermore, in view of (42) and (37),

R (
 )
#
!
 "

1

2
0
 )


  ) IP (
=
+  IPR (
(

 )
R
H

( +

H

( +

H

C

I

1

Z A

Z

=



0
C

"

+
=

k

I



(



1



 

I

Z A

)

1

Z A

I

1

A Z

C

C

I

I

#

Z A



#

)

(

C

"



Z

 

Z A

#



1

IPR2 (



I

 

I

Z A

C

n

p0

(
C

I



 

Z A

)

1

o

1

1

) IPR (

=

1

n

 

A Z

b R C

(

I

1



I



I

H

p0

 Z )

) + IPR (

It is easily seen from the last equality that the matrix R
blo
k upper triangular and therefore

A

)

C

1

(




 

A Z

 

)

1

o

H

( +

)

n

q

(
C

I



 

Z A

)

1

o

=0

IP

  R



:

 

Z A

I

Z A

IPR




A Z

) is stri
tly
1

=0

for every element b 2 U k nR m .
Conversely, let belongs to U mk and satisfy (56). By Corollary 2.5,
(58)

1

Z A

1




1

IPR2 (

IPR2 +
0



Z

I

C

Z

"

)

)

ELA
45

Interpolation for upper triangular matri
es


whi
h means that the matrix (
is stri
tly blo
k upper triangular
 )
and therefore


(
 ) (
 ) = 0
It follows from the last relation and from (33) that QR = 0, whi
h together with
the rst equality in (44) implies that
RR = ( QR ) =
The latter equality means that admits a representation (57) with b = R.
It remains to note that by (42),
n
o

  ) IP 21 


b = (0 k ) + (

  ) IPR (
R 
C

C

I

1

Z A

I

1

Z A

I

Z Z

:

I

:

H

H

; I

C

1

Z A

I

1

Z ;

A Z

I

C

whi
h implies, in view of (58), that b is blo
k upper triangular: b 2 U m k nR .
Using (55) and Lemma 4.2 we obtain the following result.
Theorem 4.3.
2 U m k
(20)
(59)
= R + b R
b
R
(50) (42)
R
U m k nR
Now we
an des
ribe the set of all solutions of the problem IPR (R ).
b 2 U m k nR
Lemma 4.4.
(59)
(60)
f b R R g = 0 and h b R R i = 0
H

All H

whi
h satisfy

H

where H
from

and
( +

H

are given by

)

H

are parametrized by the formula

H

and

, respe
tively and H is a free parameter

; H

H

; H



I

Z A

. It follows from (36) that TR
equality together with (37), (48) and (50) we obtain


R

= TR
= TR
= TR
= TR

1

2
 IPR (

Z

"

IP

 R

Z

"

1
2

I

n

k

IPR



IP 2

1

C

"

IP (

 R

Z

1
2

)



(

1

C

#



C

I

 R Z A

Z



A Z

I

 

Z A

#



I

(

C


Z A )

(



I



I



A Z

)

1



1

1

#

C+

IPR

I

1

Z A

1

= 0. Making use of this

 

) +(
  ) IP
R

 

Z A



(


C

+

)

:

1

Z IPR2 Z A
C

Proof

"

( +

is orthogonal: for every H



H

)

.

The representation

R

( +

H

) IPR
1

  )
1 
C+
"
A Z

+ TR

1

1

o

IPR IPR

IP 2 (

 R

Z



C+

1

1

I

A

0



#

C+

 Z )

1

0
It is readily seen from (34) that the proje
tion TR is blo
k diagonal. Then it follows
from the last equation that R R is stri
tly blo
k upper triangular. Thus, for every upper triangular b , the matrix b R R is stri
tly blo
k upper triangular and
:

H

H

H

H



C+

#

ELA
46

Daniel Alpay and Vladimir Bolotnikov

therefore,

n

b

H

R

o
; HR



=

whi
h in turn, implies
h b R
H

; HR

i = Tr

b

R

b

R

H

p0



H









=0

HR

HR

=0

and nishes the proof of the lemma.

All solutions H of the problem IPR (R ) are parametrized by the
b varies in U m(k+nR ) and is subje
t to
the parameter H

Theorem 4.5.

formula

(59)

n

(61)

b

o

b

H TR ; H TR

 R

C+

where TR is the orthogonal proje
tion dened in
Proof. In view of (44),
n

b

H

R

b

; H

o

R =



p0

b

H



R R b  =
H

IPR

1



C+ ;

(36).


b

b



H TR H

p0

=

whi
h together with (51), (59) and (60) leads to
f

H; H

g=

n

+ b R

HR

H

o

; HR

n

b

b

n

+ b R = f R R g + n b R
=
IPR  + b TR
H

H

; H

1

C+

o

H TR ; H TR

C+

b

; H

H

; H TR

b

o

R o

H

;

and thus, the matrix of the form (59) satises (29) if and only if the
orresponding
parameter b satises (61).
5. Left{sided problem. In this se
tion we des
ribe all
2 U mk satisfying
the
ondition (21).
H

H

H

Lemma 5.1.

Let

HL

be the blo
k upper triangular matrix given by

(
IPL 
 )
Then L satises the
ondition (21) and
(63)
[ L L ℄ = IPL 
h L L i = Tr
Proof. Making use of (62) and (41) we get
(62)

HL

=

B+

I

1

Z A

1

B

:

H

H

; H

1

B

B

;

H

; H

B

IPL

1

B




:

 
  IPL
= (
+ IPL
(64) L (
 )
 )
 
 
and sin
e the se
ond term on the right hand side in the latter equality is stri
tly blo
k
lower triangular, the
ondition (21) follows from (64). Furthermore, multiplying (64)
by IPL  from the right we get
H

B+

1

I

1

Z A

B

I

Z A

1

1

B

I

A Z

1

A Z

B



HL HL

=

B

(

I

Z A

) IPL
1

1

B



+

B

IPL

1

I

 


A Z

1

  ;

A Z B

ELA
47

Interpolation for upper triangular matri
es

whi
h implies
o
n
(
IPL  = IPL 
[ L L ℄ = ( L L) =
 )
and in parti
ular the se
ond equality in (63) is in for
e.
It will be shown that L has minimal Hilbert{S
hmidt norm among all U mk {
solutions of the problem IPL . The rest is similar to
onsiderations from the previous
se
tion: every 2 U mk satisfying the
ondition (21) is of the form
(65)
= L+
where is an element from U mk su
h that
n
o
 (
)
(66)
=0
 
H

; H

p0

H

H

B

p0

I

1

Z A

1

1

B

B

B

;

H

H

H

p0

H

B+

I

1

Z A

belongs to U mk

The matrix
admits a representation
Lemma 5.2.

(67)

:

and satises

(66) if and only if it

= L b
where L is given by (43) and b 2 U m nL k .
Proof. Let b be in U m nL k and let be of the form (67). Sin
e L is blo
k
upper triangular, 2 U mk . Furthermore, in view of (43) and (41),
L  (  "  ) 1
!
#


 IP 2 (
0
)



 
(
IPL
+  L
=
 )
 
k
H

(

H

(

H

B+

I

I

Z

=

"

Z

IP 21



L

#

+

)

"

Z

B+

#

IP 12 (I



I

Z A

L

B+

B+

(

I

1

1

Z A

B+

0
Therefore, the matrix L
n
 (
p0

+

)

1

Z A

I

+

I

) IPL

1

A Z



B+

 

I

A Z







1

B+

I

Z A

1

  IPL:

A Z





) is stri
tly blo
k lower triangular and thus,
n
o
o
b  L (
=0
=
 )
)
1

Z A

1

Z A

H

p0

B+

I

Z A

1

for every blo
k upper triangular b 2 U m nL k .
Conversely, let belongs to U mk and satisfy (66). By Corollary 2.5,
o
n
=0
 (
 )
(

H

p

B+

I

+

)

Z A

1

and thus, the operator  (
is stri
tly blo
k lower triangular. There )
fore

 (
(
 )
 ) = 0
and now it follows from (35) that  QL = 0. In view of (45),
L L = ( QL) = 
B+

B+

I

Z A

I

Z A

1

1

I

I

Z Z

:

ELA
48

Daniel Alpay and Vladimir Bolotnikov

Taking adjoints in the last equality we
on
lude that admits a representation (67)
with b := L , whi
h belongs to U p nL k , by (43).
Using (65) and Lemma 5.2 we obtain the following result.
Theorem 5.3. All
2 U mk whi
h satisfy (21) are parameterized by the formula
(68)
= L + L b
where L and L are given by (62) and (43) respe
tively and b is a parameter from
U m nL k .
Now we des
ribe the set of all solutions of the problem IPL (L).
m nL k
Lemma 5.4. The representation (68) is orthogonal: for every b 2 U
,
(69)
[L b L ℄ = 0 and hL b L i = 0
( +

H

)

H

H

H

H

H

(

+

H

)

(

H

H; H

H; H



+

)

:



. It follows from (36) that   IPL2   TL = 0 . Taking advantage
of the last equality together with (41), (49) and (62) we get
Proof

1

Z ; B+

A Z









 

(
IPL 2  m TL
IPL
 )
 

 

  IP 12   TL
 
=
IPL
 
  L
1
2
+ (
IP1L  ( nL 0) TL
 )
=
(
IPL 2  ( nL 0) TL
 )
whi
h shows (sin
e the proje
tion TL is blo
k diagonal) that L L is stri
tly blo
k
upper triangular. Therefore, for every blo
k upper triangular b , the matrix L L b
is also stri
tly blo
k upper triangular and hen
e,
[L b L ℄ = ( L L b ) = 0
In parti
ular,
hL b L i = Tr ( L L b ) = 0
whi
h ends the proof.
Theorem 5.5. All solutions
of the problem IPL (L ) are parameterized by
the formula (68) the parameter b varies in U m nL k and is subje
t to
i
h
(70)
IPL 
TL b TL b  L
Proof. In view of (44),
i
h

i
 h


b  TL b = TL b TL b
b  L L b =
L b L b =
whi
h together with (63), (68) and (69) leads to
i
i
h
h
[
℄ = L + L b L + L b = [ L L ℄ + h L b L b i
=
IPL  + TL b TL b
H

 L =

L

1

B

1

B

B

B

I

A Z

I

A Z

B+

B+

I

1

1

Z A

Z ; I

1

I

I

1

A Z

1

Z A

Z

1

Z A

Z

Z ; B+

;

I

;

I

H

H

H; H

p0

H; H

H

H

:

H

H

;

H

H

(

H

H;

H;

H; H

H

H

p0

H

H; H

H

+

)

1

B

H

p0

H

B

H

H

H

B

H;

H;

; H

1

:

B

H;

H

H

H

H

ELA
49

Interpolation for upper triangular matri
es

and thus, the matrix H of the form (68) satises (30) if and only if the
orresponding
parameter Hb satises (70).
6. Solutions of the two{sided problem. Using the results from the two previous se
tions we now des
ribe the set of all solutions of problems IP and IP
. By
Theorem 4.3 all operators H 2 U mk satisfying the
ondition (20) are given by the
formula (59). It is possible to restri
t the set of parameters Hb in (59) in su
h way
that the operator H of the form (59) would satisfy also (21) and (22). We begin with
the following auxiliary result; for the proof see [2, Lemma 6.1℄.
n k and 2 U nL m , it holds
Lemma 6.1. For every
hoi
e of matri
es  2 U L

that

n
P0

where

L

o

 
1  = P0

n

A Z

I

2 DnL k is dened by

L



= P0

 


 



1

A Z

I

1

A Z

I

o

L





 .

A matrix H of the form (59) belongs to U mk and satises
ondib belongs to U m(k+nR )
tions (21), (22) if and only if the
orresponding parameter H
and satises the
ondition
Lemma 6.2.

n

(71)

p0

where Bb



(

b B+ I
TR H

2 DnR nL is dened by

(72)

"

= TR

b
B

Z A

Z

)

1

o

= Bb
#

1
IPR 2 Z
B

Proof. Let H be of the form (59). Multiplying (59) by R on the right and using

(44) we get

b

H TR

= (H

HR

) R :

Making use of (48), we rewrite the last equality as

b TR = H C (I Z  A ) 1 IP
H
(73)


C+

(I

1
2





Z ; I k



R



1
) 1 IPR 2 Z ; 0

Z A



TR
TR :

Sin
e TR is blo
k diagonal, the se
ond term on the right hand side in (73) is stri
tly
blo
k lower triangular and therefore,

o

n
 
1 B  C+ (I Z  A ) 1 IP 12 Z  ; 0 TR = 0:
P0
I
A Z
 



+

Then it follows from (73) that
 
1 B  Hb TR =
I
A Z
P0
 

(74)

= P0

I

 

A Z
 

+

1



B+ H



C

(I

R



Z A



1
) 1 IPR 2 Z ;


Ik

TR :

ELA
Daniel Alpay and Vladimir Bolotnikov

50

Suppose that H satises (22) and the se
ond equality from (26) (whi
h is equivalent
1
1
to (21) by Remark 2.6). Then, as IPR 2 Z = Z Z IPR 2 Z , it follows from (74) that
o
n
 
1 B  Hb TR
I
A Z
P0
+

oi
n
h
1


TR
= P0 I A Z 1 B+ H C (I Z A ) 1 Z Z IPR 2 Z ; Ik


1
= Z IP 2 Z  ; B  TR :


R

Taking adjoints in the last equality, using Remark 2.6 and (72), we get (71):
p

0

n

TR b 

1o = T
Z A )
R

+ (I

H B

"

 
Z IP
R Z
1
2

#

= Bb

B

:

Conversely, let Hb satisfy (71) and let H be of the form (59). Then
o
n
o

n




(75) P0 I A Z 1 B+ H = P0 I A Z 1 B+ HR + Hb R :
Applying Lemma 6.1 to  = B+ Hb and = R and taking into a

ount (71) we
obtain
o
o
n
n
 
1 B  Hb R = P0 I A Z 
1 Bb R :
(76)
P0
I
A Z
+
 
It follows from (42), (48) and (72), that
b  R
B

=
=




Z IP
R Z ;
1
2


Z

IPR

1
2


B



Z ;

B

(77)

= B +



TR



"



1

Z

IPR 2 (I

A Z




IPR1

Z

1
2

IPR (I



A Z



1

2 
A Z IP Z ;




A )
B C

(Z

) 1 C

Ik

"

) 1 C

Ik


A + B C




R





IPR1 (I


C



"

#



R





1

Z

IPR 2 (I

A Z

) 1 C

Ik

1 C :

A Z )


Furthermore, in view of (50),
1 C
1



B+ HR = B+ C+ IP
A Z )
R (I
whi
h being added to (77) leads, on a

ount of (19), to


B+ HR + B R =


= B  + (Z A ) B  C + B+ C+ IPR1 (I


= B  + Z A Z Z IPR1 (I A Z ) 1 C 


(78)
= B  + I A Z  Z IP 1 (I A Z ) 1 C  :


#



A Z

) 1 C

#

ELA
51

Interpolation for upper triangular matri
es

Substituting (76) into (75) and using (78) we obtain
o
o
n
n
 
1 B  H = P0 I A Z 
1 B  HR + B  R

P0
I
A Z
+
+
 
n


= P0 I A Z  1 B  + P0 Z IP 1 (I




R



A Z

o
) 1 C :

Sin
e the matrix Z IPR1 (I A Z ) 1 C  is stri
tly blo
k upper triangular,
the se
ond term on the right hand side in the last equality equals to zero and thus,
o
n
n
o
 
1 B  H = P0 I A  Z 
1 B  = B 
I
A Z
P0
+
 
+
whi
h is equivalent to (21), by Remark 2.6.
The veri
ation of (22) is done in mu
h the same way: in view of (59),



  1 B+ H C (I Z A ) 1 Z
I
A Z
P0
 


(79)


 1
= P0 I A Z  B+ HR + Hb R C (I Z A ) 1 Z :




Applying Lemma 6.1 to  = B+ Hb and = R C (I Z A ) 1 Z (whi
h is stri
tly
blo
k lower triangular) and taking into a

ount (71) we obtain
n
o
 
1 B  Hb R C (I Z  A ) 1 Z 
I
A Z
P0
+


o
n


1
1
= P0 I A Z  Bb R C (I Z  A ) Z 






whi
h being substituted to (79), leads to


 1
1





I
A Z
B+ H C
(I Z A ) Z
P0


 1

b  R C (I
B+ HR + B
= P0 I A  Z 






1 Z

Z A )



:

Upon substituting (78) into this equality and using (37) we get
o
n
 
1 B  H C (I Z  A ) 1 Z 
I
A Z
P0
+


n
o


1
1
= P0 I A Z  B  C (I Z  A ) Z 
+P0
= P0

n











o
IPR1 (I A Z ) 1 C  C (I Z A ) 1 Z
o
 
1 B  C (I Z  A ) 1 Z 
A Z

Z

n

+P0



I

n

Z

(I



Z A

Sin
e the matri
es
 
1 B  C (I
I
A Z

o



) 1 Z + P0


Z A



n

Z

) 1 Z and

IPR1 (I

Z



A Z

IPR1 (I

o
) 1 A Z IPR :



A Z

) 1 A Z IPR

ELA
Daniel Alpay and Vladimir Bolotnikov

52

are respe
tively, stri
tly blo
k lower and blo
k upper triangular, the rst and the
third terms on the right hand side in the last equality are equal to zero and by (18),
P0

n

I

A Z




1

 HC
B+

Z A )

(I

1

Z

o

n

= P0 Z (I A Z ) 1 Z
= Z Z = :

o

Therefore H satises the
ondition (22) whi
h ends the proof of the lemma.
A

ording to Theorem 4.3, all operators Hb 2 U (nR +k)m whi
h satisfy (71) are
of the form
(80)

b
H

L + L h
=H

where L is dened by (43),
(81)

L
H

= B+ (I

Z A )

1

IPL 1 Bb

and h is an arbitrary operator from U (k+nR )(m+nL ) . Sin
e TR is the proje
tion, it
follows from (72) that
(82)

L TR
H

= H
L:

Substituting (80) into (59) we
ome to the following theorem.
Theorem 6.3.

formula

All solutions H of the problem

(83)

H

IP

are parameterized by the

= HR + H
L R + LhR

L ; L and R are blo
k upper triangular matri
es dened by (50), (81),
where HR ; H
(43) and (42) respe
tively and h is a free parameter from U (k+nR )(m+nL ) .

Note that equation (83) is in fa
t (31) with

(84)

Hmin

= HR + H
L R :

A

ording to Lemma 5.1 and in view of (82),

hHbL TR ; Hb L TR i

= hHb L ; Hb Li = Tr Bb IPL 1 Bb 

while Lemmas 4.4 and 5.4 ensure that the representation (83) is orthogonal with
respe
t to the inner produ
t (5). Due to (44) and (45),

hH; H i = hHR ; HR i + hHbL TR ; Hb L TR i + hTL hTR ; TL hTR i
= Tr C IPR C  + Tr Bb IPL Bb + hTL hTR ; TL hTR i
+

1

1

+

This equality leads to the des
ription of all the solutions to the problem
under the additional norm
onstraint (28).
Theorem 6.4.

(i) The problem

IP

is solvable if and only if

Tr C+ IPR1 C+ + Tr Bb IPL 1 Bb


:

IP

(i.e.,

ELA
53

Interpolation for upper triangular matri
es

(ii) All solutions of the problem IP
are parameterized by the formula (83) when the
parameter h 2 U (k+nR )(m+nL ) is su
h that

Tr C+ IPR1 C+

hTL hTR ; TL hTR i 

Tr Bb IPL 1 Bb  =

kH kHS :
min

2

Theorem 3.1 is
learly a
onsequen
e of these two last theorems.
7. The minimal norm solution. It was shown in the previous se
tion that
the minimal norm solution Hmin of the problem IP is given by the formula (84)
and
ontains as an additive term the minimal norm solution HR of the right{sided
problem (20). Of
ourse, kHminkHS is not less than kHR kHS and their dieren
e is
determined by the supplementary left{sided problem (71). From symmetry arguments
the presen
e of the minimal norm solution HL of the left{sided problem (21) as a term
in the additive representation of Hmin should be expe
ted.
Lemma 7.1. The matrix Hmin dened by (84)
an be represented as
(85)

= HL + LHb R

Hmin

where HL and L are given by (62) and (43) respe
tively,

(86)

= Cb+ IPR1 (I

bR
H

and where

"

(87)

b+
C

= TL

A Z )

1

C

#

Z IPL 2 Z
C+
1

:

Proof. First we note that the se
ond equality in (87) follows immediately from
(36). As a
onsequen
e of (87) and (49) we get


LCb+ = B+ (I

Z A )

= C+ + B+ (I

1

IPL Z ; Im

Z A )

Multiplying this equality by IPR1 (I
a

ount (50), (86), we obtain
(88)

LHb R = HR + B+ (I

Z A )

IPR

1

1
2

(I

1

IPL 1



A Z )

"


TL

C

n

b L R
H

= HL + B+ (I






#



 C+ :
B+

Z

on the right and taking into

IPL 1 I A Z
A Z ) 1 C  :
1

Furthermore, multiplying the equality (77) by B+ (I
and taking into a

ount (62), (81), we get
(89)

1

A Z

I
1

Z IPL 2 Z
C+



Z

Z A )

Z A ) 1 IPL 1
(Z A )
Z IPR1 Z (I A Z ) 1 C  :

1

 C+
B+

o

IPL 1 from the left
B C



ELA
54

Daniel Alpay and Vladimir Bolotnikov

Subtra
ting this equality from (88) and using (19), we get
LHb R

b R = HR

HL

HL

whi
h means that (84) and (85) dene the same operator Hmin.
It follows from (84) and (89) that Hmin
an be represented as
(90)

H

min(z ) = HL + HC + HR

where HL , HR are given by (62), (50) respe
tively and
1 IP 1  (Z A ) B  C IP 1 (I
HC = B + ( I
Z A )


L
R



A Z

) 1 C :

The representation (90) is not orthogonal; nevertheless it turns out that
HL

?(

HC

+ HR )

(HL + HC ) ? HR :

and

Indeed, in view of (88) and (89),
HC

+ HR = Hb LR ;

HL

+ HC = LHbR

and the
laimed orthogonalities hold by Lemmas 4.4 and 5.4.
REFERENCES
[1℄ D. Alpay and V. Bolotnikov. On tangential interpolation in reprodu
ing kernel Hilbert spa
e
modules and appli
ations. In H. Dym, B. Fritzs
he, V. Katsnelson, and B. Kirstein, editors,
Topi
s in interpolation theory, volume 95 of Operator Theory: Advan
es and Appli
ations,
pages 37{68. Birkh
auser Verlag, Basel, 1997.
[2℄ D. Alpay and V. Bolotnikov. On two-sided interpolation for upper triangular Hilbert-S
hmidt
operators. Integral Equations Operator Theory, 31:259{286, 1998.
[3℄ D. Alpay, V. Bolotnikov, B. Freydin, and Y. Peretz. On bitangential interpolation in the timevarying setting for Hilbert-S
hmidt operators: the
ontinuous time
ase. J. Math. Anal.
Appl., 228:275{292, 1998.
[4℄ D. Alpay, V. Bolotnikov, and L. Rodman. Two{sided tangential interpolation for Hilbert{
S
hmidt operator fun
tions on polydisks. In A. Dijksma, R. Kaashoek, and A. Ran, editors,
Operator Theory: Advan
es and Appli
ations, Birkh
auser Verlag, Basel, 2000. To appear.
[5℄ D. Alpay and P. Dewilde. Time-varying signal approximation and estimation. In M. Kaashoek,
J.H. van S
huppen, and A.C.M. Ran, editors, Signal pro
essing, s
attering and operator
theory, and numeri
al methods (Amsterdam, 1989), volume 5 of Progress in systems and
ontrol theory, pages 1{22. Birkh
auser Boston, Boston, MA, 1990.
[6℄ D. Alpay, P. Dewilde, and H. Dym. Lossless inverse s
attering and reprodu
ing kernels for
upper triangular operators. In I. Gohberg, editor, Extension and interpolation of linear
operators and matrix fun
tions, pages 61{135. Birkh
auser, Basel, 1990.
[7℄ D. Alpay and Y. Peretz. Quasi{
oisometri
realizations of upper triangular matri
es. Fields
Institute Publi
ations, AMS, to appear.
[8℄ D. Alpay and V. Bolotnikov. Two-sided interpolation for matrix fun
tions with entries in the
Hardy spa
e. Linear Algebra Appl., 223/224:31{56, 1995. Spe
ial issue honoring Miroslav
Fiedler and Vlastimil Ptak.
[9℄ J. Ball, I. Gohberg, and M.A. Kaashoek. Nevanlinna{Pi
k interpolation for time{varying
input{output maps: the dis
rete
ase. In I. Gohberg, editor, Time{variant systems and
interpolation, volume 56 of Operator Theory: Advan
es and Appli
ations, pages 1{51.
Birkhauser Verlag, Basel, 1992.

ELA
Interpolation for upper triangular matri
es

55

[10℄ J.A. Ball and I. Gohberg. Classi
ation of shift invariant subspa
es of matri
es with Hermitian
form and
ompletion of matri
es. In H. Bart, I Gohberg, and M.A. Kaashoek, editors,
Operator theory and systems (Amsterdam, 1985), pages 23{85. Birkh
auser, Basel, 1986.
[11℄ J.A. Ball and I. Gohberg. Pairs of shift invariant subspa
es of matri
es and non
anoni
al
fa
torization. Linear and Multilinear Algebra, 20:27{61, 1986.
[12℄ J.A. Ball and I. Gohberg. Shift invariant subspa
es, fa
torization, and interpolation for matri
es. I. The
anoni
al
ase. Linear Algebra Appl., 74:87{150, 1986.
[13℄ P. Dewilde and H. Dym. Interpolation for upper triangular operators. In I. Gohberg, editor,
Time{variant systems and interpolation, volume 56 of Operator Theory: Advan
es and
Appli
ations, pages 153{260. Birkh
auser Verlag, Basel, 1992.
[14℄ P. Dewilde and Alle-Jan van der Veen. Time-varying systems and
omputations. Kluwer
A
ademi
Publishers, Boston, MA, 1998.
[15℄ H. Dym. J
ontra
tive matrix fun
tions, reprodu
ing kernel Hilbert spa
es and interpolation.
Published for the Conferen
e Board of the Mathemati
al S
ien
es, Washington, DC, 1989.
[16℄ H. Dym and B. Freydin. Bitangential interpolation for triangular operators when the Pi
k
operator is stri
tly positive. In H. Dym, B. Fritzs
he, V. Katsnelson, and B. Kirstein,
editors, Topi
s in interpolation theory, volume 95 of Operator Theory: Advan
es and
Appli
ations, pages 143{164. Birkh
auser Verlag, Basel, 1997.
[17℄ H. Dym and B. Freydin. Bitangential interpolation for upper triangular operators. In H. Dym,
B. Fritzs
he, V. Katsnelson, and B. Kirstein, editors, Topi
s in interpolation theory, volume 95 of Operator Theory: Advan
es and Appli
ations, pages 105{142. Birkhauser Verlag,
Basel, 1997.
[18℄ S. Itoh. Reprodu
ing kernels in modules over C  {algebras and their appli
ations. Bull. Kyushu
Inst. Te
h., 37:1{20, 1990.