Matrix Representations of Linear Transformations and ...

1 Matrix Representations of Linear Transformations andChanges of Subspaces and DefinitionsAsubspaceVofRnis a subset ofRnthat contains the zero element and is closed under additionand scalar multiplication:(1)0 V(2)u,v V= u+v V(3)u Vandk R= ku VEquivalently,Vis a subspace ifau+bv Vfor alla,b Randu,v V. (You should try to provethat this is an equivalent statement to the first.)Example {(t,3t, 2t)|t R}. ThenVis a subspace ofR3:(1)0 Vbecause we can taket= 0.(2)Ifu,v V, thenu= (s,3s, 2s)andv= (t,3t, 2t)for some real numberssandt. But thenu+v= (s+t,3s+ 3t, 2s 2t) = (s+t,3(s+t), 2(s+t)) = (t ,3t , 2t ) Vwheret =s+t R.(3)Ifu V, thenu= (t,3t, 2t)for somet R, so ifk R, thenku= (kt,3(kt), 2(kt)) = (t ,3t , 2t ) Vwheret =kt R. Example unit circleS1inR2is not a subspace because it doesn t contain0= (0,0)andbecause, for example,(1,0)and(0,1)lie inSbut(1,0) + (0,1) = (1,1)does not.

2 Similarly,(1,0)lies inSbut2(1,0) = (2,0)does not. Alinear combinationof vectorsv1,..,vk Rnis thefinitesuma1v1+ +akvk( )which is a vector inRn(becauseRnis a subspace of itself, right?). Theai Rare called thecoefficientsof the Linear combination. Ifa1= =ak= 0, then the Linear combination is saidto betrivial. In particular, considering the special case of0 Rn, the zero vector , we note that0may always be represented as a Linear combination ofanyvectorsu1,..,uk Rn,0u1+ + 0uk=0 This representation is called thetrivial representation of 0byu1,..,uk. If, on the other hand,there are vectorsu1,..,uk Rnand scalarsa1,..,an Rsuch thata1u1+ +akuk=01where at least oneai6= 0, then that Linear combination is called anontrivial representation of0. Using Linear combinations we can generate subspaces, as follows. IfSis a nonempty subset ofRn, then thespanofSis given byspan(S) :={v Rn|vis a Linear combination of vectors inS}( )The span of the empty set, , is by definitionspan( ) :={0}( )Remark showed in class thatspan(S)is always a subspace ofRn(well, we showed this forSa finite collection of vectorsS={u1.)}

3 ,uk}, but you should check that it s true for anyS). LetV:= span(S) be the subspace ofRnspanned by someS Rn. ThenSis said togenerateorspanV, and to be ageneratingorspanning setforV. IfVis already known to be a subspace,then finding a spanning setSforVcan be useful, because it is often easier to work with the smallerspanning set than with the entire subspaceV, for example if we are trying to understand the behaviorof Linear Transformations the unit circle inR3which lies in thex-yplane. Thenspan(S)is the entirex-yplane. Example {(x,y,z) R3|x=y= 0,1< z <3}. Thenspan(S)is thez-axis. A nonempty subsetSof a vector spaceRnis said to belinearly independentif, taking any finitenumber of distinct vectorsu1,..,uk S, we have for alla1,..,ak Rthata1u1+a2u2+ +akuk= 0 = a1= =an= 0 That isSis linearly independent if theonlyrepresentation of0 Rnby vectors inSis the trivial this case, the vectorsu1.

4 ,ukthemselves are also said to be linearly independent. Otherwise,if there is at least one nontrivial representation of0by vectors inS, thenSis said to vectorsu= (1,2)andv= (0, 1)inR2are linearly independent, because ifau+bv=0that isa(1,2) +b(0, 1) = (0,0)then(a,2a b) = (0,0), which gives a system of equations:a= 02a b= 0or[102 1][ab]=[00]But the Matrix [102 1]is invertible, in fact it is its own inverse, so that left-multiplying both sidesby it gives[ab]=[1 00 1][ab]=[102 1]2[ab]=[102 1][00]=[00]which meansa=b= 0. 2 Example vectors(1,2,3),(4,5,6),(7,8,9) R3are not linearly independent because1(1,2,3) 2(4,5,6) + 1(7,8,9) = (0,0,0)That is, we have founda= 1,b= 2andc= 1, not all of which are zero, such thata(1,2,3) +b(4,5,6) +c(7,8,9) = (0,0,0). Given 6=S V, a nonzero vectorv Sis said to be anessentially unique Linear combinationof the vectors inSif, up to order of terms, there is one and only one way to expressvas a linearcombination ofu1.

5 ,uk S. That is, if there area1,..,an,b1,..,b` R\{0}and distinctu1,..,uk Sand distinctv1,..,v` Sdistinct, then, re-indexing thebis if necessary,v=a1u1+ +anuk=b1v1+ +b`v`}= k=`and{ai=biui=vi}for alli= 1,..,kIfVis a subspace ofRn, then a subset ofVis called abasisforVif it is linearly independentand spansV. We also say that thevectorsof form a basis forV. Equivalently, as explainedin Theorem below, is a basis if every nonzero vectorv Vis an essentially unique linearcombination of vectors in .Remark the context of inner product spacesVof inifinite dimension, there is a differencebetween a vector space basis, theHamel basisofV, and an orthonormal basis forV, theHilbertbasisforV, because though the two always exist, they are not always equal unlessdim(V)< . Thedimensionof a subspaceVofRnis the cardinality of any basis forV, the number ofelements in (which may in principle be infinite), and is denoted dim(V).

6 This is a well definedconcept, by Theorem below, since all bases have the same it isthe zero vector space{0}or if it has a basis of finite cardinality. Otherwise, if it s basis has infinitecardinality, it is calledinfinite-dimensional. In the former case, dim(V) =| |=k < for somen N, andVis said to bek-dimensional, while in the latter case, dim(V) =| |= , where is acardinal number, andVis said to be are not unique. For example, ={e1,e2}and ={(1,1),(1,0)}are bothbases forR2. IfVis finite-dimensional, say of dimensionn, then anordered basisforVa finite sequence orn-tuple (v1,..,vn) of linearly independent vectorsv1,..,vn Vsuch that{v1,..,vn}is a basisforV. IfVis infinite-dimensional but with a countable basis, then an ordered basis is a sequence(vn)n Nsuch that the set{vn|n N}is a basis Properties of BasesTheorem ,..,vk Rnare linearly independent iff noviis a Linear combinationof the :Letv1.

7 ,vk Rnbe linearly independent and suppose thatvk=c1v1+ +ck 1vk 1(we may supposevkis a Linear combination of the othervj, else we can simply re-index so that thisis the case). Thenc1v1+ +ck 1vk 1+ ( 1)vk=0 But this contradicts Linear independence, since 16= 0. Hencevkcannot be a Linear combinationof the othervk. By re-indexing theviwe can conclude this for , supposev1,..,vkare linearly dependent, there are scalarsc1,..,ck Rnot allzero such thatc1v1+ +ckvk=0 Sayci6= 0. Then,vi=( c1ci)v1+ +( ci 1ci)vi 1+( ci+1ci)vi+1+ +( ckci)vkso thatviis a Linear combination of the othervj. This is the contrapositive of the equivalentstatement, If noviis a Linear combination of the othervj, thenv1,..,vkare linearly independent. Theorem a subspace ofRn. Then a collection ={v1,..,vk}is a basis forViffevery vectorv Vhas an essentially unique expression as a Linear combination of the basis :Suppose is a basis and suppose thatvhas two Representations as a Linear combination ofthevi:v=c1v1+ +ckvk=d1v1+ +dkvkThen,0=v v= (c1 d1)v1+ + (ck dk)vkso by Linear independence we must havec1 d1= =ck dk= 0, orci=difor alli, and sovhas only one expression as a Linear combination of basis vectors, up to order of , suppose everyv Vhas an essentially unique expression as a Linear combination of thevi.

8 Then clearly is a spanning set forV, and moreover theviare linearly independent: for note,since 0v1+ + 0vk=0, by uniqueness of Representations we must havec1v1+ +ckvk=0= c1= =ck= 0. Thus is a basis. 4 Theorem (Replacement Theorem)LetVbe a subspace ofRnand letv1,..,vpandw1,..,wqbe vectors inV. Ifv1,..,vpare linearly independent andw1,..,wqspanV, thenp :LetA= [w1 wq] Rn qbe the Matrix whose columns are thewjand letB=[v1 vp] Rn pbe the Matrix whose columns are thevk. Then note that{v1,..,vp} V= span(w1,..,wq) = imAThus, there existu1,..,up Rqsuch thatAui=vi. Consequently,B= [v1 vp] = [Au1 up] =A[u1 up] =ACwhereC= [u1 up] Rq p. Now, sincev1 vpare linearly independent,c1v1+ +cpvp=0implies allci= 0, , or kerB={0}. But you will notice that kerC kerB,since ifx kerC, then the fact thatB=ACimpliesBx= (AC)x=A(Cx) =A0=0, orx kerC.

9 Since kerB={0}, this means that kerC={0}as well. But thenCmust have at leastas many rows as columns, q, because rref(C) must have the form[IpO], possibly with noOsubmatrix, but at least withIpin the top portion. Theorem a subspace forRn. Then all bases forVhave the same :By the previous theorem two bases ={v1,..,vp}and ={w1,..,wq}forVboth spanVand both are linearly independent, so we havep qandp q. Thereforep=q. Corollary bases :Notice that ={e1, ,en}forms a basis forRn: first, the elementary vectorseispanRn,since ifx= (a1,..,an) Rn, thenx= (a1,..,an) =a1(1,0,..,0) +a2(0,1,0,..,0) + +an(0,..,0,1)=a1e1+a2e2+ +anen span(e1,..,en)Also,e1,..,enare linearly independent, for if0=c1e1+ +cnen=[e1 en] =Incthenc= (c1,..,cn) kerIn={0}, soc1= =cn= 0. Since| |=n, all bases forRnsatisfy| |=nbe the previous theorem. Theorem (Characterizations of Bases)IfVis a subspace ofRnanddim(V) =k, then(1)There are at mostklinearly independent vectors inV.

10 Consequently, a basis is amaximallinearly independentset inV.(2)At leastkvectors are needed to spanV. Thus a basis is aminimal spanning set.(3)Ifkvectors inVare linearly independent, then they form a basis forV.(4)Ifkvectors spanV, then they form a basis :(1) Ifv1,..,vp Vare linearly independent andw1,..,wk Vform a basis forV, thenp kby the Replacement Theorem. (2) Ifv1,..,vp VspanVandw1,..,wk Vform abasis forVthen again we must havek pby the Replacement Theorem. (3) Ifv1,..,vk Varelinearly independent, we must show they also spanV. Pickv Vand note that by (1) the vectorsv1,..,vk,v Vare linearly dependent, because there arek+ 1 of them. (4) Ifv1,..,vk,v VspanVbut are not linearly independent, then sayvk span(v1,..,vk 1). But in this caseV= span(v1,..,vk 1), contradicting (2). Theorem Rm n, thendim(imA) = :This follows from Theorem inSystems of Linear Equations, since ifB= rref(A), thenrankA= rankB=# of columns of the formeiinB=# of nonredundant vectors inA.