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linear tranformation- VC&LA

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Kaushal Patel

The document discusses linear transformations and their properties. It defines key concepts such as the kernel, range, rank, and nullity of a linear transformation. The kernel is the set of all vectors that map to the zero vector, and is a subspace of the domain. The range is the set of images of all vectors under the transformation. The rank is the dimension of the range, and the nullity is the dimension of the kernel. A linear transformation is one-to-one if different vectors always map to different outputs, and onto if its range is equal to the codomain. An isomorphism is a linear transformation that is both one-to-one and onto, and maps spaces to spaces of the same dimension.

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Linear Transformation By Kaushal Patel
Linear Transformation  Zero transformation: VWVT  vu,,: VT  vv ,0)(  Identity transformation: VVT : VT  vvv ,)(  Properties of linear transformations WVT : 00 )((1)T )()((2) vv TT  )()()((3) vuvu TTT  )()()( )()(Then If(4) 2211 2211 2211 nn nn nn vTcvTcvTc vcvcvcTT vcvcvc       v v 2
The Kernel and Range of a Linear Transformation › Kernel of a linear transformation T: Let be a linear transformationWVT : Then the set of all vectors v in V that satisfy is called the kernel of T and is denoted by ker(T). 0)( vT },0)(|{)ker( VTT  vvv  Ex 1: (Finding the kernel of a linear transformation) ):()( 3223   MMTAAT T Sol:                      00 00 00 )ker(T 3
The kernel is a subspace of V The kernel of a linear transformation is a subspace of the domain V. )16.Theorem(0)0( TPf: VT ofsubsetnonemptyais)ker( then.ofkernelin thevectorsbeandLet Tvu 000)()()(  vuvu TTT 00)()(  ccTcT uu )ker(Tc  u )ker(T vu .ofsubspaceais)ker(Thus, VT Note: The kernel of T is sometimes called the nullspace of T. WVT : 4
 Ex 6: (Finding a basis for the kernel)                 82000 10201 01312 11021 andRiniswhere,)(bydefinedbe:Let 545 A ATRRT xxx Find a basis for ker(T) as a subspace of R5. 5
Sol:                                  000000 041000 020110 010201 082000 010201 001312 011021 0 .. EJG A s t                                                                 1 4 0 2 1 0 0 1 1 2 4 2 2 5 4 3 2 1 ts t t s ts ts x x x x x x   TB ofkernelfor thebasisone:)1,4,0,2,1(),0,0,1,1,2(  6
 Range of a linear transformation T: )(bydenotedisandTofrangethecalledisVin vectorofimagesarein W thatwvectorsallofsetThen the L.T.abe:Let Trange WVT  }|)({)( VTTrange  vv 7
.:Tnnsformatiolinear traaofrangeThe WWV foecapsbusasi  The range of T is a subspace of W Pf: )1Thm.6.(0)0( T WTrange ofsubsetnonemptyais)( TTT ofrangein thevectorbe)(and)(Let vu )()()()( TrangeTTT  vuvu )()()( TrangecTcT  uu ),( VVV  vuvu )( VcV  uu .subspaceis)(Therefore, WTrange 8
 Notes: ofsubspaceis)()1( VTKer L.T.ais: WVT  ofsubspaceis)()2( WTrange 9
 Ex 7: (Finding a basis for the range of a linear transformation)                 82000 10201 01312 11021 andiswhere,)(bydefinedbe:Let 545 A RATRRT xxx Find a basis for the range of T. 10
Sol: BA EJG                                 00000 41000 20110 10201 82000 10201 01312 11021 .. 54321 ccccc 54321 wwwww     )(forbasisais,, )(forbasisais,, 421 421 ACSccc BCSwww   Tofrangefor thebasisais)2,0,1,1(),0,0,1,2(),0,1,2,1(  11
 Rank of a linear transformation T:V→W: TTrank ofrangetheofdimensionthe)(   Nullity of a linear transformation T:V→W: TTnullity ofkerneltheofdimensionthe)(   Note: )()( )()( then,)(bygivenL.T.thebe:Let AnullityTnullity ArankTrank ATRRT mn    xx Rank and Nullity of Linear Transformation 12
then.spacevectorainto spacevectorldimensiona-nanformL.T.abe:Let W VWVT   Sum of rank and nullity Pf: AmatrixnmT anbydrepresenteisLet  )ofdomaindim()ofkerneldim()ofrangedim( )()( TTT nTnullityTrank   rArank )(Assume rArank ATTrank   )( )ofspacecolumndim()ofrangedim()((1) nrnrTnullityTrank  )()()( rn ATTnullity   )ofspacesolutiondim()ofkerneldim()()2( 13
 Ex 8: (Finding the rank and nullity of a linear transformation)              000 110 201 bydefine:L.T.theofnullityandranktheFind 33 A RRT Sol: 123)()ofdomaindim()( 2)()(   TrankTTnullity ArankTrank 14
 Ex 9: (Finding the rank and nullity of a linear transformation) }0{)(ifofranktheFind)( 4isofnullitytheifofranktheFind)( 2israngetheof dimensiontheifofkerneltheofdimensiontheFind)( n.nsformatiolinear traabe:Let 75   TKerTc TTb Ta RRT Sol: 325)ofrangedim()ofkerneldim( 5)ofdomaindim()(   TnT Ta 145)()()(  TnullitynTrankb 505)()()(  TnullitynTrankc 15
One-to-one Transformation vector.singleaofconsistsrangein theevery w ofpreimagetheifone-to-onecalledis:functionA WVT  .thatimplies )()(inV,vanduallforiffone-to-oneis vu vu   TTT one-to-one not one-to-one 16
Onto Transformation inpreimageahasin elementeveryifontobetosaidis:functionA V WVT w  (T is onto W when W is equal to the range of T.) 17
 One-to-one linear transformation }0{)(iff1-1isTThen L.T.abe:Let   TKer WVT Pf: 1-1isSupposeT 0:solutiononeonlyhavecan0)(Then  vvT }0{)(i.e. TKer )()(and}0{)(Suppose vTuTTKer  0)()()(  vTuTvuT L.T.aisT 0)(  vuTKervu 1-1isT 18
 Ex 10: (One-to-one and not one-to-one linear transformation) one.-to-oneis )(bygiven:L.T.The)( T mnnm AATMMTa   matrix.zeroonly theofconsistskernelitsBecause nm one.-to-onenotis:ationtransformzeroThe)( 33 RRTb  .ofalliskernelitsBecause 3 R 19
 Thm 6.7: (Onto linear transformation) .ofdimensionthetoequalisofranktheiffontoisThen l.dimensionafiniteiswhereL.T.,abe:Let WTT WWVT   Thm 6.8: (One-to-one and onto linear transformation) onto.isitifonlyandifone-to-oneisThen.dimension ofbothandspaceorwith vectL.T.abe:Let Tn WVWVT  Pf: 0))(dim(and}0{)(thenone,-to-oneisIf  TKerTKerT )dim())(dim())(dim( WnTKernTrange  onto.isly,Consequent T 0)ofrangedim())(dim(  nnTnTKer one.-to-oneisTherefore,T nWTT  )dim()ofrangedim(thenonto,isIf 20
 Ex 11: neither.oronto,one,-to-oneiswhetherdetermineandof rankandnullitytheFind,)(bygivenis:L.T.The TT ATRRT mn xx           100 110 021 )( Aa          00 10 21 )( Ab       110 021 )( Ac          000 110 021 )( Ad Sol: T:Rn→Rm dim(domain of T) rank(T) nullity(T) 1-1 onto (a)T:R3→R3 3 3 0 Yes Yes (b)T:R2→R3 2 2 0 Yes No (c)T:R3→R2 3 2 1 No Yes (d)T:R3→R3 3 2 1 No No 21
 Isomorphism: other.eachtoisomorphicbetosaidare andthen,tofrommisomorphisanexiststheresuch that spacesvectorareandifMoreover,m.isomorphisancalledis ontoandonetooneisthat:nnsformatiolinear traA WVWV WV WVT   Isomorphic spaces and dimension Pf: .dimensionhaswhere,toisomorphicisthatAssume nVWV onto.andonetooneisthat:L.T.aexistsThere WVT  one-to-oneisT nnTKerTT TKer   0))(dim()ofdomaindim()ofrangedim( 0))(dim( Two finite-dimensional vector space V and W are isomorphic if and only if they are of the same dimension. Bijective Transformation 22
.dimensionhavebothandthatAssume nWV onto.isT nWT  )dim()ofrangedim( nWV  )dim()dim(Thus     .ofbasisabe,,,let andV,ofbasisabe,,,Let 21 21 Wwww vvv n n   nnvcvcvc V  2211 asdrepresentebecaninvectorarbitraryanThen v nnwcwcwcT WVT   2211)( follows.as:L.T.adefinecanyouand v It can be shown that this L.T. is both 1-1 and onto. Thus V and W are isomorphic. 23
Inverse linear Transformation ineveryfors.t.L.T.are:and:If 21 nnnnn RRRTRRT v ))((and))(( 2112 vvvv  TTTT invertiblebetosaidisandofinversethecalledisThen 112 TTT  Note: If the transformation T is invertible, then the inverse is unique and denoted by T–1 . 24
Existence of an inverse transformation .equivalentareconditionfollowingThen the ,matrixstandardwithL.T.abe:Let ARRT nn   Note: If T is invertible with standard matrix A, then the standard matrix for T–1 is A–1 . (1) T is invertible. (2) T is an isomorphism. (3) A is invertible. 25
Finding the inverse of a linear transformation bydefinedis：L.T.The 33 RRT  )42,33,32(),,( 321321321321 xxxxxxxxxxxxT  Sol: 142 133 132 formatrixstandardThe           A T 321 321 321 42 33 32 xxx xxx xxx               100142 010133 001132 3IA Show that T is invertible, and find its inverse. 26
 1.. 326100 101010 011001                AIEJG 11 isformatrixstandardtheandinvertibleisTherefore  ATT             326 101 011 1 A                                  321 31 21 3 2 1 11 326326 101 011 )( xxx xx xx x x x AT vv )326,,(),,( s,other wordIn 3213121321 1 xxxxxxxxxxT  27
.formatrixstandardtheFindaxis.-xtheontoinpointeach projectingbygivenis:nnsformatiolinear traThe 2 22 TR RRT  Sol: )0,(),( xyxT           00 01 )1,0()0,1()()( 21 TTeTeTA  Notes: (1) The standard matrix for the zero transformation from Rn into Rm is the mn zero matrix. (2) The standard matrix for the zero transformation from Rn into Rn is the nn identity matrix In Finding the matrix of a linear transformation 28
Composition of T1:Rn→Rm with T2:Rm→Rp : n RTTT  vvv )),(()( 12 112 ofdomainofdomain, TTTTT    Composition of linear transformations then,andmatricesstandardwith L.T.be:and:Let 21 21 AA RRTRRT pmmn  L.T.ais)),(()(bydefined,:ncompositioThe(1) 12 vv TTTRRT pn  12productmatrixby thegivenisformatrixstandardThe)2( AAATA  29
Pf: nscalar theanybecletandinvectorsbeandLet L.T.)ais((1) n R T vu )formatrixstandardtheis)(2( 12 TAA )()())(())(( ))()(())(()( 1212 11212 vuvu vuvuvu TTTTTT TTTTTT   )())(())(())(()( 121212 vvvvv cTTcTcTTcTTcT  vvvvv )()())(()( 12121212 AAAAATTTT   Note: 1221 TTTT   30
The standard matrix of a composition s.t.intofromL.T.beandLet 33 21 RRTT ),0,2(),,(1 zxyxzyxT  ),z,(),,(2 yyxzyxT  ,'and nscompositiofor thematricesstandardtheFind 2112 TTTTTT   Sol: )formatrixstandard( 101 000 012 11 TA            )formatrixstandard( 010 100 011 22 TA             31
12formatrixstandardThe TTT  21'formatrixstandardThe TTT                             000 101 012 101 000 012 010 100 011 12 AAA                                   001 000 122 010 100 011 101 000 012 ' 21AAA 32
Matrices for Linear Transformations )43,23,2(),,()1( 32321321321 xxxxxxxxxxxT   Three reasons for matrix representation of a linear transformation:                    3 2 1 430 231 112 )()2( x x x AT xx  It is simpler to write.  It is simpler to read.  It is more easily adapted for computer use.  Two representations of the linear transformation T:R3→R3 : 33
Definition 1: A nonzero vector x is an eigenvector (or characteristic vector) of a square matrix A if there exists a scalar λ such that Ax = λx. Then λ is an eigenvalue (or characteristic value) of A. Note: The zero vector can not be an eigenvector even though A0 = λ0. But λ = 0 can be an eigenvalue. Example: Show x  2 1      isaneigenvector for A  2 4 3 6       Solution: Ax  2 4 3 6       2 1       0 0       But for   0, x  0 2 1       0 0       Thus,xisaneigenvectorof A,and  0 isaneigenvalue. Definitions 34
An n×n matrix A multiplied by n×1 vector x results in another n×1 vector y=Ax. Thus A can be considered as a transformation matrix. In general, a matrix acts on a vector by changing both its magnitude and its direction. However, a matrix may act on certain vectors by changing only their magnitude, and leaving their direction unchanged (or possibly reversing it). These vectors are the eigenvectors of the matrix. A matrix acts on an eigenvector by multiplying its magnitude by a factor, which is positive if its direction is unchanged and negative if its direction is reversed. This factor is the eigenvalue associated with that eigenvector. Geometric interpretation of Eigenvalues and Eigenvectors 35
Let x be an eigenvector of the matrix A. Then there must exist an eigenvalue λ such that Ax = λx or, equivalently, Ax - λx = 0 or (A – λI)x = 0 If we define a new matrix B = A – λI, then Bx = 0 If B has an inverse then x = B-10 = 0. But an eigenvector cannot be zero. Thus, it follows that x will be an eigenvector of A if and only if B does not have an inverse, or equivalently det(B)=0, or det(A – λI) = 0 This is called the characteristic equation of A. Its roots determine the eigenvalues of A. Eigenvalues 36
Eigenvalues: examples Example 1: Find the eigenvalues of two eigenvalues: 1,  2 Note: The roots of the characteristic equation can be repeated. That is, λ1 = λ2 =…= λk. If that happens, the eigenvalue is said to be of multiplicity k.          51 122 A )2)(1(23 12)5)(2( 51 122 2           AI 37
Eigenvectors Example 1 (cont.):                 00 41 41 123 )1(:1 AI 0, 1 4 ,404 2 1 1 2121               tt x x txtxxx x                 00 31 31 124 )2(:2 AI 0, 1 3 2 1 2              ss x x x To each distinct eigenvalue of a matrix A there will correspond at least one eigenvector which can be found by solving the appropriate set of homogenous equations. If λi is an eigenvalue then the corresponding eigenvector xi is the solution of (A – λiI)xi = 0 38
Example 2 (cont.): Find the eigenvectors of Recall that λ = 2 is an eigenvector of multiplicity 3. Solve the homogeneous linear system represented by Let . The eigenvectors of  = 2 are of the form and t not both zero.                                  0 0 0 000 000 010 )2( 3 2 1 x x x AI x txsx  31 , , 1 0 0 0 0 1 0 3 2 1                                             ts t s x x x x            200 020 012 A 39
Definition: The trace of a matrix A, designated by tr(A), is the sum of the elements on the main diagonal. Property 1: The sum of the eigenvalues of a matrix equals the trace of the matrix. Property 2: A matrix is singular if and only if it has a zero eigenvalue. Property 3: The eigenvalues of an upper (or lower) triangular matrix are the elements on the main diagonal. Property 4: If λ is an eigenvalue of A and A is invertible, then 1/λ is an eigenvalue of matrix A-1. Properties of Eigenvalues and Eigenvectors 40
Property 5: If λ is an eigenvalue of A then kλ is an eigenvalue of kA where k is any arbitrary scalar. Property 6: If λ is an eigenvalue of A then λk is an eigenvalue of Ak for any positive integer k. Property 8: If λ is an eigenvalue of A then λ is an eigenvalue of AT. Property 9: The product of the eigenvalues (counting multiplicity) of a matrix equals the determinant of the matrix. 41
Theorem: Eigenvectors corresponding to distinct (that is, different) eigenvalues are linearly independent. Theorem: If λ is an eigenvalue of multiplicity k of an n  n matrix A then the number of linearly independent eigenvectors of A associated with λ is given by m = n - r(A- λI). Furthermore, 1 ≤ m ≤ k. Example 2 (cont.): The eigenvectors of  = 2 are of the form s and t not both zero.  = 2 has two linearly independent eigenvectors , 1 0 0 0 0 1 0 3 2 1                                             ts t s x x x x Linearly independent eigenvectors 42
Diagonalization  Diagonalizable matrix: A square matrix A is called diagonalizable if there exists an invertible matrix P such that P-1AP is a diagonal matrix. (P diagonalizes A)  Notes: (1) If there exists an invertible matrix P such that , then two square matrices A and B are called similar. (2) The eigenvalue problem is related closely to the diagonalization problem. APPB 1  43
A diagonalizable matrix           200 013 031 A Sol: Characteristic equation: 0)2)(4( 200 013 031 I 2           A 2,2,4:sEigenvalue 321   Ex.5)p.403(See 0 1 1 :rEigenvecto4)1( 1            p 44
1 0 0 , 0 1 1 :igenvectorE2)2( 32                       pp                         200 020 004 100 011 011 ][ 1 321 APPpppP                                                   400 020 002 010 101 101 ][(2) 200 040 002 100 011 011 ][(1) 1 132 1 312 APPpppP APPpppP  Notes: 45
Condition for Diagonalization An nn matrix A is diagonalizable if and only if it has n linearly independent eigenvectors. Pf: ablediagonalizis)( A ),,,(and][Let diagonaliss.t.invertibleanexiststhere 2121 1 nn diagDpppP APPDP      ][ 00 00 00 ][ 2211 2 1 21 nn n n ppp pppPD                         46
][][ 2121 nn ApApAppppAAP   )ofrseigenvectoareoftorcolumn vecthe..( ,,2,1, APpei nipAp PDAP i iii      t.independenlinearlyare,,,invertibleis 21 npppP   rs.eigenvectotindependenlinearlyhas nA n npppnA    ,,seigenvalueingcorrespondhwit ,,rseigenvectotindependenlinearlyhas)( 21 21 nipAp iii ,,2,1,i.e.   ][Let 21 n pppP  47
PDppp ppp ApApAp pppAAP n n nn n n                              00 00 00 ][ ][ ][ ][ 2 1 21 2211 21 21 ablediagonalizis invertibleistindependenlinearlyare,,, 1 11 A DAPP Pppp n      Note: If n linearly independent vectors do not exist, then an nn matrix A is not diagonalizable. 48
A matrix that is not diagonalizable        10 21 able.diagonaliznotismatrixfollowingthat theShow A Sol: Characteristic equation: 0)1( 10 21 I 2         A 1:Eigenvalue 1                      0 1 :rEigenvecto 00 10 ~ 00 20 I 1pAIA A does not have two (n=2) linearly independent eigenvectors, so A is not diagonalizable. 49
Steps for diagonalizing an nn square matrix: n ,,, 21  Step 2: Let ][ 21 npppP  Step 1: Find n linearly independent eigenvectors for A with corresponding eigenvalues nppp ,,, 21  Step 3: nipApDAPP iii n ,,2,1,where, 00 00 00 2 1 1                        Note: The order of the eigenvalues used to form P will determine the order in which the eigenvalues appear on the main diagonal of D. 50
Diagonalizing a matrix diagonal.issuch thatmatrixaFind 113 131 111 1 APPP A               Sol: Characteristic equation: 0)3)(2)(2( 113 131 111 I           A 3,2,2:sEigenvalue 321   51
21                    000 010 101 ~ 313 111 111 I1 A                                             1 0 1 :rEigenvecto 1 0 1 0 1 3 2 1 pt t t x x x 22                       000 10 01 ~ 113 151 113 I 4 1 4 1 2 A                                             4 1 1 :rEigenvecto 4 1 1 24 1 4 1 4 1 3 2 1 pt t t t x x x 52
33                     000 110 101 ~ 413 101 112 I3 A                                             1 1 1 :rEigenvecto 1 1 1 3 3 2 1 pt t t t x x x                          300 020 002 141 110 111 ][Let 1 321 APP pppP 53
 Notes: k is a positive integer                           k n k k k n d d d D d d d D         00 00 00 00 00 00 )1( 2 1 2 1 1 1 1 1111 111 1 1 )()()( )())(( )( )2(               PPDA PAP APAAP APPPPPAPPAP APPAPPAPP APPD APPD kk k kk 54
Sufficient conditions for Diagonalization If an nn matrix A has n distinct eigenvalues, then the corresponding eigenvectors are linearly independent and A is diagonalizable. 55
 Determining whether a matrix is diagonalizable              300 100 121 A Sol: Because A is a triangular matrix, its eigenvalues are the main diagonal entries. 3,0,1 321   These three values are distinct, so A is diagonalizable. (Thm.7.6) 56
Finding a diagonalizing matrix for a linear transformation diagonal.istorelative formatrixthesuch thatforbasisaFind )33()( bygivennnsformatiolinear trathebeLet 3 321321321321 33 B TRB xxx,xx, xxxx,x,xxT RT:R   Sol:                113 131 111 )()()( bygivenisformatrixstandardThe 321 eTeTeTA T From Ex. 5, there are three distinct eigenvalues so A is diagonalizable. (Thm. 7.6) 3,2,2 321   57
                    300 020 002 ][][][ ][][][ ])([)]([])([ 332211 321 321 BBB BBB BBB ppp ApApAp pTpTpTD  The matrix for T relative to this basis is )}1,1,1(),4,1,1(),1,0,1{(},,{ 321  pppB Thus, the three linearly independent eigenvectors found in Ex. 5 can be used to form the basis B. That is )1,1,1(),4,1,1(),1,0,1( 321  ppp 58

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linear tranformation- VC&LA

  • 2. Linear Transformation  Zero transformation: VWVT  vu,,: VT  vv ,0)(  Identity transformation: VVT : VT  vvv ,)(  Properties of linear transformations WVT : 00 )((1)T )()((2) vv TT  )()()((3) vuvu TTT  )()()( )()(Then If(4) 2211 2211 2211 nn nn nn vTcvTcvTc vcvcvcTT vcvcvc       v v 2
  • 3. The Kernel and Range of a Linear Transformation › Kernel of a linear transformation T: Let be a linear transformationWVT : Then the set of all vectors v in V that satisfy is called the kernel of T and is denoted by ker(T). 0)( vT },0)(|{)ker( VTT  vvv  Ex 1: (Finding the kernel of a linear transformation) ):()( 3223   MMTAAT T Sol:                      00 00 00 )ker(T 3
  • 4. The kernel is a subspace of V The kernel of a linear transformation is a subspace of the domain V. )16.Theorem(0)0( TPf: VT ofsubsetnonemptyais)ker( then.ofkernelin thevectorsbeandLet Tvu 000)()()(  vuvu TTT 00)()(  ccTcT uu )ker(Tc  u )ker(T vu .ofsubspaceais)ker(Thus, VT Note: The kernel of T is sometimes called the nullspace of T. WVT : 4
  • 5.  Ex 6: (Finding a basis for the kernel)                 82000 10201 01312 11021 andRiniswhere,)(bydefinedbe:Let 545 A ATRRT xxx Find a basis for ker(T) as a subspace of R5. 5
  • 6. Sol:                                  000000 041000 020110 010201 082000 010201 001312 011021 0 .. EJG A s t                                                                 1 4 0 2 1 0 0 1 1 2 4 2 2 5 4 3 2 1 ts t t s ts ts x x x x x x   TB ofkernelfor thebasisone:)1,4,0,2,1(),0,0,1,1,2(  6
  • 7.  Range of a linear transformation T: )(bydenotedisandTofrangethecalledisVin vectorofimagesarein W thatwvectorsallofsetThen the L.T.abe:Let Trange WVT  }|)({)( VTTrange  vv 7
  • 8. .:Tnnsformatiolinear traaofrangeThe WWV foecapsbusasi  The range of T is a subspace of W Pf: )1Thm.6.(0)0( T WTrange ofsubsetnonemptyais)( TTT ofrangein thevectorbe)(and)(Let vu )()()()( TrangeTTT  vuvu )()()( TrangecTcT  uu ),( VVV  vuvu )( VcV  uu .subspaceis)(Therefore, WTrange 8
  • 9.  Notes: ofsubspaceis)()1( VTKer L.T.ais: WVT  ofsubspaceis)()2( WTrange 9
  • 10.  Ex 7: (Finding a basis for the range of a linear transformation)                 82000 10201 01312 11021 andiswhere,)(bydefinedbe:Let 545 A RATRRT xxx Find a basis for the range of T. 10
  • 11. Sol: BA EJG                                 00000 41000 20110 10201 82000 10201 01312 11021 .. 54321 ccccc 54321 wwwww     )(forbasisais,, )(forbasisais,, 421 421 ACSccc BCSwww   Tofrangefor thebasisais)2,0,1,1(),0,0,1,2(),0,1,2,1(  11
  • 12.  Rank of a linear transformation T:V→W: TTrank ofrangetheofdimensionthe)(   Nullity of a linear transformation T:V→W: TTnullity ofkerneltheofdimensionthe)(   Note: )()( )()( then,)(bygivenL.T.thebe:Let AnullityTnullity ArankTrank ATRRT mn    xx Rank and Nullity of Linear Transformation 12
  • 13. then.spacevectorainto spacevectorldimensiona-nanformL.T.abe:Let W VWVT   Sum of rank and nullity Pf: AmatrixnmT anbydrepresenteisLet  )ofdomaindim()ofkerneldim()ofrangedim( )()( TTT nTnullityTrank   rArank )(Assume rArank ATTrank   )( )ofspacecolumndim()ofrangedim()((1) nrnrTnullityTrank  )()()( rn ATTnullity   )ofspacesolutiondim()ofkerneldim()()2( 13
  • 14.  Ex 8: (Finding the rank and nullity of a linear transformation)              000 110 201 bydefine:L.T.theofnullityandranktheFind 33 A RRT Sol: 123)()ofdomaindim()( 2)()(   TrankTTnullity ArankTrank 14
  • 15.  Ex 9: (Finding the rank and nullity of a linear transformation) }0{)(ifofranktheFind)( 4isofnullitytheifofranktheFind)( 2israngetheof dimensiontheifofkerneltheofdimensiontheFind)( n.nsformatiolinear traabe:Let 75   TKerTc TTb Ta RRT Sol: 325)ofrangedim()ofkerneldim( 5)ofdomaindim()(   TnT Ta 145)()()(  TnullitynTrankb 505)()()(  TnullitynTrankc 15
  • 16. One-to-one Transformation vector.singleaofconsistsrangein theevery w ofpreimagetheifone-to-onecalledis:functionA WVT  .thatimplies )()(inV,vanduallforiffone-to-oneis vu vu   TTT one-to-one not one-to-one 16
  • 18.  One-to-one linear transformation }0{)(iff1-1isTThen L.T.abe:Let   TKer WVT Pf: 1-1isSupposeT 0:solutiononeonlyhavecan0)(Then  vvT }0{)(i.e. TKer )()(and}0{)(Suppose vTuTTKer  0)()()(  vTuTvuT L.T.aisT 0)(  vuTKervu 1-1isT 18
  • 19.  Ex 10: (One-to-one and not one-to-one linear transformation) one.-to-oneis )(bygiven:L.T.The)( T mnnm AATMMTa   matrix.zeroonly theofconsistskernelitsBecause nm one.-to-onenotis:ationtransformzeroThe)( 33 RRTb  .ofalliskernelitsBecause 3 R 19
  • 20.  Thm 6.7: (Onto linear transformation) .ofdimensionthetoequalisofranktheiffontoisThen l.dimensionafiniteiswhereL.T.,abe:Let WTT WWVT   Thm 6.8: (One-to-one and onto linear transformation) onto.isitifonlyandifone-to-oneisThen.dimension ofbothandspaceorwith vectL.T.abe:Let Tn WVWVT  Pf: 0))(dim(and}0{)(thenone,-to-oneisIf  TKerTKerT )dim())(dim())(dim( WnTKernTrange  onto.isly,Consequent T 0)ofrangedim())(dim(  nnTnTKer one.-to-oneisTherefore,T nWTT  )dim()ofrangedim(thenonto,isIf 20
  • 21.  Ex 11: neither.oronto,one,-to-oneiswhetherdetermineandof rankandnullitytheFind,)(bygivenis:L.T.The TT ATRRT mn xx           100 110 021 )( Aa          00 10 21 )( Ab       110 021 )( Ac          000 110 021 )( Ad Sol: T:Rn→Rm dim(domain of T) rank(T) nullity(T) 1-1 onto (a)T:R3→R3 3 3 0 Yes Yes (b)T:R2→R3 2 2 0 Yes No (c)T:R3→R2 3 2 1 No Yes (d)T:R3→R3 3 2 1 No No 21
  • 22.  Isomorphism: other.eachtoisomorphicbetosaidare andthen,tofrommisomorphisanexiststheresuch that spacesvectorareandifMoreover,m.isomorphisancalledis ontoandonetooneisthat:nnsformatiolinear traA WVWV WV WVT   Isomorphic spaces and dimension Pf: .dimensionhaswhere,toisomorphicisthatAssume nVWV onto.andonetooneisthat:L.T.aexistsThere WVT  one-to-oneisT nnTKerTT TKer   0))(dim()ofdomaindim()ofrangedim( 0))(dim( Two finite-dimensional vector space V and W are isomorphic if and only if they are of the same dimension. Bijective Transformation 22
  • 23. .dimensionhavebothandthatAssume nWV onto.isT nWT  )dim()ofrangedim( nWV  )dim()dim(Thus     .ofbasisabe,,,let andV,ofbasisabe,,,Let 21 21 Wwww vvv n n   nnvcvcvc V  2211 asdrepresentebecaninvectorarbitraryanThen v nnwcwcwcT WVT   2211)( follows.as:L.T.adefinecanyouand v It can be shown that this L.T. is both 1-1 and onto. Thus V and W are isomorphic. 23
  • 24. Inverse linear Transformation ineveryfors.t.L.T.are:and:If 21 nnnnn RRRTRRT v ))((and))(( 2112 vvvv  TTTT invertiblebetosaidisandofinversethecalledisThen 112 TTT  Note: If the transformation T is invertible, then the inverse is unique and denoted by T–1 . 24
  • 25. Existence of an inverse transformation .equivalentareconditionfollowingThen the ,matrixstandardwithL.T.abe:Let ARRT nn   Note: If T is invertible with standard matrix A, then the standard matrix for T–1 is A–1 . (1) T is invertible. (2) T is an isomorphism. (3) A is invertible. 25
  • 26. Finding the inverse of a linear transformation bydefinedis：L.T.The 33 RRT  )42,33,32(),,( 321321321321 xxxxxxxxxxxxT  Sol: 142 133 132 formatrixstandardThe           A T 321 321 321 42 33 32 xxx xxx xxx               100142 010133 001132 3IA Show that T is invertible, and find its inverse. 26
  • 27.  1.. 326100 101010 011001                AIEJG 11 isformatrixstandardtheandinvertibleisTherefore  ATT             326 101 011 1 A                                  321 31 21 3 2 1 11 326326 101 011 )( xxx xx xx x x x AT vv )326,,(),,( s,other wordIn 3213121321 1 xxxxxxxxxxT  27
  • 28. .formatrixstandardtheFindaxis.-xtheontoinpointeach projectingbygivenis:nnsformatiolinear traThe 2 22 TR RRT  Sol: )0,(),( xyxT           00 01 )1,0()0,1()()( 21 TTeTeTA  Notes: (1) The standard matrix for the zero transformation from Rn into Rm is the mn zero matrix. (2) The standard matrix for the zero transformation from Rn into Rn is the nn identity matrix In Finding the matrix of a linear transformation 28
  • 29. Composition of T1:Rn→Rm with T2:Rm→Rp : n RTTT  vvv )),(()( 12 112 ofdomainofdomain, TTTTT    Composition of linear transformations then,andmatricesstandardwith L.T.be:and:Let 21 21 AA RRTRRT pmmn  L.T.ais)),(()(bydefined,:ncompositioThe(1) 12 vv TTTRRT pn  12productmatrixby thegivenisformatrixstandardThe)2( AAATA  29
  • 30. Pf: nscalar theanybecletandinvectorsbeandLet L.T.)ais((1) n R T vu )formatrixstandardtheis)(2( 12 TAA )()())(())(( ))()(())(()( 1212 11212 vuvu vuvuvu TTTTTT TTTTTT   )())(())(())(()( 121212 vvvvv cTTcTcTTcTTcT  vvvvv )()())(()( 12121212 AAAAATTTT   Note: 1221 TTTT   30
  • 31. The standard matrix of a composition s.t.intofromL.T.beandLet 33 21 RRTT ),0,2(),,(1 zxyxzyxT  ),z,(),,(2 yyxzyxT  ,'and nscompositiofor thematricesstandardtheFind 2112 TTTTTT   Sol: )formatrixstandard( 101 000 012 11 TA            )formatrixstandard( 010 100 011 22 TA             31
  • 32. 12formatrixstandardThe TTT  21'formatrixstandardThe TTT                             000 101 012 101 000 012 010 100 011 12 AAA                                   001 000 122 010 100 011 101 000 012 ' 21AAA 32
  • 33. Matrices for Linear Transformations )43,23,2(),,()1( 32321321321 xxxxxxxxxxxT   Three reasons for matrix representation of a linear transformation:                    3 2 1 430 231 112 )()2( x x x AT xx  It is simpler to write.  It is simpler to read.  It is more easily adapted for computer use.  Two representations of the linear transformation T:R3→R3 : 33
  • 34. Definition 1: A nonzero vector x is an eigenvector (or characteristic vector) of a square matrix A if there exists a scalar λ such that Ax = λx. Then λ is an eigenvalue (or characteristic value) of A. Note: The zero vector can not be an eigenvector even though A0 = λ0. But λ = 0 can be an eigenvalue. Example: Show x  2 1      isaneigenvector for A  2 4 3 6       Solution: Ax  2 4 3 6       2 1       0 0       But for   0, x  0 2 1       0 0       Thus,xisaneigenvectorof A,and  0 isaneigenvalue. Definitions 34
  • 35. An n×n matrix A multiplied by n×1 vector x results in another n×1 vector y=Ax. Thus A can be considered as a transformation matrix. In general, a matrix acts on a vector by changing both its magnitude and its direction. However, a matrix may act on certain vectors by changing only their magnitude, and leaving their direction unchanged (or possibly reversing it). These vectors are the eigenvectors of the matrix. A matrix acts on an eigenvector by multiplying its magnitude by a factor, which is positive if its direction is unchanged and negative if its direction is reversed. This factor is the eigenvalue associated with that eigenvector. Geometric interpretation of Eigenvalues and Eigenvectors 35
  • 36. Let x be an eigenvector of the matrix A. Then there must exist an eigenvalue λ such that Ax = λx or, equivalently, Ax - λx = 0 or (A – λI)x = 0 If we define a new matrix B = A – λI, then Bx = 0 If B has an inverse then x = B-10 = 0. But an eigenvector cannot be zero. Thus, it follows that x will be an eigenvector of A if and only if B does not have an inverse, or equivalently det(B)=0, or det(A – λI) = 0 This is called the characteristic equation of A. Its roots determine the eigenvalues of A. Eigenvalues 36
  • 37. Eigenvalues: examples Example 1: Find the eigenvalues of two eigenvalues: 1,  2 Note: The roots of the characteristic equation can be repeated. That is, λ1 = λ2 =…= λk. If that happens, the eigenvalue is said to be of multiplicity k.          51 122 A )2)(1(23 12)5)(2( 51 122 2           AI 37
  • 38. Eigenvectors Example 1 (cont.):                 00 41 41 123 )1(:1 AI 0, 1 4 ,404 2 1 1 2121               tt x x txtxxx x                 00 31 31 124 )2(:2 AI 0, 1 3 2 1 2              ss x x x To each distinct eigenvalue of a matrix A there will correspond at least one eigenvector which can be found by solving the appropriate set of homogenous equations. If λi is an eigenvalue then the corresponding eigenvector xi is the solution of (A – λiI)xi = 0 38
  • 39. Example 2 (cont.): Find the eigenvectors of Recall that λ = 2 is an eigenvector of multiplicity 3. Solve the homogeneous linear system represented by Let . The eigenvectors of  = 2 are of the form and t not both zero.                                  0 0 0 000 000 010 )2( 3 2 1 x x x AI x txsx  31 , , 1 0 0 0 0 1 0 3 2 1                                             ts t s x x x x            200 020 012 A 39
  • 40. Definition: The trace of a matrix A, designated by tr(A), is the sum of the elements on the main diagonal. Property 1: The sum of the eigenvalues of a matrix equals the trace of the matrix. Property 2: A matrix is singular if and only if it has a zero eigenvalue. Property 3: The eigenvalues of an upper (or lower) triangular matrix are the elements on the main diagonal. Property 4: If λ is an eigenvalue of A and A is invertible, then 1/λ is an eigenvalue of matrix A-1. Properties of Eigenvalues and Eigenvectors 40
  • 41. Property 5: If λ is an eigenvalue of A then kλ is an eigenvalue of kA where k is any arbitrary scalar. Property 6: If λ is an eigenvalue of A then λk is an eigenvalue of Ak for any positive integer k. Property 8: If λ is an eigenvalue of A then λ is an eigenvalue of AT. Property 9: The product of the eigenvalues (counting multiplicity) of a matrix equals the determinant of the matrix. 41
  • 42. Theorem: Eigenvectors corresponding to distinct (that is, different) eigenvalues are linearly independent. Theorem: If λ is an eigenvalue of multiplicity k of an n  n matrix A then the number of linearly independent eigenvectors of A associated with λ is given by m = n - r(A- λI). Furthermore, 1 ≤ m ≤ k. Example 2 (cont.): The eigenvectors of  = 2 are of the form s and t not both zero.  = 2 has two linearly independent eigenvectors , 1 0 0 0 0 1 0 3 2 1                                             ts t s x x x x Linearly independent eigenvectors 42
  • 43. Diagonalization  Diagonalizable matrix: A square matrix A is called diagonalizable if there exists an invertible matrix P such that P-1AP is a diagonal matrix. (P diagonalizes A)  Notes: (1) If there exists an invertible matrix P such that , then two square matrices A and B are called similar. (2) The eigenvalue problem is related closely to the diagonalization problem. APPB 1  43
  • 44. A diagonalizable matrix           200 013 031 A Sol: Characteristic equation: 0)2)(4( 200 013 031 I 2           A 2,2,4:sEigenvalue 321   Ex.5)p.403(See 0 1 1 :rEigenvecto4)1( 1            p 44
  • 45. 1 0 0 , 0 1 1 :igenvectorE2)2( 32                       pp                         200 020 004 100 011 011 ][ 1 321 APPpppP                                                   400 020 002 010 101 101 ][(2) 200 040 002 100 011 011 ][(1) 1 132 1 312 APPpppP APPpppP  Notes: 45
  • 46. Condition for Diagonalization An nn matrix A is diagonalizable if and only if it has n linearly independent eigenvectors. Pf: ablediagonalizis)( A ),,,(and][Let diagonaliss.t.invertibleanexiststhere 2121 1 nn diagDpppP APPDP      ][ 00 00 00 ][ 2211 2 1 21 nn n n ppp pppPD                         46
  • 47. ][][ 2121 nn ApApAppppAAP   )ofrseigenvectoareoftorcolumn vecthe..( ,,2,1, APpei nipAp PDAP i iii      t.independenlinearlyare,,,invertibleis 21 npppP   rs.eigenvectotindependenlinearlyhas nA n npppnA    ,,seigenvalueingcorrespondhwit ,,rseigenvectotindependenlinearlyhas)( 21 21 nipAp iii ,,2,1,i.e.   ][Let 21 n pppP  47
  • 49. A matrix that is not diagonalizable        10 21 able.diagonaliznotismatrixfollowingthat theShow A Sol: Characteristic equation: 0)1( 10 21 I 2         A 1:Eigenvalue 1                      0 1 :rEigenvecto 00 10 ~ 00 20 I 1pAIA A does not have two (n=2) linearly independent eigenvectors, so A is not diagonalizable. 49
  • 50. Steps for diagonalizing an nn square matrix: n ,,, 21  Step 2: Let ][ 21 npppP  Step 1: Find n linearly independent eigenvectors for A with corresponding eigenvalues nppp ,,, 21  Step 3: nipApDAPP iii n ,,2,1,where, 00 00 00 2 1 1                        Note: The order of the eigenvalues used to form P will determine the order in which the eigenvalues appear on the main diagonal of D. 50
  • 51. Diagonalizing a matrix diagonal.issuch thatmatrixaFind 113 131 111 1 APPP A               Sol: Characteristic equation: 0)3)(2)(2( 113 131 111 I           A 3,2,2:sEigenvalue 321   51
  • 52. 21                    000 010 101 ~ 313 111 111 I1 A                                             1 0 1 :rEigenvecto 1 0 1 0 1 3 2 1 pt t t x x x 22                       000 10 01 ~ 113 151 113 I 4 1 4 1 2 A                                             4 1 1 :rEigenvecto 4 1 1 24 1 4 1 4 1 3 2 1 pt t t t x x x 52
  • 54.  Notes: k is a positive integer                           k n k k k n d d d D d d d D         00 00 00 00 00 00 )1( 2 1 2 1 1 1 1 1111 111 1 1 )()()( )())(( )( )2(               PPDA PAP APAAP APPPPPAPPAP APPAPPAPP APPD APPD kk k kk 54
  • 55. Sufficient conditions for Diagonalization If an nn matrix A has n distinct eigenvalues, then the corresponding eigenvectors are linearly independent and A is diagonalizable. 55
  • 56.  Determining whether a matrix is diagonalizable              300 100 121 A Sol: Because A is a triangular matrix, its eigenvalues are the main diagonal entries. 3,0,1 321   These three values are distinct, so A is diagonalizable. (Thm.7.6) 56
  • 57. Finding a diagonalizing matrix for a linear transformation diagonal.istorelative formatrixthesuch thatforbasisaFind )33()( bygivennnsformatiolinear trathebeLet 3 321321321321 33 B TRB xxx,xx, xxxx,x,xxT RT:R   Sol:                113 131 111 )()()( bygivenisformatrixstandardThe 321 eTeTeTA T From Ex. 5, there are three distinct eigenvalues so A is diagonalizable. (Thm. 7.6) 3,2,2 321   57
  • 58.                     300 020 002 ][][][ ][][][ ])([)]([])([ 332211 321 321 BBB BBB BBB ppp ApApAp pTpTpTD  The matrix for T relative to this basis is )}1,1,1(),4,1,1(),1,0,1{(},,{ 321  pppB Thus, the three linearly independent eigenvectors found in Ex. 5 can be used to form the basis B. That is )1,1,1(),4,1,1(),1,0,1( 321  ppp 58