Toán học - Chapter 5: Linear transformation

Math Dept, Faculty of Applied Science, HCM University of Technology ------------------------------------------------------------------------------------- Subject: Linear Algebra Chapter 5: Linear transformation • Instructor Dr. Dang Van Vinh (12/2006) CONTENTS --------------------------------------------------------------------------------------------------------------------------- I – Definition and examples III – The Matrix of a linear transformation IV –Transition Matrix and change of basis V –Similarity II – The Kernel and Image of linear transformation I. The Definition and Examples --------------------------------------------------------------------------------------------------------------------------- Given two sets X and Y, if we assign to each element x of X one and only one element y of Y, then the rule of assignment is called a function (or a mapping or a transformation). Definition of mapping (Transformation) :f X Y ( )x f x y The set X is called the domain of the function f. The set Y is called the co-domain of the function f. For an element x of the domain, the assigned element f(x) is called the image of x. The set of all images, which is a subset of the co-domain Y, is called the range of the function. I. The Definition and Examples --------------------------------------------------------------------------------------------------------------------------- A mapping is said to be onto Y if each y in Y is the image of at least one x in X. :f X Y A mapping is said to be one-to-one (or 1:1) if each y in Y is the image of at most one x in X. :f X Y I. The Definition and Examples --------------------------------------------------------------------------------------------------------------------------- Definition of Linear transformation 1. f(v1 + v2) = f(v1) + f(v2) The operator f from V into W is said to be linear transformation if for any elements v1 and v2 of the space V and any scalar , the following relations are satisfied:  Let V and W be a linear spaces whose dimensions are n and m respectively. 2. f( v) = f(v)   If the space W coincides with the space V, then the linear operator acting in this case from V into V is also called a linear transformation of the space V. Remark I. The Definition and Examples --------------------------------------------------------------------------------------------------------------------------- Let be a linear transformation. WVf : Let E ={e1, e2, , en} be a basis (or spanning set) of V. Assume that f(e1), f(e2), , f(en) are given. 1 1 2 2 n nx V x x e x e x e       1 1 2 2( ) ( )n nf x f x e x e x e    1 1 2 2( ) ( ) ( ) ( )n nf x f x e f x e f x e    1 1 2 2( ) ( ) ( ) ( )n nf x x f e x f e x f e    I. The Definition and Examples --------------------------------------------------------------------------------------------------------------------------- 23: RRf  )2,32()();,,( 31321321 xxxxxxfxxxx  Show that the map given by Example is a linear transformation. Let V be a vector space. Show that the map given by is a linear transformation. 0)(:  xfVx Example RVf : I. The Definition and Examples --------------------------------------------------------------------------------------------------------------------------- Example Let be a linear transformation. 3 3:f R R Assume that (1,1,1) (1,2,1);f  (1,1,2) (2,1,0);f  (1,2,1) (1,0,3);f  1. Find f (5,3,6) 2. Find f (x) I. The Definition and Examples --------------------------------------------------------------------------------------------------------------------------- Which of the following map is a linear transformation? Example 1. ),32(),(;: 1212122 xxxxxfRRf  2. )0,2(),(;: 212122 xxxxfRRf  3. )1,2(),(;: 1212122  xxxxxfRRf 4. ),1(),(;: 212122 xxxxfRRf  5. ),(),(;: 2 1212122 xxxxxfRRf  6 ),(),(;: 122122 xxxxfRRf  I. The Definition and Examples --------------------------------------------------------------------------------------------------------------------------- Example Let be a linear mapping. Set 3 2:f R R (1,1,1) (1,2),f (1,1,0) (2, 1),f   (1,0,1) ( 1,1);f   1. Find f (3,1,5) 2. Calculate f (x) 1. Assume (3,1,5) (1,1,0) (1,1,1) (1,0,1)     3 1 5                  2, 3, 2       (3,1,5) ( (1,1,0) (1,1,1) (1,0,1))f f       (3,1,5) (1,1,0) (1,1,1) (1,0,1)f f f f      (3,1,5) 2(2, 1) 3(1,2) 2( 1,1)f       ( 3,10)  The mapping f is defined if we know the images of one basis for R3. Example A linear mapping f is a rotation around the z-axis through the angle 30o counterclockwise. Find f(x). Linear mapping 3 3:f R R o y z x (0,0,1) (0,0,1)f  3 1 (1,0,0) ( , ,0) 2 2 f  1 3 (0,1,0) ( , ,0) 2 2 f   1 2 1 2 3 3 1 1 3 ( ) ( , , ) 2 2 2 2 f x x x x x x    The mapping f is completely define if we know an images of one basis for R3. Example A linear transformation f is a symmetric with respect to the plane . Find f(x). A linear mapping 3 3:f R R (2, 1,3) ( 2,1, 3)f     (1,2,0) (1,2,0)f  (0,3,1) (0,3,1)f  ( )f x 2 3 0x y z   A basis {(1,0,0), (0,1,0), (0,0,1)}is not good for calculating. Choose a basis {(1,2,0), (0,3,1), (2,-1,3)}. II. Kernel and image --------------------------------------------------------------------------------------------------------------------------- V W 0Kerf Let be a linear transformation. The kernel of a linear transformation f is the set of all the element x of the space V for which f(x) = 0. Definition of the Kernel WVf :  0)(|  xfVxKerf II. Kernel and image --------------------------------------------------------------------------------------------------------------------------- V W Imf Let be a linear transformation. The image of a linear transformation f is the set of all the element y of the space W which can be represented as y = f(x). Definition of the Image WVf :  )(:|Im xfyVxWyf  II. Kernel and image --------------------------------------------------------------------------------------------------------------------------- Theorem Let be a linear transformation. WVf : 1. The kernel of a linear transformation f is a subspace of V. 2. The image of a linear transformation f is a subspace of W. 3. dim(kerf) +dim(Imf) = dim (V) Proof. II. Kernel and image --------------------------------------------------------------------------------------------------------------------------- 3. dim(kerf) +dim(Imf) = dim (V) Proof. Suppose dim(Kerf) = m. Basis for Kerf 1 2, ,...,{ }mE e e e Extend E to the basis for V: 1 1 1,..., , ,..., }{ m nE e e v v Show that is a basis for Imf. 2 1( ),..., ( ){ }nE f v f v Im : ( )y f x V y f x      1 1 1 1( ... ... )m m n ny f e e v v          1 1 1 1( ) ... ( ) ( ) ... ( )m m n ny f e f e f v f v          1 1( ) ... ( ).n ny f v f v     E2 is a spanning set for Imf. 1) II. Kernel and image --------------------------------------------------------------------------------------------------------------------------- 2) Show that E2 is a linear independent set. 1 1( ... ) 0n nf v v     1 1 ... .ern nv v K f     1 1 1 1... ...n n m mv v e e         1 1 1 1... ... 0n n m mv v e e          E1 is a independent: 1 2 ... 0m      E2 is a linear independent. 1 1( ) ... ( ) 0n nf v f v   Assume Thus E2 is a basis for Imf. dim(Imf ) = n. dim(Imf ) + dim(Kerf ) = m + n = dim(V). is a basis ( or spanning set) for Imf. Theorem If is a basis (spanning set) for V, then Proof. Let be a spanning set for V. 1 2, ,...,{ }nE e e e Imy f  : ( )x V y f x   x is a linear combination of E. 1 1 2 2( ... )n ny f x e x e x e    The mapping f is linear mapping 1 1 2 2( ) ( ) ... ( )n ny x f e x f e x f e    1 2( ), ( ),..., ( ){ }nF f e f e f e spans y. 1 2Im ( ), ( ),..., ( )nf f e f e f e   1 2, ,...,{ }nE e e e 1 2( ), ( ),..., ({ )}nf e f e f e II. Kernel and image --------------------------------------------------------------------------------------------------------------------------- Steps for finding an Image of a linear transformation 1. Select a basis for V: 1 2, ,...,{ }nE e e e 3. 1 2Im ( ), ( ),..., ( )nf f e f e f e  2. Find 1 2( ), ( ),..., ( )nf e f e f e II. Kernel and image --------------------------------------------------------------------------------------------------------------------------- Example Let be a linear transformation. Given 3 3:f R R 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 ( , , ) : ( ) ( , , ) ( ,2 3 ,3 5 ) x x x x R f x f x x x x x x x x x x x x            1. Find a basis and dimension for Kerf. 1 2 3( , , ) Kerx x x x f   ( ) 0f x  1 2 3 1 2 3 1 2 3( ,2 3 ,3 5 ) (0,0,0)x x x x x x x x x        1 2 3 1 2 3 1 2 3 0 2 3 0 3 5 0 x x x x x x x x x             1 2 32 ; ;x x x       (2 , , )x      (2, 1,1)x    E={(2,-1,1)} is a spanning set and basis for Kerf dim(Kerf) = 1. Example 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 ( , , ) : ( ) ( , , ) ( ,2 3 ,3 5 ) x x x x R f x f x x x x x x x x x x x x            2. Find a basis and dimension of Imf. Select a basis for R3: (1,0,0), (0,1,0), (0,0,1){ }E  Im (1,0,0), (0,1,0), (0,0,1)f f f f  Im (1,2,3), (1,3,5),( 1, 1, 1)f      dim(Imf) = rank {(1,2,3), (1,3,5), (-1,-1,-1)} dim(Im ) 2f  Basis: E={(1,1,1), (0,1,2)} Let be a linear transformation. Given 3 3:f R R Example Let be a linear transformation. Set 3 3:f R R 1. Find a basis and dimension for Kerf. First solusion. (1,1,1) (1,2,1);f  (1,1,2) (2,1, 1);f   (1,2,1) (5,4, 1);f   1 2 3 3( , , )x x x x R   1 2 3( , , ) (1,1,1) (1,1,2) (1,2,1)x x x x       1 2 3 2 2 x x x                      1 2 3 3 1 2 1 3x x x x x x x              1 2 3 1 2 3 1 2 3( ) ( 4 4 , 2 ,5 2 2 )f x x x x x x x x x x         er1 2 3( , , )x x x x K f   ( ) 0f x  AX = 0. (2 , ,4 )x     (2,1,4)x   Basis for Kerf E={(2,1,4)}, dim(Kerf) = 1. (1,1,1),(1,1,2),(1,2,1){ }E Second solution. Select a basis Kerx f  ( ) 0f x  1 2 3 [ ]E x x x x           1 2 3(1,1,1) (1,1,2) (1,2,1)x x x x    1 2 3( ) (1,1,1) (1,1,2) (1,2,1)f x x f x f x f    1 2 3 1 2 3 1 2 3( ) ( 2 5 ,2 4 , )       f x x x x x x x x x x AX = 0. ( ) 0f x  1 2 3, 2 ,x x x       2[ ]               Ex (1,1,1) 2 (1,1,2) (1,2,1)x        ( 2 , , 4 ) (2,1,4)x           Basis of Kerf: E={(2,1,4)}, dim(Kerf) = 1. II. Kernel and image --------------------------------------------------------------------------------------------------------------------------- Example Let be a linear transformation. 3 2:f R R Assume that (1,1,1) (1,1);f  (1,1,0) (3,1);f  (1,0,1) (0,1);f  1. Find a basis and dimension of Kerf. 2. Find a basis and dimension of Imf. Example Let be a linear transformation. Set 3 3:f R R 2. Find a basis and dimension for Imf. (1,1,1) (1,2,1);f  (1,1,2) (2,1, 1);f   (1,2,1) (5,4, 1);f   Select a basis for R3: (1,1,1), (1,1,2), (1,2,1){ }E  Im (1,1,1), (1,1,2), (1,2,1)f f f f  Im (1,2,1),(2,1, 1), (5,4, 1)f     dim(Imf) = Rank {(1,2,1),(2,1,-1),(5,4,-1)} dim(Im ) 2f  Basis: E={(1,2,1), (0,1,1)} Example A linear mapping f is a rotation around z-axis an angle 300 counterclockwise. Find a basis and dimension of the kernel and the image of f. o y z x There is only one zero vector has an image is zero vector. dim(kerf) + dim(Imf) = dim (R3). Thus dim(Imf) = 3 Imf = R3. The Kerf contains only one zero vector. dim(Kerf) = 0. Example Find a linear mapping , such that 4 3:f R R 1 2Im (1,1,1), (1,2,1)f f f    1 2er (1,1,1,0), (2,1,0,1)K f e e    1(1,1,1,0)e  2 (2,1,0,1)e  3(0,0,1,1)e  4 (0,0,0,1)e  (0,0,0) 1(1,1,1)f 2(1,2,1)f 1 2( ) ( ) 0f e f e  3 4( ) (1,1,1), ( ) (1,2,1)f e f e  ( )f x II. Kernel and image --------------------------------------------------------------------------------------------------------------------------- Example Let be a linear transformation. 3 3:f R R Assume that (1,1,1) (1,2,1);f  (1,1,2) (2,1,0);f  (1,2,1) (0,1, 1);f   1. Find a basis and dimension of Kerf. 2. Find a basis and dimension of Imf. II. Kernel and image --------------------------------------------------------------------------------------------------------------------------- Example Let be a linear transformation. 3 3:f R R Assume that 1. Find a basis and dimension of Kerf. 2. Find a basis and dimension of Imf. 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 ( , , ) : ( ) ( , , ) ( ,2 3 ,3 5 ) x x x x R f x f x x x x x x x x x x x x            III. The Matrix of a linear transformation --------------------------------------------------------------------------------------------------------------------------- Definition. Then Let E = {e1, e2, , en} be a basis for V. F = {f1, f2, , fm} be a basis for W. , 1 2[ ( )] [ ( )] [ ( )]E F F F n FA f e f e f e            Let be a linear mapping. WVf : is the matrix representation of f with respect to E and F. Let be a linear mapping. Assume that 23: RRf  1 2 3 1 2 3 1 3( , , ); ( ) ( 2 3 ,2 )     x x x x f x x x x x x Example Find a matrix of f with respect to E and F. { }; { }(1,1,1),(1,0,1),(1,1,0) (1,1),(1,2)E F  (1,1,1) (0,3)f  [ ](1, 3 1,1) 3 Ff          (1,0,1) ( 2,3)f   [ ](1, 7 0,1) 5 Ff          (1,1,0) (3,2)f  [ ](1, 4 1,0) 1 Ff         7 5 3 3 4 1 A           III. The Matrix of a linear transformation --------------------------------------------------------------------------------------------------------------------------- 1. For any linear mapping there exists one and only one matrix AE,F such that :f V W ,[ ( )] [ ]F E F Ef x A x where E and F are two bases for V and W respectively. Theorem 2. Let be a matrix over K. There exists one and only one linear mapping such that ( )ij m nA a  : n mf K K ,[ ( )] [ ]F E F Ef x A x Example E = {(1,1,1); (1,0,1); (1,1,0)} and F = {(1,1); (2,1)} Let be a linear mapping and 3 2:f R R be a matrix representation of f with respect to two bases , 2 1 3 0 3 4E F A        Step1. Find coordinates of (3,1,5) : Step 2. Use a formula ,[ ( )] [ ]F E F Ef x A x Step 3. Find coordinates of f(3,1,5) with respect to natural basis. 3 (3,1,5) 2 2 [ ]E          3 2 1 3 14 [ (3,1,5)] 2 0 3 4 2 2 Ff                   (3,1,5) 14(1,1) 2(2,1) (10,12)f    1. Find f (3,1,5) Example Let be a linear mapping and 1. Calculate f (2,3,-1) 2. Find Kerf. , 1 1 1 2 3 3 1 2 4 E EA           3 3:f R R ker ( ) 0x f f x   Suppose 1 2 3 [ ]E x x x x           [ ( )] 0Ef x  , .[ ] 0E E EA x  be a matrix representation of f with respect to the basis E = {(1,1,1); (1,0,1); (1,1,0)}. 1 1 2 2 3 3x x e x e x e    1 2 3 1 1 1 2 3 3 0 1 2 4 x x x                  6 [ ] 5Ex                6 (1,1,1) 5 (1,0,1) (1,1,0)x       (2 ,7 , )x     (2,7,1) is a basis for Kerf. (2,7,1){ }E  dim( er ) =1K f 1 2 3( , , ) x x x x (1,1,1) (1,0,1) (1,1,0)     1 2 3 1 2 1 3; ;x x x x x x x           [ ] 1 2 3 1 2 1 3 E x x x x x x x x              Example E = {(1,1,1); (1,0,1); (1,1,0)} and F = {(1,1); (2,1)} Let be a linear mapping and 3 2:f R R be a matrix representation of f with respect to two bases , 2 1 3 0 3 4E F A        2. Find f (x) [ ] 1 2 3 1 2 1 3 2 1 3 ( ) 0 3 4 F x x x f x x x x x               Using the formula: [ ] [ ],( ) .F E F Ef x A x [ ] 1 2 3 1 2 3 4 5 ( ) 7 3 4F x x x f x x x x           1 2 3 1 2 3( ) ( 4 5 )(1,1) (7 3 4 )(2,1)f x x x x x x x        1 2 3 1 2 3( ) (10 5 3 ,3 2 )f x x x x x x x      III. Similarity --------------------------------------------------------------------------------------------------------------------------- E F E’ F’ A P Q Q-1AP Schema: Notice: Q is the change of matrix from F to F’, then Q is invertible. III. Matrix of a linear transformation --------------------------------------------------------------------------------------------------------------------------- Example Let be a linear transformation. 3 3:f R R Assume that 1. Find f(2,1,5). 2. Find a matrix representation of f with respect to the basis E = {(1,1,1); (1,1,2); (1,2,1)}. 1 2 3 1 2 3 1 2 3 1 2 3( ) ( , , ) ( ,2 ,3 4 )f x f x x x x x x x x x x x x        3. Find f(2,1,5) using item 2), compare the result with item 1). Let be a linear mapping and III. The Matrix of a linear transformation --------------------------------------------------------------------------------------------------------------------------- 3 3:f R R Example 1. Calculate f (4,3, 5) , 1 0 1 2 1 4 1 1 3 E EA           2. Find a basis and dimension of Imf. be a matrix representation of f with respect to the basis E = {(1,1,1); (1,1,0); (1,0,0)} III. Similarity --------------------------------------------------------------------------------------------------------------------------- { }; { }' ' ' '1 2 1 2, ,..., , ,...,n nE e e e E e e e Two bases of V: Given a linear transform W:f V  { }; { }' ' ' '1 2 1 2, ,..., , ,...,m mF f f f F f f f Two bases of W: Let P be the change of basis matrix from E to E’ (the change –of- coordinates matrix) A be the matrix representation of f with respect to E and F. Then is a matrix representation of f with respect to E’ and F’. 1 EFQ A P  [ ( )] [ ]F EF Ef x A x ' '[ ( )] [ ]EFF EQ f x A P x  ' ' 1[ ( )] [ ]EFF Ef x Q A P x   Let Q be the change of basis matrix from F to F’. III. Similarity --------------------------------------------------------------------------------------------------------------------------- Let A and B be two nxn matrices over K. Definition A is similar to B if there is an invertible matrix P such thant P-1 A P = B. Theorem A is a matrix representation of f with respect to E, E. Let be a linear mapping. : Vf V  B is a matrix representation of f with respect to F, F. Then A and B are similar. Let be a linear mapping, and the matrix representation of f with respect to : f V V Example , 2 1 3 1 2 0 1 1 1          E EA Find the matrix representation of f with respect to the basis P is the change of basis matrix from E to F The matrix repesentation of f with respect to F is 1B P AP 1 2 2 0 1 0 0 1 1 P          { , ,2 }1 2 3 1 2 3 1 2 32E e e e e e e e e e       { , , }1 2 3 1 2 2 3F e e e e e e e     III. Similarity --------------------------------------------------------------------------------------------------------------------------- { }; { }' ' ' '1 2 1 2, ,..., , ,...,n nE e e e E e e e Given two bases of V: Given a linear mapping V:f V  P is the change of basis matrix from E to E’. A is the matrix representation of f with respect to E. Then is the matrix representation of f with respect to E’. 1P AP ' ' 1[ ( )] [ ] E E f x P AP x  E E E’ E’ A P P P-1AP Let be a linear mapping, and the matrix representation of f with respect to basis E = {(1,2,1); (1,1,2); (1,1,1)} is 3 3:f R R Example , 1 0 1 2 1 4 1 1 3 E EA           Find the matrix representation of f with respect to the standard basis. Standard basis: { }(1,0,0),(0,1,0),(0,0,1)F  P: the change of matrix from E to F. is the matrix representation of f with respect to F 1B P AP Instead of calculating P, we find P-1 is the change of basis matrix from standard basis to E 1P  1 1 1 1 2 1 1 1 2 1 P           