a Xi :ma h/9809102 1 [ma h.OA] 18 Sep 1998
SOME LINEAR PRESERVER PROBLEMS ON B(H)
CONCERNING RANK AND CORANK
LAJOS MOLN´
AR
Abs ac . As a con inua ion o he wo k on linea maps be ween op-
e a o algeb as which p ese e ce ain subse s o ope a o s wi h ini e
ank, o ini e co ank, he e we conside he p oblem inbe ween, ha
is, we ea he ques ion o p ese ing ope a o s wi h in ini e ank and
in ini e co ank. Since, as i u ns ou , in his gene ali y ou p ese e s
canno be w i en in a nice o m wha we ha e go used o when dealing
wi h linea p ese e p oblems, hence we es ic ou a en ion o ce -
ain impo an classes o ope a o s like idempo en s, o p ojec ions, o
pa ial isome ies. We conclude he pape wi h a esul on he o m o
linea maps which p ese e he le ideals in B(H).
1. In oduc ion
Linea p ese e p oblems ep esen one o he mos ac i e esea ch opics
in ma ix heo y (see he su ey pape [8]). In he las decade conside able
a en ion has been also paid o simila ques ions in in ini e dimension, ha
is, o linea p ese e p oblems on ope a o algeb as (see he su ey pape
[2]). In bo h cases, he p oblem is o cha ac e ize hose linea maps on
he algeb a in ques ion which lea e in a ian a gi en subse , o ela ion, o
unc ion. One o he mos impo an such ques ions conce ns he ank. This
is because in many cases p ese e p oblems can be educed o he p oblem
o ank p ese e s. The e o e, i is no su p ising ha a lo o wo k has been
done on such p ese e s (see, o example, [1, 5] o he ini e dimensional
case and [7, 11] o he in ini e dimensional case as well as he e e ences
he ein). In ou ecen pape [6], we conside ed, among o he hings, he
e y simila p oblem o co ank p ese e s which p oblem dese es a en ion,
o cou se, only in he in ini e dimensional case. I His a (complex) in ini e
dimensional Hilbe space, deno e by B(H) he algeb a o all bounded linea
ope a o s ac ing on H. The esul [6, Theo em 3] eads as ollows. Le
1991 Ma hema ics Subjec Classi ica ion. P ima y: 47B49.
Key wo ds and ph ases. Linea p ese e s, pa ial isome ies, idempo en s, p ojec ions,
one-sided ideals.
This esea ch was suppo ed om he ollowing sou ces:
1) Join Hunga ian-Slo ene esea ch p ojec suppo ed by OMFB in Hunga y and he
Minis y o Science and Technology in Slo enia, Reg. No. SLO-2/96,
2) Hunga ian Na ional Founda ion o Scien i ic Resea ch (OTKA), G an No. T–016846
F–019322,
3) A g an om he Minis y o Educa ion, Hunga y, Reg. No. FKFP 0304/1997.
1
2 LAJOS MOLN´
AR
φ:B(H)→B(H) be a bijec i e linea map which is weakly con inuous on
no m bounded se s. I φp ese es he co ank-kope a o s in bo h di ec ions,
hen he e exis in e ible ope a o s A, B ∈B(H) such ha φis o he o m
φ(T) = ATB (T∈B(H)).
Now, i seems o be a na u al ques ion o conside he p oblem o such
p ese e s which a e ”inbe ween” ank p ese e s and co ank p ese e s,
ha is, o de e mine hose linea maps which p ese e he ope a o s wi h
in ini e ank and in ini e co ank. We say ha an ope a o A∈B(H) has
in ini e ank and in ini e co ank i he (Hilbe space) dimensions o ng A
and ng A⊥a e bo h in ini e. He e, ng As ands o he ange o A. We
conside sepa able Hilbe spaces since in his case he e is only one so
o in ini e dimension. Un o una ely, he p ese e s abo e do no ha e such
a nice o m which we ha e go used o when dealing wi h linea p ese e
p oblems. Namely, he e exis p ese e s o he abo e kind which canno
be exp essed in e ms o mul iplica ions by ixed ope a o s and, possibly, by
ansposi ion. To see his, le ψ:B(H)→B(H) be a linea map wi h no m
less han 1 whose ange consis s o ini e ank ope a o s. Then i ollows
om a basic Banach algeb a ac ha he linea map φde ined by
φ(T) = T−ψ(T) (T∈B(H))
is a bijec ion o B(H) on o i sel , and i is easy o check ha φp ese es he
ope a o s wi h in ini e ank and in ini e co ank in bo h di ec ions (obse e
ha his map p ese es he F edholm index as well which p ese e p ob-
lem migh also seem o be na u al a e discussing co ank p ese e s). So,
in o de o ha e one o he desi ed nice o ms o ou p ese e s we should
somehow modi y he p oblem by, o example, es ic ing he se o ope a-
o s wi h in ini e ank and in ini e co ank which we wan p ese e. This is
exac ly wha we a e doing he e conside ing he impo an se s o idempo-
en s, p ojec ions and pa ial isome ies, espec i ely. In he las esul o
he pape we desc ibe he linea bijec ions o B(H) which p ese e he le
ideals in bo h di ec ions. As i will be clea om he p oo , his p oblem is
also connec ed wi h he p oblem o ank p ese e s.
Le us ix he concep s and no a ion ha we shall use h oughou . By a
p ojec ion we mean a sel -adjoin idempo en in B(H). An elemen W∈
B(H) is called a pa ial isome y i i is an isome y on a closed subspace o H
and 0 on i s o hogonal complemen . Algeb aically, Wcan be cha ac e ized
by he equa ion W W ∗W=W. We say ha he ope a o s A, B ∈B(H)
a e o hogonal o each o he i A∗B=AB∗= 0. This means ha he
anges o Aand Bas well as he o hogonal complemen s o hei ke nels
a e o hogonal o each o he . I x, y ∈H, hen x⊗ydeno es he ope a o
de ined by (x⊗y)z=hz, yix(z∈H). In wha ollows F(H) s ands o he
ideal o all ini e ank ope a o s in B(H).
SOME LINEAR PRESERVER PROBLEMS ON B(H) 3
2. Resul s
We begin wi h he desc ip ion o all linea bijec ions φo B(H) which
p ese e he pa ial isome ies o in ini e ank and in ini e co ank in bo h
di ec ions ( his means ha Wis a pa ial isome y wi h in ini e ank and
in ini e co ank i and only i so is φ(A)).
Theo em 1. Le Hbe a sepa able in ini e dimensional Hilbe space. Le φ:
B(H)→B(H)be a linea bijec ion which p ese es he pa ial isome ies o
in ini e ank and in ini e co ank in bo h di ec ions. Then he e exis uni a y
ope a o s U, V ∈B(H)such ha φis ei he o he o m
φ(T) = UTV (T∈B(H))
o o he o m
φ(T) = UT V(T∈B(H))
whe e deno es he anspose wi h espec o an a bi a y bu ixed comple e
o hono mal sequence in H.
In he p oo we shall use he ollowing wo auxilia y esul s.
Lemma 1. Le T, S ∈B(H)be pa ial isome ies wi h S=ST ∗S. Then
we ha e TT∗S=Sand ST∗T=S.
P oo . Deno e Q=TS∗. Since SS∗and T∗Ta e p ojec ions, we compu e
SS∗=ST∗SS∗T S∗=Q∗(SS∗)Q≤Q∗Q=S(T∗T)S∗≤SS∗.
This implies Q∗Q=SS∗. In pa icula , we ob ain kQk ≤ 1 (in ac , he
no m o Qis ei he 0 o 1). Bu Qis an idempo en . Indeed, we ha e
Q2=TS∗TS∗=T(ST∗S)∗=TS∗=Q.
So, Qis a con ac i e idempo en . I is easy o see ha his implies ha
Qis a sel -adjoin idempo en , ha is, a p ojec ion. To e i y his, pick
a bi a y elemen s x∈ke Qand y∈ ng Q. Then we ha e
kyk2≤ kµx +yk2(µ∈C).
An elemen a y a gumen shows ha his implies ha x⊥y. Hence he
ke nel and he ange o Qa e o hogonal o each o he and his e i ies ha
Qis a p ojec ion. Now, om Q∗Q=SS∗we ob ain Q=SS∗. The e o e,
TS∗=SS∗and, as SS∗is he p ojec ion on o ng S, i ollows ha he ange
o Sis included in ha o T. Since TT∗is he p ojec ion on o ng T, we
ha e T T ∗S=S. Simila ly, om he equali y S∗=S∗TS∗one can deduce
T∗TS∗=S∗which is equi alen o ST∗T=S.
Lemma 2. Suppose ha T, S ∈B(H)a e pa ial isome ies. The ope a o
T+λS is a pa ial isome y o e e y λ∈Cwi h |λ|= 1 i and only i T
and Sa e o hogonal o each o he .
4 LAJOS MOLN´
AR
P oo . Suppose i s ha
(T+λS)(T+λS)∗(T+λS) = T+λS
holds o e e y λ∈Cwi h |λ|= 1. Using he ac ha T, S a e pa ial
isome ies, one can conclude ha
0 = λ2ST ∗S+λ(TT∗S+ST∗T) + ¯
λTS∗T+SS∗T+TS∗S.
Since his is alid o e e y λ∈Co modulus 1, choosing he pa icula
alues λ= 1,−1, i, −i, i is easy o deduce ha
ST∗S= 0(1)
TT∗S+ST ∗T= 0(2)
TS∗T= 0(3)
SS∗T+T S∗S= 0.(4)
Mul iplying (2) by T∗ om he le and aking (3) in o accoun , we ob ain
T∗S= 0. Simila ly, mul iplying (4) by S∗ om he igh and aking (1)
in o accoun , we ha e TS∗= 0. So, Tand Sa e o hogonal. As o he
e e se implica ion, i T, S a e mu ually o hogonal pa ial isome ies, hen
i is jus a simple calcula ion ha T+λS is a pa ial isome y o e e y
λ∈Co modulus 1.
P oo o Theo em 1. Le {x1,... ,xk} ⊆ Hand {y1,... ,yk} ⊆ Hbe wo
sys ems o pai wise o hogonal uni ec o s. We claim ha he image o
he ini e ank pa ial isome y R=Pk
j=1 xj⊗yjunde φis also a i-
ni e ank pa ial isome y. Le (en) be an o hono mal sequence in he
o hogonal complemen o {x1,... ,xk}which gene a es a closed subspace
o in ini e codimension. Simila ly, le ( n) be an o hono mal sequence in
{y1,... ,yk}⊥. Deno e U=Pnen⊗ nand le V=U+R. Clea ly, Uand V
a e pa ial isome ies o in ini e ank and in ini e co ank. Mo eo e , o e -
e y λ∈Co modulus 1, he ope a o R+λU = (V−U)+λU is also a pa ial
isome y o in ini e ank and in ini e co ank. The e o e, φ(V) + (λ−1)φ(U)
is a pa ial isome y o e e y λ∈Cwi h |λ|= 1. This means ha wi h he
no a ion V′=φ(V), U′=φ(U) we ha e
(V′+ (λ−1)U′)(V′+ (λ−1)U′)∗(V′+ (λ−1)U′) = (V′+ (λ−1)U′)
o e e y λ∈Co modulus 1. Pe o ming he abo e ope a ions we ob ain a
polynomial in λ, ¯
λwi h ope a o coe icien s which equals 0 on he pe ime e
o he uni disc in he complex plane. Jus as in he p oo o Lemma 2,
choosing he pa icula alues λ= 1,−1, i, −iwe ind ha he coe icien s
o he polynomial in ques ion a e all 0. The e o e, we ha e
−U′+U′V′∗U′= 0,(5)
2U′+V′V′∗U′+U′V′∗V′−U′U′∗V′−2U′V′∗U′−V′U′∗U′= 0,(6)
SOME LINEAR PRESERVER PROBLEMS ON B(H) 5
U′+V′U′∗V′−U′U′∗V′−V′U′∗U′= 0,(7)
and
(8) −2U′−V′V′∗U′−V′U′∗V′−U′V′∗V′+
2U′U′∗V′+U′V′∗U′+ 2V′U′∗U′= 0,
whe e he le hand sides o (5), (6), (7), (8) a e he coe icien s o λ2,λ,¯
λ
and 1, espec i ely. F om (5) and (6) we deduce
V′V′∗U′+U′V′∗V′=U′U′∗V′+V′U′∗U′.(9)
We p o e ha φ(R) = V′−U′is a pa ial isome y. Indeed, we compu e
(V′−U′)(V′−U′)∗(V′−U′) =(10)
V′−U′−V′V′∗U′−V′U′∗V′−U′V′∗V′+U′U′∗V′+U′V′∗U′+V′U′∗U′.
F om (9) we know ha
−V′V′∗U′−U′V′∗V′+U′U′∗V′+V′U′∗U′= 0.
So, we ha e o show ha V′−U′−V′U′∗V′+U′V′∗U′=V′−U′. By (5)
we ha e U′V′∗U′=U′. I emains o e i y ha V′U′∗V′=U′. F om (7)
and (9) we in e ha
U′+V′U′∗V′−V′V′∗U′−U′V′∗V′= 0.(11)
By Lemma 1 i ollows ha V′V′∗U′=U′and U′V′∗V′=U′. Now, (11)
gi es V′U′∗V′=U′. Consequen ly, he igh hand side o he equa ion (10)
is equal o V′−U′which e i ies ha φ(R) is a pa ial isome y.
We nex p o e ha φ(R) has ini e ank. By he p ese e p ope y o
φi is su icien o p o e ha φ(R) has in ini e co ank. We ha e seen ha
o e e y λ∈Co modulus 1, he ope a o R+λU is a pa ial isome y
o in ini e ank and in ini e co ank. This implies ha o R′=φ(R), he
ope a o R′+λU′is a pa ial isome y o e e y λ∈Cwi h |λ|= 1.
Acco ding o Lemma 2, we ob ain ha R′and U′a e o hogonal o each
o he . Since he ange o U′is in ini e dimensional, i ollows ha R′is o
in ini e co ank which implies ha R′is a ini e ank pa ial isome y.
We nex p o e ha φp ese es he pa ial isome ies in gene al. To
see his, le Wbe a pa ial isome y. I i is o ini e ank, hen he e
is now no hing o p o e. So, le Wbe o in ini e ank. In ha case we
ha e an o hogonal sequence (Wn) o pa ial isome ies o in ini e ank and
in ini e co ank whose sum is W. By he p ese e p ope y o φi ollows
ha he ope a o s An=φ(W)−Pn+1
k=1 φ(Wk) = φ(W−Pn+1
k=1 Wk) and
Bn=Pn
k=1 φ(Wk) = φ(Pn
k=1 Wk) a e pa ial isome ies. Because o he
same eason, An+λBnis a pa ial isome y o e e y λ∈Co modulus 1.
By Lemma 2 his implies ha Anand Bna e o hogonal o each o he . The
s a emen [10, Lemma 1.3] ells us ha he se ies o pai wise o hogonal
pa ial isome ies is con e gen in he s ong ope a o opology and i s sum
is also a pa ial isome y. Conside he ope a o s A=φ(W)−Pnφ(Wn)

6 LAJOS MOLN´
AR
and B=Pnφ(Wn). By he jus men ioned esul Pφ(Wn)∗is s ongly
con e gen as well, and Pnφ(Wn)∗=B∗. We hen also ha e φ(W)∗−
Pnφ(Wn)∗=A∗. Since Anand Bna e o hogonal o e e y n∈N, i
is now easy o e i y ha Ais o hogonal o B. The ope a o Bis a
pa ial isome y. As o A, we know ha (An) s ongly con e ges o A
and, as we ha e seen, (A∗
n) s ongly con e ges o A∗. I is well-known ha
he mul iplica ion is s ongly coun inuous on he no m-bounded subse s o
B(H). Consequen ly, we in e ha (AnA∗
n) s ongly con e ges o AA∗and
hen ha (AnA∗
nAn) s ongly con e ges o AA∗A. Since Anis a pa ial
isome y o e e y n, we ob ain ha Ais also a pa ial isome y. Now,
since φ(W) is he sum o he mu ually o hogonal pa ial isome ies Aand
B, i ollows ha φ(W) is a pa ial isome y as well. We ha e assumed
ha φ−1has he same p ese e p ope ies as φ. The e o e, φp ese es
he pa ial isome ies in bo h di ec ions. Suppose ha Wis a maximal
pa ial isome y, ha is, suppose ha Wis a pa ial isome y and he e
is no nonze o pa ial isome y which is o hogonal o W. I V∈B(H) is
a nonze o pa ial isome y which is o hogonal o φ(W), hen V+λφ(W)
is a pa ial isome y o e e y λ∈Cwi h |λ|= 1. This gi es us ha
φ−1(V) + λW is also a pa ial isome y o e e y λ∈Co modulus 1.
By Lemma 2 his esul s in he o hogonali y o φ−1(V) and Wwhich is a
con adic ion. Consequen ly, we ob ain ha φp ese es he maximal pa ial
isome ies which a e p ecisely he isome ies and he coisome ies. I is well-
known ha he se o all ex eme poin s o he uni ball o B(H) consis s
o hese ope a o s exac ly. So, φis a linea map on B(H) which p ese es
he ex eme poin s o he uni ball. The o m o all linea maps wi h his
p ope y ac ing on a on Neumann ac o was de e mined in [9]. The esul
[9, Theo em 1] says ha he e is a uni a y ope a o U∈B(H) such ha
ei he he e exis s a *-homomo phism ψ:B(H)→B(H) such ha
φ(T) = Uψ(T) (T∈B(H))
o he e exis s a *-an ihomomo phism ψ′:B(H)→B(H) such ha
φ(T) = Uψ′(T) (T∈B(H)).
Since ou map φis bijec i e, he same mus hold o he co esponding
mo phism ψo ψ′abo e. Now, e e ing o olk esul s on he o m o *-
au omo phisms and *-an iau omo phisms o B(H), we conclude he p oo .
We con inue wi h a esul o he same spi i on idempo en p ese e s.
Theo em 2. Le Hbe a sepa able in ini e dimensional Hilbe space. Sup-
pose ha φ:B(H)→B(H)is a linea bijec ion which p ese es he idem-
po en s o in ini e ank and in ini e co ank in bo h di ec ions. Then he e is
an in e ible ope a o A∈B(H)such ha φis ei he o he o m
φ(T) = ATA−1(T∈B(H))
SOME LINEAR PRESERVER PROBLEMS ON B(H) 7
o o he o m
φ(T) = AT A−1(T∈B(H)).
In he p oo we shall use he ollowing lemma which is ce ainly well-
known and is included he e only o he sake o comple eness.
Lemma 3. I P, Q ∈B(H)a e idempo en s, hen
(i)P+Qis an idempo en i and only i PQ =QP = 0;
(ii)P−Qis an idempo en i and only i PQ =QP =Q.
P oo . I ollows om elemen a y algeb aic compu a ions.
P oo o Theo em 2. I P, Q ∈B(H) a e idempo en s, hen we w i e P≤Q
i PQ =QP =P. Clea ly, his is equi alen o he condi ion ha ng P⊆
ng Qand ke Q⊆ke P. Le us say ha an idempo en P∈B(H) is egula
i i has in ini e ank and in ini e co ank. We p o e ha o any wo egula
idempo en s P, Q we ha e P≤Qi and only i o e e y egula idempo en
R∈B(H), i Q+Ris a egula idempo en , hen so is P+R. The necessi y
is almos e iden . To he su iciency suppose i s ha ng P* ng Q. Le
x∈Hbe such ha Px =xand Qx 6=x. Choose a egula idempo en
R≤I−Q o which Q+Ris a egula idempo en and Rx 6= 0 (obse e
ha (I−Q)x6= 0). Since P+Ris an idempo en , we ha e P R =RP = 0.
I ollows ha 0 = RP x =Rx which is a con adic ion. Hence, we ha e
ng P⊆ ng Q. The ela ion ke Q⊆ke Pcan be p o ed in a simila
manne . Using he abo e cha ac e iza ion and he p ese e p ope y o
φ, we ob ain ha φp ese es he ela ion ≤be ween egula idempo en s.
Now, i Ris a ini e ank idempo en , hen Rcan be w i en in he o m
R=Q−Pwi h some egula idempo en s P≤Q. Since φ(P)≤φ(Q),
i ollows ha φ(R) = φ(Q)−φ(P) is also an idempo en . We p o e ha
φ(R) is o ini e ank. Choosing a egula idempo en Pwi h R≤P, i
ollows ha P−Ris a egula idempo en and hence φ(P)−φ(R) is also an
idempo en . By Lemma 3 (ii) his implies ha φ(R)≤φ(P). I φ(R) is no
o ini e ank, hen i is egula which implies ha Ris also egula and his
is a con adi ion. The e o e, using he p ese e p ope ies o φand φ−1we
ob ain ha φp ese es he ini e ank idempo en s in bo h di ec ions. I is
now easy o see ha φis a linea bijec ion o F(H) on o i sel . By Lemma 3
(i), o any idempo en s R, R′∈F(H) we ha e RR′=R′R= 0 i and only
i φ(R)φ(R′) = φ(R′)φ(R) = 0. Using his p ope y i is easy o e i y ha
φp ese es he ank-one idempo en s in bo h di ec ions. By [11, Theo em
4.4] we in e ha he e is an in e ible bounded linea ope a o A∈B(H)
such ha φis ei he o he o m
φ(T) = ATA−1(T∈F(H))
o o he o m
φ(T) = AT A−1(T∈F(H)).
8 LAJOS MOLN´
AR
Wi hou loss o gene ali y we may assume ha φis o he i s o m and hen
ha A=I. We in end o show ha φ(T) = T(T∈B(H)). Le P∈B(H)
be a egula idempo en . I Ris any ini e ank idempo en wi h R≤P,
hen jus as abo e, we ob ain R=φ(R)≤φ(P). Since R≤Pwas a bi a y,
i now ollows ha P≤φ(P). Since φ−1has he same p ese e p ope y
as φ, i ollows ha P≤φ−1(P). Bu φp ese es he o de be ween he
egula idempo en s. Hence, we ha e φ(P)≤P. The e o e, φ(P) = P
o e e y egula idempo en P. Since e e y idempo en o ini e co ank is
he sum o wo egula idempo en s, we ob ain ha φ(P) = Pholds o
e e y idempo en P∈B(H). Since e e y elemen o B(H) is a ini e linea
combina ion o p ojec ions [4, Theo em 2], we conclude ha φ(T) = Tis
alid o e e y T∈B(H). This comple es he p oo .
In a simila ashion one can e i y he ollowing esul conce ning p ojec-
ion p ese e s.
Theo em 3. Le Hbe a sepa able in ini e dimensional Hilbe space. Sup-
pose ha φ:B(H)→B(H)is a linea bijec ion which p ese es he p ojec-
ions o in ini e ank and in ini e co ank in bo h di ec ions. Then he e is a
uni a y ope a o U∈B(H)such ha φis ei he o he o m
φ(T) = UTU∗(T∈B(H))
o o he o m
φ(T) = UT U∗(T∈B(H)).
Ou inal esul desc ibes he linea bijec ions φo B(H) which p ese e
he le ideals in bo h di ec ions ( his means ha L ⊆ B(H) is a le ideal
i and only i φ(L) is a le ideal). As i will be clea om he p oo , his
p oblem is also connec ed wi h he p oblem o ank p ese e s.
Theo em 4. Le Hbe a Hilbe space. Suppose ha φ:B(H)→B(H)is
a linea bijec ion p ese ing he le ideals o B(H)in bo h di ec ions. Then
he e a e in e ible ope a o s A, B ∈B(H)such ha φis o he o m
φ(T) = ATB (T∈B(H)).
P oo . The minimal le ideals o B(H) a e p ecisely he se s {x⊗y:x∈H}
o nonze o y∈H. Since φclea ly p ese es he minimal le ideals o B(H)
in bo h di ec ions, we easily deduce ha φis a linea bijec ion o F(H) on o
i sel which p ese es he ank-one ope a o s. By [11, Theo em 3.3] (see also
[7]) i ollows ha he e a e linea bijec ions A, B :H→Hsuch ha φis
ei he o he o m
φ(x⊗y) = Ax ⊗By (x, y ∈H)(12)
o o he o m
φ(x⊗y) = Ay ⊗Bx (x, y ∈H).
Since φis le ideal p ese ing, he second possibili y abo e ob iously canno
occu .
SOME LINEAR PRESERVER PROBLEMS ON B(H) 9
We p o e ha φ(I) is in e ible. Fi s we no e he ollowing. I is ue
in any algeb a wi h uni ha an elemen ails o ha e a le in e se i and
only i his elemen is included in a maximal le ideal ( ecall ha e e y
p ope le ideal is included in a maximal le ideal). The e o e, φp ese es
he le in e ible elemen s o B(H) in bo h di ec ions. We ecall ha an
ope a o Sin B(H) is le in e ible i and only i Sis injec i e and S
has closed ange. Now, le x, y ∈Hbe a bi a y nonze o ec o s. Le
λ∈C. By F edholm al e na i e x⊗y−λI is injec i e i and only i i
is su jec i e. This gi es us ha x⊗y−λI is le in e ible i and only i
i is in e ible. Since he spec um o any elemen in B(H) is nonemp y,
we in e ha he e is a λ∈C o which x⊗y−λI is no le in e ible.
Suppose ha x⊗yis no quasinilpo en , ha is, hx, yi 6= 0. Then he scala
λabo e can be chosen o be nonze o. I ollows ha Ax ⊗By −λφ(I)
is no le in e ible. On he o he hand, φ(I) is le in e ible and hence
i is a le F edholm ope a o (see [3, 2.3. De ini ion, p. 356]). Bu any
compac pe u ba ion o a le F edholm ope a o has closed ange [3, 2.5.
Theo em, p. 356]. So, he ope a o Ax ⊗By −λφ(I) is no le in e ible
bu i has closed ange. The e o e, his ope a o is no injec i e, ha is,
he e exis s a nonze o ec o z∈Hsuch ha λφ(I)z=hz, ByiAx. Clea ly,
his implies ha Ax ∈ ng φ(I). Since x∈Hwas a bi a y, we conclude
ha H= ng A⊆φ(I) which means ha φ(I) is su jec i e. This gi es us
ha φ(I) is in e ible.
We nex show ha he linea ope a o s A, B in (12) a e bounded. Le
x, y ∈H. We ha e seen abo e ha x⊗y−λI is no le in e ible i and only
i λ∈σ(x⊗y), whe e σ(.) deno es he spec um. Simila ly, by F edholm
al e na i e again, φ(I)−1(Ax ⊗By −λφ(I)) is no le in e ible i and only
i λ∈σ(φ(I)−1Ax ⊗By). Since φp ese es he le in e ible ope a o s in
bo h di ec ions, we ob ain
σ(x⊗y) = σ(φ(I)−1Ax ⊗By).
By he spec al adius o mula we ha e
|hx, yi| =|hφ(I)−1Ax, Byi| (x, y ∈H).
Now, an easy applica ion o he closed g aph heo em shows ha A, B a e
con inuous.
E iden ly, we may suppose wi hou any loss o gene ali y ha A=B=I.
Le S∈B(H) be in e ible and w i e C=φ(S). We claim ha C=S. Le
x∈Hbe an a bi a y uni ec o . Then S(I−λx ⊗x) has a le in e se i
and only i λ6= 1. Consequen ly, he ope a o C−λSx ⊗xis injec i e o
e e y λ6= 1 and o e e y uni ec o x∈H. Le z∈Hbe a nonze o ec o .
Le y=S−1Cz which is also nonze o since C=φ(S) is le in e ible. I
hz, yi 6= 0, hen choosing λ=kyk2/hz, yiwe see ha
Cz −λ(1/kyk2Sy ⊗y)(z) = Sy −Sy = 0.
Since zis nonze o, we deduce λ= 1 which means kyk2=hz, yi. The e-
o e, o e e y nonze o ec o z∈Hwe ha e wo possibili ies. Ei he