Chapter 5 Linear Transformations and Operators

1 Chapter 5 Linear Transformations The Algebra of Linear TransformationsTheorem vector spaces over the fieldF. LetTandUbetwo Linear Transformations fromVintoW. The function(T+U)defined pointwiseby(T+U) (v) =Tv+Uvis a Linear transformation fromVintoW. Furthermore, ifs F, the function(sT)defined by(sT) (v) =s(Tv)is also a Linear transformation fromVintoW. The set of all Linear transformationfromVintoW, together with the addition and scalar multiplication defined above,is a vector space over the thatTandUare Linear transformation fromVintoW.

2 For(T+U)defined above, we have(T+U) (sv+w) =T(sv+w) +U(sv+w)=s(Tv) +Tw+s(Uv) +Uw=s(Tv+Uv) + (Tw+Uw)=s(T+U)v+ (T+U)w,8384 Chapter 5. Linear Transformations AND Operators which shows that(T+U)is a Linear transformation . Similarly, we have(rT) (sv+w) =r(T(sv+w))=r(s(Tv) + (Tw))=rs(Tv) +r(Tw)=s(r(Tv)) +rT(w)=s((rT)v) + (rT)wwhich shows that(rT)is a Linear verify that the set of Linear Transformations fromVintoWtogether with theoperations defined above is a vector space, one must directly check the conditionsof Definition These are straightforward to verify, and we leave this exerciseto the denote the space of Linear Transformations fromVintoWbyL(V,W).

3 NotethatL(V,W)is defined only whenVandWare vector spaces over the same ann-dimensional vector space over the fieldF, and letWbe anm-dimensional vector space overF. Then the spaceL(V,W)is finite-dimensional and has ,W, andZbe vector spaces over a fieldF. LetT L(V,W)andU L(W,Z). Then the composed functionUTdefined by(UT) (v) =U(T(v))is a Linear transformation ,v2 Vands F. Then, we have(UT) (sv1+v2) =U(T(sv1+v2))=U(sTv1+Tv2)=sU(Tv1) +U(Tv2)=s(UT) (v1) + (UT) (v2),as a vector space over the fieldF, alinear operatoronVisa Linear transformation THE ALGEBRA OF Linear TRANSFORMATIONS85 Definition Linear transformationTfromVintoWis calledinvertibleifthere exists a functionUfromWtoVsuch thatUTis the identity function onVandTUis the identity function onW.

4 IfTis invertible, the functionUis uniqueand is denoted byT 1. Furthermore,Tis invertible if and only one-to-one:Tv1=Tv2= v1= onto: the range the vector spaceVof semi-infinite real sequencesR wherev= (v1,v2,v3,..) Vwithvn Rforn N. LetL:V Vbe theleft-shift Linear transformation defined byLv= (v2,v3,v4,..)andR:V Vbe the right-shift Linear transformation defined byRv= (0,v1,v2,..).Notice thatLis onto but not one-to-one andRis one-to-one but not onto. Therefore,neither transformation is the normed vector spaceVof semi-infinite real sequencesR with the standard Schauder basis{e1,e2.}

5 }. LetT:V Vbe the lineartransformation that satisfiesTei=i 1eifori= 1,2,.. Let the Linear transfor-mationU:V VsatisfyUei=ieifori= 1,2,.. It is easy to verify thatU=T 1andUT=TU= example should actually bother you somewhat. SinceTreduces vectorcomponents arbitrarily, its inverse must enlarge them arbitrarily. Clearly, this is nota desirable property. Later, we will introduce a norm for Linear transforms whichquantifies this vector spaces over the fieldFand letTbe alinear transformation fromVintoW. IfTis invertible, then the inverse functionT 1is a Linear transformation vectors inWand lets F.

6 Definevj=T 1wj, forj= 1,2. SinceTis a Linear transformation , we haveT(sv1+v2) =sT(v1) +T(v2) =sw1+ 5. Linear Transformations AND OPERATORSThat is,sv1+v2is the unique vector inVthat maps tosw1+w2underT. It followsthatT 1(sw1+w2) =sv1+v2=s(T 1w1)+T 1w2andT 1is a Linear a mapping between algebraic structures which preservesall relevant structure. Anisomorphismis a homomorphism which is also invert-ible. For vector spaces, the relevant structure is given by vector addition and scalarmultiplication. Since a Linear transformation preserves both of these operation, it isalso avector space homomorphism.

7 Likewise, an invertible Linear transformation isavector space Linear Functionals on Vector SpacesDefinition a vector space over a fieldF. A Linear transformationffromVinto the scalar fieldFis called alinear is,fis a functional onVsuch thatf(sv1+v2) =sf(v1) +f(v2)for allv1,v2 Vands a field and lets1,..,snbe scalars inF. Then the func-tionalfonFndefined byf(v1,..,vn) =s1v1+ +snvnis a Linear functional. It is the Linear functional which is represented by the matrix[s1s2 sn]relative to the standard ordered basis forFn.

8 Every Linear functional onFnis ofthis form, for some scalarss1,.., a positive integer andFa field. IfAis ann nmatrixwith entries inF, thetraceofAis the scalartr(A) =A11+A22+ + Linear FUNCTIONALS ON VECTOR SPACES87 Example trace function is a Linear functional on the matrix spaceFn nsincetr(sA+B) =n i=1(sAii+Bii)=sn i=1 Aii+n i=1 Bii=str(A) + tr(B).Example [a,b]be a closed interval on the real line and letC([a,b])bethe space of continuous real-valued functions on[a,b]. ThenL(g) = bag(t)dtdefines a Linear functionalLonC([a,b]).

9 Definition a vector space. The collection of all Linear functionalsonV, denotedL(V,F), forms a vector space. We also denote this space byV andcall it thedual following theorem shows that, ifVis finite dimensional, thendimV = this case, one actually finds thatVis isomorphic toV . Therefore, the two spacescan be identified with each other so thatV=V for finite a finite-dimensional vector space over the fieldF, andletB=v1,..,vnbe a basis forV. There is a unique dual basisB =f1,..,fnforV such thatfj(vi) = ij. For each Linear functional onV, we havef=n i=1f(vi)fiand for each vectorvinV, we havev=n i=1fi(v) 5.

10 Linear Transformations AND ,..,vnbe a basis forV. According to Theorem , there is aunique Linear functionalfionVsuch thatfi(vj)= , we obtain fromBa set ofndistinct Linear functionalsf1,..,fnonV. Thesefunctionals are linearly independent; suppose thatf=n i=1sifi,thenf(vj)=n i=1sifi(vj)=n i=1si ij= particular, iffis the zero functional,f(vj)= 0forj= 1,..,nand hence thescalars{sj}must all equal0. It follows that the functionalsf1,..,fnare linearlyindependent. SincedimV =n, we conclude thatB =f1,..,fnforms a basisforV , thedual , we want to show that there is a unique basis which is dual toB.