Chapter 6 Linear Transformations 6.1 Introduction to Linear Transformations 6.2 The Kernel and Range of a Linear Transformation 6.3 Matrices for Linear Transformations 6.4 Transition Matrices and Similarity Elementary Linear Algebra R. Larsen et al. (6 Edition) 投影片設計編製者 淡江大學 電機系 翁慶昌 教授 6.1 Introduction to Linear Transformations Function T that maps a vector space V into a vector space W: T : V mapping W , V ,W : vectorspace V: the domain of T W: the codomain of T Elementary Linear Algebra: Section 6.1, pp.361-362 1/78 Image of v under T: If v is in V and w is in W such that T ( v) w Then w is called the image of v under T . the range of T: The set of all images of vectors in V. the preimage of w: The set of all v in V such that T(v)=w. Elementary Linear Algebra: Section 6.1, p.361 2/78 Ex 1: (A function from R2 into R2 ) T : R 2 R 2 v (v1 , v2 ) R 2 T (v1 , v2 ) (v1 v2 , v1 2v2 ) (a) Find the image of v=(-1,2). (b) Find the preimage of w=(-1,11) Sol: (a ) v (1, 2) T ( v ) T (1, 2) (1 2, 1 2(2)) (3, 3) (b) T ( v) w (1, 11) T (v1 , v2 ) (v1 v2 , v1 2v2 ) (1, 11) v1 v2 1 v1 2v2 11 v1 3, v2 4 Thus {(3, 4)} is the preimage of w=(-1, 11). Elementary Linear Algebra: Section 6.1, p.362 3/78 Linear Transformation (L.T.): V ,W: vect orspace T : V W: V toW linear transformation (1) T (u v) T (u) T ( v), u, v V (2) T (cu) cT (u), c R Elementary Linear Algebra: Section 6.1, p.362 4/78 Notes: (1) A linear transformation is said to be operation preserving. T (u v) T (u) T ( v) Addition in V Addition in W T (cu) cT (u) Scalar multiplication in V Scalar multiplication in W (2) A linear transformation T : V V from a vector space into itself is called a linear operator. Elementary Linear Algebra: Section 6.1, p.363 5/78 Ex 2: (Verifying a linear transformation T from R2 into R2) T (v1 , v2 ) (v1 v2 , v1 2v2 ) Pf: u (u1, u2 ), v (v1, v2 ) : vectorin R2 , c : any real number (1)Vectoraddition : u v (u1 , u2 ) (v1 , v2 ) (u1 v1 , u2 v2 ) T (u v ) T (u1 v1 , u 2 v2 ) ((u1 v1 ) (u2 v2 ), (u1 v1 ) 2(u2 v2 )) ((u1 u2 ) (v1 v2 ), (u1 2u2 ) (v1 2v2 )) (u1 u2 , u1 2u2 ) (v1 v2 , v1 2v2 ) T (u) T ( v ) Elementary Linear Algebra: Section 6.1, p.363 6/78 (2) Scalar multiplication cu c(u1 , u2 ) (cu1 , cu 2 ) T (cu) T (cu1 , cu 2 ) (cu1 cu 2 , cu1 2cu 2 ) c(u1 u 2 , u1 2u 2 ) cT (u) Therefore, T is a linear transformation. Elementary Linear Algebra: Section 6.1, p.363 7/78 Ex 3: (Functions that are not linear transformations) (a) f ( x) sin x sin(x1 x2 ) sin(x1 ) sin(x2 ) f ( x) sin x is not sin(2 3 ) sin(2 ) sin(3 ) (b) f ( x) x 2 ( x1 x2 )2 x12 x22 (1 2) 1 2 2 2 2 linear tra nsformatio n f ( x) x 2 is not linear tra nsformatio n (c) f ( x) x 1 f ( x1 x2 ) x1 x2 1 f ( x1 ) f ( x2 ) ( x1 1) ( x2 1) x1 x2 2 f ( x1 x2 ) f ( x1 ) f ( x2 ) f ( x) x 1 is not Elementary Linear Algebra: Section 6.1, p.363 linear tra nsformatio n 8/78 Notes: Two uses of the term “linear”. (1) f ( x) x 1 is called a linear function because its graph is a line. (2) f ( x) x 1 is not a linear transformation from a vector space R into R because it preserves neither vector addition nor scalar multiplication. Elementary Linear Algebra: Section 6.1, p.364 9/78 Zero transformation: T :V W Identity transformation: T :V V T ( v) 0, v V T ( v) v, v V Thm 6.1: (Properties of linear transformations) T : V W , u, v V (1)T (0) 0 (2)T ( v) T ( v) (3)T (u v) T (u) T ( v) (4) If v c1v1 c2v2 cn vn T henT ( v) T (c1v1 c2v2 cn vn ) c1T (v1 ) c2T (v2 ) cnT (vn ) Elementary Linear Algebra: Section 6.1, p.365 10/78 Ex 4: (Linear transformations and bases) Let T : R3 R3 be a linear transformation such that T (1,0,0) (2,1,4) T (0,1,0) (1,5,2) T (0,0,1) (0,3,1) Find T(2, 3, -2). Sol: (2,3,2) 2(1,0,0) 3(0,1,0) 2(0,0,1) T (2,3,2) 2T (1,0,0) 3T (0,1,0) 2T (0,0,1) 2(2,1,4) 3(1,5,2) 2T (0,3,1) (7,7,0) Elementary Linear Algebra: Section 6.1, p.365 (T is a L.T.) 11/78 Ex 5: (A linear transformation defined by a matrix) 0 3 v1 2 3 1 The function T : R R is defined as T ( v) Av 2 v 2 1 2 (a) Find T ( v) , where v (2,1) (b) Show that T is a linear transformation form R 2 into R3 Sol: (a) v (2,1) R 2 vector R 3 vector 0 3 6 2 T ( v) Av 2 1 3 1 1 2 0 T (2,1) (6,3,0) (b) T (u v) A(u v) Au Av T (u) T ( v) T (cu) A(cu) c( Au) cT (u) Elementary Linear Algebra: Section 6.1, p.366 (vector addition) (scalar multiplication) 12/78 Thm 6.2: (The linear transformation given by a matrix) Let A be an mn matrix. The function T defined by T ( v) Av is a linear transformation from Rn into Rm. Note: R n vector a11 a21 Av am1 R m vector a12 a1n v1 a11v1 a12 v2 a1n vn a22 a2 n v2 a21v1 a22 v2 a2 n vn am 2 amn vn am1v1 am 2 v2 amn vn T ( v) Av T : Rn R m Elementary Linear Algebra: Section 6.1, p.367 13/78 Ex 7: (Rotation in the plane) Show that the L.T. T : R 2 R 2 given by the matrix cos A sin sin cos has the property that it rotates every vector in R2 counterclockwise about the origin through the angle . Sol: v ( x, y) (r cos , r sin ) (polar coordinates) r: the length of v :the angle from the positive x-axis counterclockwise to the vector v Elementary Linear Algebra: Section 6.1, p.368 14/78 cos sin x cos T ( v) Av sin cos y sin r cos cos r sin sin r sin cos r cos sin r cos( ) r sin( ) sin r cos cos r sin r:the length of T(v) +:the angle from the positive x-axis counterclockwise to the vector T(v) Thus, T(v) is the vector that results from rotating the vector v counterclockwise through the angle . Elementary Linear Algebra: Section 6.1, p.368 15/78 Ex 8: (A projection in R3) The linear transformation T : R 3 R 3 is given by 1 0 0 A 0 1 0 0 0 0 is called a projection in R3. Elementary Linear Algebra: Section 6.1, p.369 16/78 Ex 9: (A linear transformation from Mmn into Mn m ) T ( A) AT (T : M mn M nm ) Show that T is a linear transformation. Sol: A, B M mn T ( A B) ( A B)T AT BT T ( A) T ( B) T (cA) (cA)T cAT cT ( A) Therefore, T is a linear transformation from Mmn into Mn m. Elementary Linear Algebra: Section 6.1, p.369 17/78 Keywords in Section 6.1: function: 函數 domain: 論域 codomain: 對應論域 image of v under T: 在T映射下v的像 range of T: T的值域 preimage of w: w的反像 linear transformation: 線性轉換 linear operator: 線性運算子 zero transformation: 零轉換 identity transformation: 相等轉換 18/78 6.2 The Kernel and Range of a Linear Transformation Kernel of a linear transformation T: Let T : V W be a linear transformation Then the set of all vectors v in V that satisfy T ( v) 0 is called the kernel of T and is denoted by ker(T). ker(T ) {v | T ( v) 0, v V } Ex 1: (Finding the kernel of a linear transformation) T ( A) AT (T : M32 M 23 ) Sol: 0 0 ker(T ) 0 0 0 0 Elementary Linear Algebra: Section 6.2, p.375 19/78 Ex 2: (The kernel of the zero and identity transformations) (a) T(v)=0 (the zero transformation T : V W ) ker(T ) V (b) T(v)=v (the identity transformation T : V V ) ker(T ) {0} Ex 3: (Finding the kernel of a linear transformation) T ( x, y, z) ( x, y,0) (T : R3 R3 ) ker(T ) ? Sol: ker(T ) {(0,0, z) | z is a real number} Elementary Linear Algebra: Section 6.2, p.375 20/78 Ex 5: (Finding the kernel of a linear transformation) x1 1 1 2 T (x) Ax x2 3 1 2 x3 ker(T ) ? (T : R 3 R 2 ) Sol: ker(T ) {( x1, x2 , x3 ) | T ( x1, x2 , x3 ) (0,0), x ( x1, x2 , x3 ) R3} T ( x1 , x2 , x3 ) (0,0) x1 1 1 2 0 x2 1 2 3 0 x3 Elementary Linear Algebra: Section 6.2, p.376 21/78 1 1 2 0 G. J .E 1 0 1 0 1 2 3 0 0 1 1 0 x1 t 1 x2 t t 1 x3 t 1 ker(T ) {t (1,1,1) | t is a real number} span{(1,1,1)} Elementary Linear Algebra: Section 6.2, p.377 22/78 Thm 6.3: (The kernel is a subspace of V) The kernel of a linear transformation T : V W is a subspace of the domain V. T (0) 0 (T heorem6.1) ker(T ) is a nonemptysubset of V Pf: Let u and v be vectors in the kernel of T . then T (u v) T (u) T ( v) 0 0 0 T (cu) cT (u) c0 0 Thus, ker(T ) is a subspace of V . u v ker(T ) cu ker(T ) Note: The kernel of T is sometimes called the nullspace of T. Elementary Linear Algebra: Section 6.2, p.377 23/78 Ex 6: (Finding a basis for the kernel) Let T : R 5 R 4 be defined by T (x) Ax, where x is in R 5 and 1 2 A 1 0 1 1 1 3 1 0 0 2 0 1 0 0 2 8 2 0 Find a basis for ker(T) as a subspace of R5. Elementary Linear Algebra: Section 6.2, p.377 24/78 Sol: A 1 2 1 0 0 2 0 1 3 0 2 0 0 1 1 1 0 0 1 2 8 0 1 0 G . J . E 0 0 0 0 0 0 2 1 1 0 0 0 0 s 0 1 0 2 1 4 0 0 t 0 0 0 0 x1 2s t 2 1 x2 s 2t 1 2 x x3 s s 1 t 0 x4 4t 0 4 x5 t 0 1 B (2, 1, 1, 0, 0), (1, 2, 0, 4, 1): one basis for thekernelof T Elementary Linear Algebra: Section 6.2, p.378 25/78 Corollary to Thm 6.3: Let T : R n R m be theL.T given by T (x) Ax T hen thekernelof T is equal to thesolutionspace of Ax 0 T (x) Ax (a linear tr ansformati on T : R n R m ) Ker (T ) NS ( A) x | Ax 0, x R m (subspace of R m ) Range of a linear transformation T: Let T : V W be a L.T . T hen theset of all vectorsw in W thatare images of vector in V is called therange of T and is denotedby range(T ) range(T ) {T ( v) | v V } Elementary Linear Algebra: Section 6.2, p.378 26/78 Thm 6.4: (The range of T is a subspace of W) T herange of a linear transformation T : V W is a subspace of W . Pf: T (0) 0 (T hm.6.1) range(T ) is a nonemptysubset of W Let T (u) andT ( v) be vectorin therange of T T (u v) T (u) T ( v) range(T ) (u V , v V u v V ) T (cu) cT (u) range(T ) (u V cu V ) T herefore,range(T ) is W subspace. Elementary Linear Algebra: Section 6.2, p.379 27/78 Notes: T : V W is a L.T. (1) Ker(T ) is subspace of V (2)range(T ) is subspace of W Corollary to Thm 6.4: Let T : R n R m be theL.T .given by T (x) Ax T hen therange of T is equal to thecolumnspace of A range(T ) CS ( A) Elementary Linear Algebra: Section 6.2, p.379 28/78 Ex 7: (Finding a basis for the range of a linear transformation) Let T : R 5 R 4 be defined by T (x) Ax, where x is R 5 and 1 2 A 1 0 1 1 1 3 1 0 0 2 0 1 0 0 2 8 2 0 Find a basis for the range of T. Elementary Linear Algebra: Section 6.2, p.379 29/78 Sol: 1 2 A 1 0 2 0 1 3 0 2 0 0 1 1 1 1 0 G . J . E 0 0 1 0 2 8 0 c1 c2 c3 c4 c5 0 2 1 1 0 0 0 0 0 1 0 2 B 1 4 0 0 w1 w2 w3 w4 w5 w1 , w2 , w4 is a basis for CS ( B) c , c , c is a basis for CS ( A) 1 2 4 (1, 2, 1, 0), (2, 1, 0, 0), (1, 1, 0, 2)is a basis for therangeof T Elementary Linear Algebra: Section 6.2, pp.379-380 30/78 Rank of a linear transformation T:V→W: rank(T ) thedimensionof therange of T Nullity of a linear transformation T:V→W: nullity(T ) thedimensionof thekernelof T Note: Let T : R n R m be theL.T .given by T (x) Ax, then rank(T ) rank( A) nullity(T ) nullity( A) Elementary Linear Algebra: Section 6.2, p.380 31/78 Thm 6.5: (Sum of rank and nullity) Let T : V W be a L.T .forman n - dimensional vectorspaceV intoa vectorspaceW . then rank(T ) nullity(T ) n dim(range of T ) dim(kernelof T ) dim(domain of T ) Pf: Let T is represented by an m n matrix A Assume rank( A) r (1)rank (T ) dim( range of T ) dim( column space of A) rank ( A) r (2)nullity(T ) dim(kernelof T ) dim(solutionspace of A) nr rank(T ) nullity(T ) r (n r ) n Elementary Linear Algebra: Section 6.2, p.380 32/78 Ex 8: (Finding the rank and nullity of a linear transformation) Find t herank and nullit yof t heL.T .T : R 3 R 3 define by 1 0 2 A 0 1 1 0 0 0 Sol: rank(T ) rank( A) 2 nullity(T ) dim(domain of T ) rank(T ) 3 2 1 Elementary Linear Algebra: Section 6.2, p.381 33/78 Ex 9: (Finding the rank and nullity of a linear transformation) Let T : R 5 R 7 be a linear tra nsformatio n. (a ) Find the dimension of the kernel of T if the dimension of the range is 2 (b) Find the rank of T if the nullity of T is 4 (c) Find the rank of T if Ker (T ) {0} Sol: (a ) dim( domain of T ) 5 dim( kernel of T ) n dim( range of T ) 5 2 3 (b)rank(T ) n nullity(T ) 5 4 1 (c)rank(T ) n nullity(T ) 5 0 5 Elementary Linear Algebra: Section 6.2, p.381 34/78 One-to-one: A functionT : V W is called one- to - oneif thepreimageof every win therange consistsof a single vector. T is one- to - oneiff for all u and v inV, T (u) T ( v) implies thatu v. one-to-one Elementary Linear Algebra: Section 6.2, p.382 not one-to-one 35/78 Onto: A functionT : V W is said to be ontoif everyelement in w has a preimagein V (T is onto W when W is equal to the range of T.) Elementary Linear Algebra: Section 6.2, p.382 36/78 Thm 6.6: (One-to-one linear transformation) Let T : V W be a L.T . T henT is 1 - 1 iff Ker (T ) {0} Pf: SupposeT is 1 -1 T henT (v) 0 can haveonlyonesolution: v 0 i.e. Ker(T ) {0} Suppose Ker(T ) {0} andT (u) T (v) T (u v) T (u) T (v) 0 T is a L.T. u v Ker(T ) u v 0 T is 1 - 1 Elementary Linear Algebra: Section 6.2, p.382 37/78 Ex 10: (One-to-one and not one-to-one linear transformation) (a ) The L.T. T : M mn M nm given by T ( A) AT is one - to - one. Because its kernel consists of only the m n zero matrix. (b) The zero transformation T : R 3 R 3 is not one - to - one. Because its kernel is all of R3 . Elementary Linear Algebra: Section 6.2, p.382 38/78 Thm 6.7: (Onto linear transformation) Let T : V W be a L.T., where W is finite dimensiona l. Then T is onto iff the rank of T is equal to the dimension of W . Thm 6.8: (One-to-one and onto linear transformation) Let T : V W be a L.T. with vect or space V and W both of dimension n. Then T is one - to - one if and only if it is onto. Pf: If T is one - to - one, then Ker (T ) {0} and dim( Ker (T )) 0 dim(range(T )) n dim(Ker(T )) n dim(W ) Consequent ly, T is onto. If T is onto, then dim( range of T ) dim(W ) n dim( Ker (T )) n dim( range of T ) n n 0 Therefore, T is one - to - one. Elementary Linear Algebra: Section 6.2, p.383 39/78 Ex 11: The L.T. T : R n R m is given by T (x) Ax, Find the nullity and rank of T and determine whether T is one - to - one, onto, or neither. 1 2 0 (a) A 0 1 1 0 0 1 1 2 0 (c ) A 0 1 1 Sol: 1 2 (b) A 0 1 0 0 1 2 ( d ) A 0 1 0 0 0 1 0 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 Elementary Linear Algebra: Section 6.2, p.383 40/78 Isomorphism: A linear transformation T : V W that is one to one and onto is called an isomorphis m. Moreover, if V and W are vector spaces such that there exists an isomorphis m from V to W , then V and W are said to be isomorphic to each other. Thm 6.9: (Isomorphic spaces and dimension) Two finite-dimensional vector space V and W are isomorphic if and only if they are of the same dimension. Pf: Assume that V is isomorphic to W , where V has dimension n. There exists a L.T. T : V W that is one to one and onto. T is one - to - one dim( Ker (T )) 0 dim( range of T ) dim( domain of T ) dim( Ker (T )) n 0 n Elementary Linear Algebra: Section 6.2, p.384 41/78 T is onto. dim( range of T ) dim(W ) n Thus dim(V ) dim(W ) n Assume that V and W both have dimension n. Let v1 , v2 , , vn be a basis of V, and let w1 , w2 , , wn be a basis of W . Then an arbitrary vector in V can be represente d as v c1v1 c2 v2 cn vn and you can define a L.T. T : V W as follows. T ( v ) c1w1 c2 w2 cn wn It can be shown that this L.T. is both 1-1 and onto. Thus V and W are isomorphic. Elementary Linear Algebra: Section 6.2, p.384 42/78 Ex 12: (Isomorphic vector spaces) The following vector spaces are isomorphic to each other. (a) R4 4 - space (b)M 41 space of all 4 1 matrices (c)M 22 space of all 2 2 matrices (d ) P3 ( x) spaceof all polynomial s of degree 3 or less (e)V {( x1, x2 , x3 , x4 , 0), xi is a real number}(subspace of R5 ) Elementary Linear Algebra: Section 6.2, p.385 43/78 Keywords in Section 6.2: kernel of a linear transformation T: 線性轉換T的核空間 range of a linear transformation T: 線性轉換T的值域 rank of a linear transformation T: 線性轉換T的秩 nullity of a linear transformation T: 線性轉換T的核次數 one-to-one: 一對一 onto: 映成 isomorphism(one-to-one and onto): 同構 isomorphic space: 同構的空間 44/78 6.3 Matrices for Linear Transformations Two representations of the linear transformation T:R3→R3 : (1)T ( x1, x2 , x3 ) (2x1 x2 x3 , x1 3x2 2x3 ,3x2 4x3 ) 2 1 1 x1 (2)T (x) Ax 1 3 2 x2 0 3 4 x3 Three reasons for matrix representation of a linear transformation: It is simpler to write. It is simpler to read. It is more easily adapted for computer use. Elementary Linear Algebra: Section 6.3, p.387 45/78 Thm 6.10: (Standard matrix for a linear transformation) Let T : R n R m be a linear trtansformation such that a11 a12 a1n a21 a22 a2 n T (e1 ) , T (e2 ) , , T (en ) , am1 am 2 amn Then them n matrix whose n columnscorrespondtoT (ei ) a11 a12 a1n a21 a22 a2 n A T (e1 ) T (e2 ) T (en ) am1 am 2 amn is such thatT ( v) Av for every v in R n . A is called thestandardmatrixfor T . Elementary Linear Algebra: Section 6.3, p.388 46/78 Pf: v1 v2 v v1e1 v2 e2 vn en vn T is a L.T . T (v ) T (v1e1 v2 e2 vn en ) T (v1e1 ) T (v2 e2 ) T (vn en ) v1T (e1 ) v2T (e2 ) vnT (en ) a11 a21 Av am1 a12 a22 am 2 a1n v1 a11v1 a12 v2 a1n vn a2 n v2 a21v1 a22 v2 a2 n vn amn vn am1v1 am 2 v2 amn vn Elementary Linear Algebra: Section 6.3, p.388 47/78 a11 a12 a1n a21 a22 a2 n v1 v2 vn am1 am 2 amn v1T (e1 ) v2T (e2 ) vnT (en ) T herefore,T ( v) Av for each v in Rn Elementary Linear Algebra: Section 6.3, p.389 48/78 Ex 1: (Finding the standard matrix of a linear transformation) Find thestandardmatrixfor theL.T .T : R3 R2 define by T ( x, y, z) ( x 2 y, 2 x y) Sol: Vector Notation T (e1 ) T (1, 0, 0) (1, 2) T (e2 ) T (0, 1, 0) (2, 1) T (e3 ) T (0, 0, 1) (0, 0) Elementary Linear Algebra: Section 6.3, p.389 Matrix Notation 1 1 T (e1 ) T ( 0 ) 2 0 0 2 T (e2 ) T ( 1 ) 1 0 0 0 T (e3 ) T ( 0 ) 0 1 49/78 A T (e1 ) T (e2 ) T (e3 ) 1 2 0 2 1 0 Check: x x 1 2 0 x 2 y A y y 2 1 0 2 x y z z i.e.T ( x, y, z ) ( x 2 y,2 x y) Note: 1 2 0 1x 2 y 0 z A 2 1 0 2 x 1y 0 z Elementary Linear Algebra: Section 6.3, p.389 50/78 Ex 2: (Finding the standard matrix of a linear transformation) T helinear transformation T : R 2 R 2 is given by projecting each pointin R 2 ontothe x - axis. Find thestandardmatrixfor T . Sol: T ( x, y) ( x, 0) 1 0 A T (e1 ) T (e2 ) T (1, 0) T (0, 1) 0 0 Notes: (1) The standard matrix for the zero transformation from Rn into Rm is the mn zero matrix. (2) The standard matrix for the zero transformation from Rn into Rn is the nn identity matrix In Elementary Linear Algebra: Section 6.3, p.390 51/78 Composition of T1:Rn→Rm with T2:Rm→Rp : T ( v) T2 (T1 ( v)), v Rn T T2 T1 , domainof T domainof T1 Thm 6.11: (Composition of linear transformations) Let T1 : R n R m and T2 : R m R p be L.T . with standardmatricesA1 and A2 , then (1)Thecomposition T : Rn R p , definedby T ( v) T2 (T1 ( v)), is a L.T. (2) T hestandardmatrixA forT is given by thematrixproduct A A2 A1 Elementary Linear Algebra: Section 6.3, p.391 52/78 Pf: (1)( T is a L.T .) Let u and v be vectorsin R n and let c be any scalar then T (u v) T2 (T1 (u v)) T2 (T1 (u) T1 ( v)) T2 (T1 (u)) T2 (T1 ( v)) T (u) T ( v) T (cv) T2 (T1 (cv)) T2 (cT1 ( v)) cT2 (T1 ( v)) cT ( v) (2)( A2 A1 is thestandardmatrixfor T ) T ( v) T2 (T1 ( v)) T2 ( A1v) A2 A1v ( A2 A1 ) v Note: T1 T2 T2 T1 Elementary Linear Algebra: Section 6.3, p.391 53/78 Ex 3: (The standard matrix of a composition) Let T1 andT2 be L.T.fromR3 into R3 s.t. Sol: T1 ( x, y, z) (2 x y, 0, x z) T2 ( x, y, z) ( x y, z, y) Find thestandardmatricesfor thecompositions T T2 T1 and T ' T1 T2 , 2 A1 0 1 1 A2 0 0 1 0 0 0 (st andardmat rixfor T1 ) 0 1 1 0 0 1 (st andardmat rixfor T2 ) 1 0 Elementary Linear Algebra: Section 6.3, p.392 54/78 T hestandardmatrixfor T T2 T1 1 1 0 2 1 0 2 1 0 A A2 A1 0 0 1 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 T hestandardmatrixforT ' T1 T2 2 1 0 1 1 0 2 2 1 A' A1 A2 0 0 0 0 0 1 0 0 0 1 0 1 0 1 0 1 0 0 Elementary Linear Algebra: Section 6.3, p.392 55/78 Inverse linear transformation: If T1 : Rn Rn andT2 : Rn Rn are L.T .s.t.for everyv inRn T2 (T1 ( v)) v and T1 (T2 ( v)) v T henT2 is called theinverseof T1 andT1 is said to be invertible Note: If the transformation T is invertible, then the inverse is unique and denoted by T–1 . Elementary Linear Algebra: Section 6.3, p.392 56/78 Thm 6.12: (Existence of an inverse transformation) Let T : R n R n be a L.T .with standardmatrixA, T hen thefollowingconditionare equivalent. (1) T is invertible. (2) T is an isomorphism. (3) A is invertible. Note: If T is invertible with standard matrix A, then the standard matrix for T–1 is A–1 . Elementary Linear Algebra: Section 6.3, p.393 57/78 Ex 4: (Finding the inverse of a linear transformation) TheL.T.T: R3 R3 is definedby T ( x1, x2 , x3 ) (2x1 3x2 x3 , 3x1 3x2 x3 , 2x1 4x2 x3 ) Show that T is invertible, and find its inverse. Sol: T hestandardmatrixfor T 2 3 1 A 3 3 1 2 4 1 2 x1 3x2 x3 3x1 3x2 x3 2 x1 4 x2 x3 2 3 1 1 0 0 A I 3 3 3 1 0 1 0 2 4 1 0 0 1 Elementary Linear Algebra: Section 6.3, p.393 58/78 0 1 0 0 1 1 . J .E G 0 1 0 1 0 1 I 0 0 1 6 2 3 A1 T hereforeT is invertibleand thestandardmatrixfor T 1 is A1 0 1 1 A1 1 0 1 6 2 3 0 x1 x1 x2 1 1 T 1 ( v) A1 v 1 0 1 x2 x1 x3 6 2 3 x 6 x 2 x 3 x 2 3 3 1 In other words, T 1 ( x1 , x2 , x3 ) ( x1 x2 , x1 x3 , 6 x1 2 x2 3x3 ) Elementary Linear Algebra: Section 6.3, p.394 59/78 the matrix of T relative to the bases B and B': T :V W B {v1 , v2 ,, vn } (a L.T ). (a basis forV ) B' {w1 , w2 ,, wm } (a basis forW ) Thus, the matrix of T relative to the bases B and B' is A T (v1 )B' , T (v2 )B' ,, T (vn )B' M mn Elementary Linear Algebra: Section 6.3, p.394 60/78 Transformation matrix for nonstandard bases: Let V andW be finite- dimensional vectorspaces with basis B and B' , respectively,where B {v1 , v2 ,, vn } If T : V W is a L.T. s.t. T (v1 )B ' a11 a12 a1n a21 a22 a2 n , T (v2 )B ' , , T (vn )B ' am1 am 2 amn then them n matrix whose n columnscorrespondtoT (vi )B' Elementary Linear Algebra: Section 6.3, p.395 61/78 a11 a21 A T (e1 ) T (e2 ) T (en ) am1 a12 a22 am 2 a1n a2 n amn is such thatT ( v)B' A[ v]B for everyv in V . Elementary Linear Algebra: Section 6.3, p.395 62/78 Ex 5: (Finding a matrix relative to nonstandard bases) Let T: R2 R2 be a L.T .definedby T ( x1 , x2 ) ( x1 x2 , 2 x1 x2 ) Find thematrixof T relativeto thebasis B {(1, 2), (1, 1)} and B' {(1, 0), (0, 1)} Sol: T (1, 2) (3, 0) 3(1, 0) 0(0, 1) T (1, 1) (0, 3) 0(1, 0) 3(0, 1) T (1, 2)B' 03, T (1, 1)B' 03 the matrix for T relative to B and B' 3 0 A T (1, 2)B ' T (1, 2)B ' 0 3 Elementary Linear Algebra: Section 6.3, p.395 63/78 Ex 6: For theL.T .T: R 2 R 2 given in Example5, use thematrix A to findT ( v), where v (2, 1) Sol: v (2, 1) 1(1, 2) 1(1, 1) B {(1, 2), (1, 1)} 1 v B 1 3 0 1 3 T ( v)B ' AvB 0 3 1 3 T ( v) 3(1, 0) 3(0, 1) (3, 3) B' {(1, 0), (0, 1)} Check: T (2, 1) (2 1, 2(2) 1) (3, 3) Elementary Linear Algebra: Section 6.3, p.395 64/78 Notes: (1)In thespecialcase whereV W and B B' , thematrix A is alled thematrixof T relativeto thebasis B (2)T : V V : theidentity ransformat t ion B {v1 , v2 , , vn } : a basis forV thematrixof T relativeto thebasis B 1 0 0 0 1 0 In A T (v1 )B , T (v2 )B , , T (vn )B 0 0 1 Elementary Linear Algebra, Section 6.3, p.396 65/78 Keywords in Section 6.3: standard matrix for T: T 的標準矩陣 composition of linear transformations: 線性轉換的合成 inverse linear transformation: 反線性轉換 matrix of T relative to the bases B and B' : T對應於基底B到 B'的矩陣 matrix of T relative to the basis B: T對應於基底B的矩陣 66/78 6.4 Transition Matrices and Similarity T :V V B {v1 , v2 ,, vn } ( a L.T ). ( a basis of V ) B' {w1 , w2 ,, wn } (a basis of V ) A T (v1 )B , T (v2 )B ,, T (vn )B A' T (w1 )B' , T (w2 )B' ,, T (wn )B' P w1 B , w2 B ,, wn B P1 v1 B' , v2 B' ,, vn B' ( matrixof T relativeto B) (matrixof T relativeto B' ) ( transition matrixfromB' to B ) ( transition matrixfromB to B' ) vB PvB' , vB' P 1vB T ( v)B AvB T ( v)B' A' vB' Elementary Linear Algebra: Section 6.4, p.399 67/78 Two ways to get from v B ' to T ( v)B ': (1)(direct) indirect A'[ v]B ' [T ( v)]B ' (2)(indirect) P 1 AP[ v]B ' [T ( v)]B ' 1 A' P AP Elementary Linear Algebra: Section 6.4, pp.399-400 direct 68/78 Ex 1: (Finding a matrix for a linear transformation) Find thematrixA' for T: R 2 R 2 T ( x1 , x2 ) (2x1 2 x2 , x1 3x2 ) reletiveto thebasis B' {(1, 0), (1, 1)} Sol: (I) A' T (1, 0)B' T (1, 1)B' 3 T (1, 0) (2, 1) 3(1, 0) 1(1, 1) T (1, 0)B ' 1 2 T (1, 1) (0, 2) 2(1, 0) 2(1, 1) T (1, 1)B ' 2 3 2 A' T (1, 0)B ' T (1, 1)B ' 1 2 Elementary Linear Algebra: Section 6.4, p.400 69/78 (II)standardmatrixfor T ( matrixof T relativeto B {(1, 0), (0, 1)}) 2 2 A T (1, 0) T (0, 1) 1 3 transitionmatrixfrom B' to B 1 1 P (1, 0)B (1, 1)B 0 1 transitionmatrixfrom B to B' 1 1 P 0 1 matrixof T relativeB' 1 1 1 2 2 1 1 3 2 A' P AP 0 1 1 3 0 1 1 2 1 Elementary Linear Algebra: Section 6.4, p.400 70/78 Ex 2: (Finding a matrix for a linear transformation) Let B {( 3, 2), (4, 2)} and B' {( 1, 2), (2, 2)} be basis for R 2 , 2 7 2 2 and let A be the matrix for T : R R relativeto B. 3 7 Find thematrixof T relativeto B'. Sol: 3 2 transitionmatrixfrom B' to B : P (1, 2)B (2, 2)B 2 1 1 2 1 transitionmatrixfrom B to B': P (3, 2)B ' (4, 2)B ' 2 3 matrix of T relative to B': 1 2 2 7 3 2 2 1 A' P AP 2 3 3 7 2 1 1 3 1 Elementary Linear Algebra: Section 6.4, p.401 71/78 Ex 3: (Finding a matrix for a linear transformation) For thelinear transformation T : R 2 R 2 given in Ex.2,find v B、 T ( v)B and T ( v)B ' , for thevectorv whose coordinatematrixis 3 vB ' 1 Sol: 3 2 3 7 vB PvB' 2 1 1 5 2 7 7 21 T ( v)B AvB 3 7 5 14 1 2 21 7 1 T ( v)B' P T ( v)B 2 3 14 0 or T ( v)B ' A' vB ' 2 1 3 7 1 3 1 0 Elementary Linear Algebra: Section 6.4, p.401 72/78 Similar matrix: For square matrices A and A‘ of order n, A‘ is said to be similar to A if there exist an invertible matrix P s.t. A' P 1 AP Thm 6.13: (Properties of similar matrices) Let A, B, and C be square matrices of order n. Then the following properties are true. (1) A is similar to A. (2) If A is similar to B, then B is similar to A. (3) If A is similar to B and B is similar to C, then A is similar to C. Pf: (1) A I n AIn (2) A P 1 BP PAP1 P( P 1BP) P 1 PAP1 B Q 1 AQ B (Q P 1 ) Elementary Linear Algebra: Section 6.4, p.402 73/78 Ex 4: (Similar matrices) 2 2 3 2 (a) A and A' are similar 1 3 1 2 1 1 1 because A' P AP, where P 0 1 2 7 2 1 (b) A and A' are similar 3 7 1 3 3 2 because A' P AP, where P 2 1 1 Elementary Linear Algebra: Section 6.4, p.403 74/78 Ex 5: (A comparison of two matrices for a linear transformation) 1 3 0 Suppose A 3 1 0 is thematrixfor T : R 3 R 3 relative 0 0 2 to thestandardbasis. Find thematrixfor T relativeto thebasis B' {(1, 1, 0), (1, 1, 0), (0, 0, 1)} Sol: The transitio n matrix from B' to the standard matrix P (1, 1, 0)B (1, 1, 0)B 1 12 2 1 1 P 2 12 0 0 0 0 1 Elementary Linear Algebra: Section 6.4, p.403 1 1 0 (0, 0, 1)B 1 1 0 0 0 1 75/78 matrixof T relativeto B': 12 12 A' P 1 AP 12 12 0 0 0 4 0 0 2 0 0 0 2 Elementary Linear Algebra: Section 6.4, p.403 0 1 3 0 1 1 0 0 3 1 0 1 1 0 1 0 0 2 0 0 1 76/78 Notes: Computational advantages of diagonal matrices: d1k 0 k (1) D 0 0 d 2k 0 0 0 d nk 0 d1 0 D 0 0 d2 0 0 0 d n (2) DT D d11 0 (3) D 1 0 1 d2 0 Elementary Linear Algebra: Section 6.4, p.404 0 0 , d i 0 1 dn 77/78 Keywords in Section 6.4: matrix of T relative to B: T 相對於B的矩陣 matrix of T relative to B' : T 相對於B'的矩陣 transition matrix from B' to B : 從B'到B的轉移矩陣 transition matrix from B to B' : 從B到B'的轉移矩陣 similar matrix: 相似矩陣 78/78