The other identities are proven the exact same way Comment If we work with one

The other identities are proven the exact same way

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The other identities are proven the exact same way.Comment.If we work with one Hilbert spaceH, then the above result givesthe fact thatB(H) is aunital involutive Banach algebra.(See Section 5 for theterminology.) This will be crucial for the development of the theory.Another important set of results deals with the kernel and the range.
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304CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSISProposition 7.2(Kernel-Range Identities).LetH1andH2be Hilbert spaces.For any operatorTB(H1,H2), one has the equalities(i) KerT= (RanT);(ii)RanT= (KerT).Proof.(i). If we start with some vectorηKerT, then for everyξH1,we have(η|)H2= (T η|ξ)H1= 0,thus proving thatη,ξH1, i.e.η(RanT). This proves the inclusionKerT= (RanT).To prove the inclusion in the other direction, we start with some vectorηKerT= (RanT), and we prove thatT η= 0.This is however pretty clearsince we haveη(TT η), i.e.0 = (η|TT η)H2= (T η|T η)H1=T η2,which forcesT η= 0.(ii). This follows immediately from part (i) applied toT:RanT=([RanT])= (KerT).Example 7.1.Letn1 be some integer, and consider the Hilbert spaceCn,whose inner product is the standard one:(ξ|η) =nk=1¯ξkηk,ξ= (ξ1, . . . , ξn),η= (η1, . . . , ηn)Cn.The Banach algebraB(Cn) is obviously identified with the algebra Matn×n(C) ofn×nmatrices with complex coefficients. The adjoint operation then correspondsto a (familiar) operation in linear algebra. ForAMatn×n(C), sayA= [ajk]nj,k=1,one takesA= [bjk]nj,k=1to be theconjugate transpose ofA, i.e.theb’s aredefined asbjk= ¯akj.A similar identification works withB(Cm,Cn), identifiedwith Matn×m(C).The adjoint operation is used for defining certain types of operators.Definitions.LetHbe a Hilbert space.A. We say that an operatorTB(H) isnormal, ifT T=TT.B. We say that an operatorTB(H) isself-adjoint, ifT=T.C. We say that an operatorTB(H) ispositive, if(|ξ)H0,ξH.Remarks 7.2.A. Every self-adjoint operatorTB(H) is normal.B. The set{TB(H) :Tnormal}is closed inB(H), in the norm topology.Indeed, if we start with a sequence (Tn)n=1of normal operators, which converges(in norm) to someTB(H), then (Tn)n=1converges toT, and since the multi-plication mapB(H)×B(H)(X, Y)-→XYB(H)is continuous, haveT T= limn→∞TnTnandTT= limn→∞TnTn, so we imme-diately getT T=TT.C. ForTB(H), the following are equivalent (see Remark 3.1):Tis self-adjoint;
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§7.Operator Theory on Hilbert spaces305the sesquilinear mapφT:H×H(ξ, η)-→(|η)HCis sesqui-symmetric, i.e. (|η) =(|ξ),ξ, ηH;(|ξ)R,ξH.In particular, we see that every positive operatorTis self-adjoint.
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