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# Class Note for MATH 290 with Professor Mandal at KU

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Chapter 6 Linear Transformation 61 Intro to Linear Transformation Homework Textbook7 61 EX 37 57 97 237 257 337 377 39 7537 557 577 61217b7 63 page 371 In this section we discuss linear transformations 189 190 CHAPTER 6 LINEAR TRANSFORMATION Recall7 from calculus courses7 a funtion f X a Y from a set X to a set Y associates to each x E X a unique element f E Y Following is some commonly used terminologies 1 X is called the domain of f 2 Y is called the codomain of f 3 If f y then we say y is the image of x The preimage of y is pretmagey x E X f 4 The range of f is the set of images of elements in X In this section we deal with functions from a vector sapce V to another vector space W that respect the vector space structures Such a function will be called a linear transformation7 de ned as follows De nition 611 Let V and W be two vector spaces A function T V a W is called a linear transformation of V into W if following two prper ties are true for all uV E V and scalars c 1 Tuv TuTV We say that T preserves additim39ty 2 Teu cTu We say that T preserves scalar multiplica tz39orz Reading assignment Read Textbook7 Examples 1 37 p 362 Trivial Examples Following are two easy exampes 61 INTRO TO LINEAR TRANSFORMATION 191 1 Let V W be two vector spaces De ne T V a W as TV 0 for all V E V Then T is a linear transforrnation7 to be called the zero trans formation 2 Let V be a vector space De ne T V a V as TV V for all V E V Then T is a linear transforrnation7 to be called the identity transformation of V 611 Properties of linear transformations Theorem 612 Let V and W be two vector spaces Suppose T V a W is a linear transformation Then 2 TV TV for all V E V 3 Tu 7 V Tu 7 TV for all uV E V V01V102V2CnVn then TV T61V1 CgVZ CnVn 61TV102TV2 CnT Vn Proof By property 2 of the de nition 6117 we have T0 T00 0T0 0 192 CHAPTER 6 LINEAR TRANSFORMATION So7 1 is proved Similarly7 So7 2 is proved Then7 by property 1 of the de nition 6117 we have Tu 7 V Tu 71V Tu T71V Tu 7 TV The last equality follows from So7 3 is proved To prove 47 we use induction7 on n For n 1 we have Tclvl 01TV1 by property 2 of the de nition 611 For n 2 by the two properties of de nition 6117 we have Tclvl Cgvz Tclvl TCgvz 61TV1 62TV2 So7 4 is prove for n 2 Now7 we assume that the formula 4 is valid for n 7 1 vectors and prove it for n We have T CiVi Csz Cnvn T 01V1 Csz Cn71Vn71T Cnvn 01TV1 CgT V2 on1T Vn1 chVn So7 the proof is complete I 612 Linear transformations given by matrices Theorem 613 Suppose A is a matrix of size m gtlt 71 Given a vector 01 1 1 V v2 6 R define TV AV A W Un n Then T is a linear transformation from R to R 61 INTRO TO LINEAR TRANSFORMATION 193 Proof From properties of matrix multiplication for uV E R and scalar c we have Tu V Au V Au AV Tu TV and Tcu Acu cAu cTu The proof is complete I Remark Most or all of our examples of linear transformations come from matrices as in this theorem Reading assignment Read Textbook Examples 2 10 p 365 613 Projections along a vector in R Projections in R is a good class of examples of linear transformations We de ne projection along a vector Recall the de nition 526 of orthogonal projection in the context of Euclidean spaces R De nition 614 Suppose V E R is a vector Then for u E R define projvu v V 1 Then projv R a R is a linear transformation Proof This is because for another vector W E R and a scalar c it is easy to check projvuw projvupr0jvw and projvcu c projvu 2 The point of such projections is that any vector u E R can be written uniquely as a sum of a vector along V and another one perpendicular to V u projvu u projvu 194 CHAPTER 6 LINEAR TRANSFORMATION It is easy to Check that u 7 projvu 1 projvu Exercise 615 Ex 4 p 371 Let T111213 2111 112 212 7 301111 7 Ug 1 Compute T745 1 Solution T745 1 2445 254344 4441 4322 45 2 Compute the preimage of W 41 71 Solution Suppose 010203 is in the preimage of 4171 Then 2111 02202 7 311101 7 03 4 1 71 So7 2111 112 4 212 73113 1 01 7113 71 The augmented matrix of this system is 2 1 0 4 1 0 0 5714 0 2 73 1 its Gauss7J0rdan form 0 1 0 285714 1 0 71 71 0 0 1 15714 So 7 Pretmage4171 571428571415714 61 INTRO TO LINEAR TRANSFORMATION 195 Exercise 616 Ex 10 p 371 Determine whether the function T R2 a R2 Ty x2y is linear Solution We have T w 2710 T96 211 w 96 0271 w 7e 96211 227w T w T27w So T does not preserve additivity So T is not linear Alternately you could check that T does not preserve scalar multi plication Alternately you could check this failures numerically For example T1 1 20 T31 91 7 T1 1 T20 Exercise 617 Ex 24 p 371 Let T R3 a R3 be a linear trans formation such that T10 0 24 71 T0 10 13 72 T001 0722 Compute T72 4 71 Solution We have 724 71 7210 0 40 10 7 001 So T724 i1 72T1004T0107T00 1 24 7113720 722 3571 Remark A linear transformation T V a V can be de ned simply by assigning values TVi for any basis V1V2 Vn of V In this case of the our problem values were assigned for the standard basis e1e2e3 of R3 196 CHAPTER 6 LINEAR TRANSFORMATION Exercise 618 Ex 38 p 372 Let 712134 002710 Let TR5aR2 be the linear transformation TX AX 1 Compute T10 713 0 Solution 0 71 2 1 3 4 7 T107130 71 0 0 2 71 0 i5 2 Compute preimage7 under T of 718 Solution The preimage consists of the solutions of the linear system 1 2 712134 71 3 002710 8 4 5 The augmented matrix of this system is 71 2 1 3 4 71 0 0 2 71 0 8 39 The Gauss Jordan form is 1720735745 0 0175 0439 61 INTRO TO LINEAR TRANSFORMATION 197 We use parameters 2 tx4 85 u and the solotions are given by 1 52t3554ux2 t3 455z4 85 u So the preimage T 171852t3554u t 455 s u tsu R Exercise 619 Ex 54 edited p 372 Let T R2 a R2 be the linear transformation such that T1 1 02 and T1 71 20 1 Compute T14 Solution We have to write 14 a11b171 Solving 14 2511715171 So 7 T14 25T11715T171 250271520 735 2 Compute T721 Solution We have to write 721 a11b171 Solving 721 7511715171 So 7 T721 75T11715T171 750271520 7371 Exercise 6110 Ex 61 edited p 372 Let T R3 a R3 the projection Tu projvu where V 111 198 CHAPTER 6 LINEAR TRANSFORMATION 1 Find Tzyz Solution See de nition 614 V 95252 PTOJvWalaz WV 17171397y72 1117171 H2 3 i yz yz yz 3 3 3 39 111 111 2 Compute T5 05 Solution We have zyz xyz zyz 101010 57075 39 3 Compute the matrix of T Solution The matrix is given by b H l I cab t MH nah A Ah t Ah t Ah t Ah t Ah t Ah t because new new new 39 N 39 mh t cap A och t Ah t Ah t Ah t 62 KERNEL AND RANGE OF LINEAR TRANSFORMATION199 62 Kernel and Range of linear Transfor mation We will essentially skip this section Before we do that let us give a few de nitions De nition 621 Let VW be two vector spaces and T V a W a linear transformation 1 Then the kernel of T denoted by kerT is the set of V E V such that TV 0 Notationally herT V E V TV 0 It is easy to see the herT is a subspace ofV 2 Recall range of T denoted by rangeT is given by rangeT V E W W TV for some V E V It is easy to see the rangeT is a subspace oflV 3 We say the T is isomorphism if T is one to one and onto It fol lows that T is an isomorphism if herT 0 and rangeT W 200 CHAPTER 6 LINEAR TRANSFORMATION 63 Matrices for Linear Transformations Homework Teatbooh 63 E55 5 7 11 Z3 Z7 Z9 21 23 25 29 3 33 35ab 37ab 39 43 45 47 p 397 Optional Homework Temtbooh 63 E55 57 591 398 We will not grade them In this section to each linear transformation we associate a matrim Linear transformations and matrices have very deep relationships In fact study of linear transformations can be reduced to the study of matrices and conversely First we will study this relationship for linear transformations T R a R and later study the same for linear transformations T V a W of general vector spaces In this section we will denote the vectors in R as column matrices Recall written as columns the standard basis of R is given by l l l Ol l Ol lollol lll Theorem 631 Let T R a R be a linear transformation Write an 012 aln a2n Te1 Te2 Ten aml amZ amn 63 MATRICES FOR LINEAR TRANSFORMATIONS 201 These columns Te1 Te2 Ten form a mgtltn matrix A as follows7 a11 a12 aln A a21 a22 0 aml amZ amn This matrix A has the property that TV AV for all V E R This matrix A is called the standard matrix of T Proof We can write V E R as 1 1 112 V Uiei0262 nen W71 We have7 an a12 aln 1 1 a11 a12 aln a21 a22 am 112 a21 a22 0 AV v1 112 vn L aml amZ amn J L n J L aml amZ amn 111Te112Te2 vnTen T 1161 Ugez linen TV The proof is complete I Reading assignment Read Textbook7 Examples 172 page 389 390 In our context of linear transformations7 we recall the following de nition of composition of functions De nition 632 Let T1R aRm T2RmaR 202 CHAPTER 6 LINEAR TRANSFORMATION be two linear transformations De ne the composition T R a R of T1 with T2 as TV T2T1V for V E R The composition T is denoted by T T20T1 Diagramatically7 Rn T1 Rm T2 TT20T1 RP Theorem 633 Suppose T1R Rm T2Rm R are two linear transformations 1 Then7 the composition T T20T1 R a RF is a linear transfor mation 2 Suppose A1 is the standard matrix of T1 and A2 is the standard matrix of T2 Then7 the standard matrix of the composition T T20T1 is the product A A2A1 Proof For uV E R and scalars c we have TUV T2T1UV T2T1uT1V T2T1uT2T1V TuTV and Tcu T2T1cu T2CT1U CT2T1U cTu So7 T preserves addition and scalar multiplication Therefore T is a linear transformation and 1 is proved To prove 27 we have Tu T2T1u T2A1u A2A1u A2A1u 63 MATRICES FOR LINEAR TRANSFORMATIONS 203 Therefore Te1 is the rst column of A2A1 Te2 is the second column of A2A1 and so on Therefore7 the standard matrix of T is A2A1 The proof is complete I Reading assignment Read Textbook7 Examples 3 page 392 De nition 634 Let T1R R T2Rn R be two linear transformations such that for every V E R we have T2T1V V and T1T2V V then we say that T2 is the inverse of T1 and we say that T1 is invert ible Such an inverse T2 of T1 is unique and is denoted by Tfl Remark Let EndR denote the set of all linear transformations T R a R Then EndlR has a binary operation by composition The identity operation I R a R acts as the identitiy under this composition operation The de nition of inverse of T1 above just cor responds to the inverse under this composition operation Theorem 635 Let T R a R be a linear transformations and let A be the standard matrix of T Then7 the following are equivalent7 1 T is invertible 2 T is an isomorphism 3 A is invertible And7 if T is invertible7 then the standard matrix of T 1 is A l 204 CHAPTER 6 LINEAR TRANSFORMATION Proof First recall de nition 62 that T is isomorphism ifT is 1 t0 1 and onto Proof of 1 i 2 Suppose T is invertible and T2 be the inverse Suppose Tu TV Then7 u T2Tu T2TV V So7 T is 1 to 1 Also7 given u E R we have uTT2u So T onto R So7 T is an isomorphism Proof of 2 i 3 So7 assume T is an isomorphism Then7 AX0gtTXT0gtX0 So7 Ax 0 has an unique solution Therefore A is invertible and 3 follows from Proof of 3 i 1 Suppose A is invertible Let T2X A lx then T2 is a linear transformation and it is easily checked that T2 is the inverse of T So7 1 follows from The proof is complete I Reading assignment Read Textbook7 Examples 4 page 393 631 Nonstandard bases and general vector spaces The above discussion about standard matrices of linear transforma tions T had to be restricted to linear transformations T R a K This wa sbecause R has a standard basis e1e2 em that we could use Suppose T V a W is a linear transformation between two abstract vector spaces VW Since V and W has no standard bases7 we cannot associate a matrix to T But7 if we x a basis B of V and B of W we can associate a matrix to T We do it as follows 63 MATRICES FOR LINEAR TRANSFORMATIONS 205 Theorem 636 Suppose T V a W is a linear transformation be tween two vector spaces V W Let B V1V2 Vn be basis of V and Bw1wzwm be basis of w We can write Writing similar equations for TV2 TVn we get all 112 am TV1 TV2 TVn W1 W2 Wm 121 122 azn am am am Them for V lvl 2V2 nVn We have all an am 11 TV w1 W2 Wm 021 022 aZn 952 am amz am n Write A calj Then7 T is determined by A with respect to bases BB Exercise 637 Ex 6 p 397 Let Txyz 5x 7 3y z 22 4y5z By 206 CHAPTER 6 LINEAR TRANSFORMATION What is the standard matrix of T Solution We have T R3 a R3 We write vectors x E R3 as columns X y instead of 71172 Recall the standard basis 1 0 0 e1 0 e2 1 e3 0 of R3 0 0 1 We have 5 73 1 Tel 0 a Tez 4 7 Te3 5 3 0 So7 the standard matrix of T is A0 Exercise 638 Ex 12 p 397 Let Write down the standard matrix of T and use it to nd T0171 Solution In this case7 T R3 a R2 With standard basis e1e2e3 as in exercise 6377 we have 63 MATRICES FOR LINEAR TRANSFORMATIONS 207 So the standard matrix of T is 2 1 0 A 0 3 71 Therefore 0 2 1 0 0 1 T0171A 1 1 71 0 3 i1 4 We will write our answer in as a row T0 1 71 14 Exercise 639 Ex 18 p 397 Let T be the re ection in the line y z in R2 So Ty 1 Write down the standard matrix of T Solution In this case T R2 a R2 With standard basis e1 1 0Te2 01T we have 1 T e lt 1 1 0 1 So the standard matrix of T is 0 1 A 1 0 2 Use the standard matrix to compute T3 4 0 Tel 1 Solution Of course we know T3 4 43 They want use to use the standard matrix to get the same answer We have T34Aif 0r T3443 208 CHAPTER 6 LINEAR TRANSFORMATION Lemma 6310 Suppose T R2 a R2 is the counterclockwise rotation by a cced angle 6 Then cos 6 7 sin 6 z TQM sin6 cos 6 y Proof We can write zrcos yrsin where r z2y2 tan y By de nition my rcosw 6gtrsmlt 6 Using trig formulas rcos 6 rcosrocos67rsinrosin6ccos67ysin6 and rsin 6 rsin cos6 rcos sin6 ycos6 xsin6 So7 Two 2213 323 3 The proof is complete I Exercise 6311 Ex 22 p 397 Let T be the counterclockwise ro tation in R2 by angle 1200 1 Write down the standard matrix of T Solution We use lemma 63107 with 6 1200 So7 the standard matrix of T is cos6 isin6 cos 120 isin 120 75 7 A sin6 cos6 sin 120 cos 120 g 7 U E 63 MATRICES FOR LINEAR TRANSFORMATIONS 209 2 Compute T2 2 Solution We have 1 7 2 75 7V 2 7 T2 2A2 2 52 71x3 We write our answer as rows T2 2 717 xg 71 Exercise 6312 Ex 32 p 397 Let T be the projection on to the vector W 715 in R2 Tu projwu 1 Find the standard matrix Solution See de nition 614 W Ly 175 Ly x75y TWay PTOJwWay WW W0 5 26 1 5 7 x75y 75z25y 26 26 39 So7 with e1 10Te2 01T we have writethink every thing as columns 2 Compute T23 Solution We have 2 i 72756 2 7 75 T2 3Al3ll We write our answer in row form T23 75 210 CHAPTER 6 LINEAR TRANSFORMATION Exercise 6313 Ex 36 p 397 Let Txyz 3x 7 2y 2 2x 7 3yy 7 42 1 Write down the standard matrix of T Solution with e1 100Te2 010Te3 001T we have writethink everything as columns 3 72 1 Tel 2 a Tez 3 7 Tez 0 0 1 74 3 i2 11 A 2 73 0 L 0 1 74 J 2 Compute T2 71 71 Solution We have 2 3 72 1 2 T27171 A 71 2 73 0 71 71 0 1 74 71 Exercise 6314 Ex 40 p 397 Let T1R2 7 R2 T1zy 7 967 2y2z3y and T2 R2 R2 T27y Compute the standard matrices of T T20T1 and T T1T2 Solution We solve it in three steps 63 MATRICES FOR LINEAR TRANSFORMATIONS 211 Step 1 First7 compute the standard matrix of T1 With e1 1 0T e2 01T we have writethink everything as columns a T1ez T So7 the standard matrix of T1 is 172 2 3 Step 2 Now7 compute the standard matrix of T2 With e1 1 0T e2 01T we have 1 a T2ez 0 So7 the standard matrix of T2 is 31 Step 3 By theorem 6337 the standard matrix of T T2T1 is 0 1 1 72 2 3 0 0 2 3 0 0 1 T161 2 A1 0 T261 0 A2 S0 Tzy2z3y0 Similarly7 the standard matrix of T T1T2 is 172 01 AA 12 2 3H00 Exercise 6315 Ex 46 p 397 Determine whether 01 02 SO WM 9211 9571195 21196 7 2y 212 CHAPTER 6 LINEAR TRANSFORMATION is invertible or not Solution Because of theorem 635 we will Check whether the stan dard matrix of T is invertible or not With e1 1 0Te2 01T we have mu Mil So the standard matrix of T is 1 2 A 1 2 Note detA 74 31 0 So T is invertible and hence T is invertible Exercise 6316 Ex 58 p 397 Determine whether 9571195 7 15096 11 Use B V1 12V2 11 as basis of the domain R2 and B W1 111W2 1 1 0W3 011 as basis ofcodomain R3 Compute matrix of T with respect to BB Solution We use theorem 636 We have Tu1 T12 7103 Tu2 T11 002 We solve the equation 7103 awl bwz CW3 a111 b1 1 0 C011 and we have 2 717073 W1 W2 W3 7393 1 64 TRANSITION MATRICES AND SIMILARITY 213 Similarly we solve 002 awl bwz CW3 a11 1 b1 1 0 C011 and we have 2 0022111721100011w1 w2 W3 0 So the matrix of T with respect to the bases B B is AJ jl 11 01 64 Transition Matrices and Similarity We will skip this section I will just explain the section heading You know what are Transition matrices of a linear transformation T V a W They are a matrices described in theorem 636 De nition 641 Suppose A B are two square matrices of size n gtlt n We say A B are similar if A P lBP for some invertible matrix P 214 CHAPTER 6 LINEAR TRANSFORMATION 65 Applications of Linear Trans Homework Textbook7 65 Ex 11 a7 13 a7 257 277 297 357 377 397 437 497 517 537 557 637 65 page 414 415 In this section we discuss geometric interpretations of linear trans formations represented by 2 gtlt 2 elementary matrices Proposition 651 Let A be a 2 gtlt 2 matrix and TzyA y We will write the right hand side as a row7 which is an abuse ofnatation 1 If 71 0 z A 0 1 then TzyAyxyT represents the re ection in yiaxis See Textbook7 Example 1 a7 p407 for the diagram 2 If 1 0 z A0 1 then TzyAyxyT represents the re ection in ziaxis See Textbook7 Example 1 b7 p407 for the diagram 65 APPLICATIONS OF LINEAR TRANS 3 215 If x A01 then TzyA 10 y 1196T represents the re ection in line y z See Textbook Example 1 c7 p407 for the diagram Al If k gt 1 then T represents expansion in horizontal direction and If k 0 th T A 01 671 Ml z kzyT y 0 lt k lt 1 then T represents contraction in horizontal direction See Textbook Example 2 Fig 612 7 p409 for diagrams lf AAJ 2 then TzyAz zkyT 9 If k gt 1 then T represents expansion in vertical direction and 0 lt k lt 1 then T represents contraction in vertical direction See Textbook Example 2 Fig 6137 p409 for diagrams Al Then T represents horizontal shear Assume k gt 0 The upper If 1 k z th T A 0 1 J en zy 96 kiwiV y half plane are sheared to right and lower half plane are sheared to left The points on the ziaxis reamain xed See Textbook Example 37 g 6147 p409 for diagrams 216 CHAPTER 6 LINEAR TRANSFORMATION 1 0 z A 2th T A k 1 671 Ml J x kx yT 9 Then T represents vertical shear Assume k gt 0 The right half plane are sheared to upward and left half plane are sheared to downward The points on the yiaxis reamain xed See Textbook7 Example 37 g 6157 p409 for diagrams 651 Computer Graphics Linear transformations are used in computer graphics to move gures on the computer screens I am sure all kinds of linear and nonlinear transformations are used Here7 we will only deal with rotations by an angle 0 around 1 ziaxis7 2 yiaxis and 3 ziaxis as follows Proposition 652 Suppose 6 is an angle Suppose we want to ro tate the point Lyvz counterclockwise about ziaxis through an an gle 0 Let us denote this transformation by T and write Txyz x y z T Then using lemma 63107 we have x cos0 isin0 0 z cos0 iysint Txyz y sint9 cost 0 y xsin6ycos0 2 0 0 1 z z Similarly7 we can write down the linear transformations corresponding to rotation around xiaxis and yiaxis We write down the transition matrices for these three matrices as follows 1 The standard matrix for this transformation of counterclockwise 65 APPLICATIONS OF LINEAR TRANS 217 rotation by an angle 6 about ziaxis is 1 0 0 0 cos6 isin6 0 sin 6 cos 6 2 The standard matrix for this transformation of counterclockwise rotation by an angle 6 about yiaxis is cos6 0 sin6 0 1 0 l isin6 0 cos6 3 The standard matrix for this transformation of counterclockwise rotation by an angle 6 about ziaxis is cos 6 7 sin 6 0 sin6 cos6 0 0 0 1 Reading assignment Read Textbook7 Examples 4 and 5 page 410 412 Exercise 653 Ex 26 p 414 Let Ty z3y This is a vertical expansion Sketch the image of the unit square with vertices 00 10 11 01 Solution We have T0000 T1010 T1113 T0103 218 CHAPTER 6 LINEAR TRANSFORMATION Diagram Here the solid arrows represent the original rectangle and the brocken arrows represent the image Exercise 654 Ex 30 p 414 Let T be the re ection in the line y x Sketch the image ofthe rectangle with vertices 0 0 0 2 1 2 1 0 Solution Recall see Proposition 651 that Ty We have T00 00T02 20T12 21T10 01 Diagram Ho an 77777 A l l 77gt Here the solid arrows represent the original rectangle and the brocken arrows represent the image 65 APPLICATIONS OF LINEAR TRANS 219 Exercise 655 Ex 38 p 414 Suppose T is the expansion and contraction represented by Ty 2s Sketch the image of the rectangle with vertices 00 02 12 10 Solution Recall see Proposition 651 that my yzWe have T00 00T02 01T12 21T10 20 69 Diagram a gt l l l as l l l l a Here the solid arrows represent the original rectangle and thebrocken arrows represent the image Exercise 656 Ex 44 p 414 Give the geometric description of the linear transformation de ned by the elementary matrix AM Solution By proposition 651 6 this is a horizontal shear Here 9571195 3911 220 CHAPTER 6 LINEAR TRANSFORMATION Exercise 657 Ex 50 and 54 p 415 Find the matrix of the trans formation T that will produce a 600 rotation about the z7axis Then compute the image T111 Solution By proposition 652 1 the matrix is given by A 1 0 0 1 0 0 1 0 0 0 cos6 7sin6 0 cos 600 7sin 600 0 g 0 sin 6 cos 6 0 sin 600 cos 600 0 g So 1 1 0 0 1 1 T11 1 A 1 0 g 1 1 o a 1 Exercise 658 Ex 64 p 415 Determine the matrixthat will pro duce a 450 rotation about the y7axis followed by 900 rotation about the 27axis 000 to 111 Solution We will do it in three or four steps Then also compute the image of the line segment from Step 1 Let T1 be the rotation by 450 about the y7axis By proposition 652 2 the matrix of T1 is given by A 7 cos6 0 sin6 cos 450 0 sin 450 f 0 1 0 0 1 0 0 0 0 1 L7s1n6 0 cos6J L7s1n45 0 cos45 J 7 Step 2 Let T2 be the rotation by 900 rotation about the 27axis proposition 652 3 the matrix of T2 is given by B cos 6 7 sin 6 0 cos 900 7 sin 900 0 0 7 sin 6 cos 6 0 sin 900 cos 900 0 1 0 0 1 0 0 1 0 0 1 0 By 65 APPLICATIONS OF LINEAR TRANS 221 Step 3 So7 thematrix ofthe composite transformation T T2T1 is matrix 1 1 0 71 0 W 0 E 0 71 0 7 7 1 1 BA7 1 0 0 0 7 W 0 E 1 1 1 1 0 0 E E E 0 E The Last Part So7 1 0 71 0 1 71 7 7 1 1 7 T1117BA 1 7 E 0 E 7 1 1 1 7E E 1 0 222 CHAPTER 6 LINEAR TRANSFORMATION

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