Some linear preserver problems on B(H) concerning rank and corank

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