Chapter 4 linear transformation

Linear transformation Phan Thi Khanh Van E-mail: khanhvanphan@hcmut.edu.vn May 13, 2021 (Phan Thi Khanh Van) Linear transformation May 13, 2021 / 27 Table of contents Linear transformation The transformation matrix Change of basis matrix and transformation matrix Kernel and Image of a linear transformation (Phan Thi Khanh Van) Linear transformation May 13, 2021 / 27 Linear transformation Let U, V be vector spaces over the field K(R, C) f : U → V is called a linear transformation if: ∀u1 , u2 ∈ U, f (u1 + u2 ) = f (u1 ) + f (u2 ) ∀u ∈ U, α ∈ K : f (αu) = αf (u) Example Let f : R2 → R2 be a linear transformation satisfying: f (x1 , x2 ) = (2x1 + x2 , x1 − 3x2 ) Find f (3, 4) f (3, 4) = (2.3 + 4, − 3.4) = (10, −9) Represent f as matrix multiplication: 2x1 + x2 x1 [f (x)] = = = A[x] x1 − 3x2 −3 x2 (Phan Thi Khanh Van) Linear transformation May 13, 2021 / 27 Example Let f : R3 → R2 be a linear transformation satisfying: f (1, 1, 3) = (1, 0), f (1, −1, 0) = (2, −3), f (2, 0, 0) = (1, 1) Find a) f (2, 0, 3) b) f (−3, −1, 6) = f (2(1, 1, 3) + 3(1, −1, 0) − 4(2, 0, 0)) c) f (2, 3, −4) (2, 0, 3) = (1, 1, 3) + (1, −1, 0) ⇒ f (2, 0, 3) = f (1, 1, 3) + f (1, −1, 0) = (1, 0) + (2, −3) = (3, −3) b) f (−3, −1, 6) = f (2(1, 1, 3) + 3(1, −1, 0) − 4(2, 0, 0)) = 2f (1, 1, 3) + 3f (1, −1, 0) − 4f (2, 0, 0) = 2(1, 0) + 3(2, −3) − 4(1, 1) = (4, −13) a) (Phan Thi Khanh Van) Linear transformation May 13, 2021 / 27 c) We have that E = {(1; 1; 3), (1; −1; 0), (2; 0; 0)} is a basis of R (det(E ) = 0)  −1     −3 1 2  [(2, 3, −4)]E = 1 −1 0   = − 13 23 0 −4 Or (2, 3, −4) = − 43 (1, 1, 3) − 13 (1, −1, 0) + 23 (2, 0, 0) Consequently, 23 [f (2, 3, −4)] = − 43 [f (1, 1, 3)] − 13 [f (1, −1, 0)] + [f (2, 0, 0)] 37 − = − 34 − 13 + 23 = 1016 −3 (Phan Thi Khanh Van) Linear transformation May 13, 2021 / 27 Example Let f : R3 → R2 be a linear transformation satisfying: f (1, 1, 3) = (1, 0), f (1, −1, 0) = (2, −3), f (2, 0, 0) = (1, 1) Find f (x1 , x2 , x3 ) E = {(1; 1; 3),  −1; 0), (2; 0; 0)} is a basis of R  (1; α Put [x]E = β  = E −1 [x] γ [x] = α[e1 ] + β[e2 ] + γ[e3 ] ⇒ [f (x)] = α[f (e1 )] + β[f(e2 )] + γ[f (e3 )] α  = f (e1 ) f (e2 ) f (e3 ) β  = f (e1 ) f (e2 ) f (e3 ) E −1 [x] γ  −1   1 x1 x1  + −3x2 + 2x3 = −1 0 x2  = x21 7x22 −4x33 −3 + + x3 0 Therefore, f (x1 , x2 , x3 ) = x21 − 3x22 + 2x33 , x21 + 7x22 − 4x33 [f (x)] = A[x] = fE E −1 [x] (Phan Thi Khanh Van) Linear transformation May 13, 2021 / 27 Example Given a triangle ABC : A(0, 1), B(2, 3), C (4, 2) Find the image of the triangle under the transformation: a) b) c) (Phan Thi Khanh Van) Linear transformation Rotation about the origin 60o counterclockwise Reflection in the line y = −3x Shear transformation with the fixed line x−axis: S(x, y ) = (x + 3y , y ) May 13, 2021 / 27 Rotation about the origin 60o counterclockwise: π3 The standard basis of R2 : E = {i = (1, 0), j = (0, 1)T } The matrix of coordinates of the triangle: [OA OB OC ] = a) √ cos( π3 ) sin( π3 ) − sin( π3 ) f (j) = cos( π3 ) Rotation matrix: f (i) = R = f (E ).E −1 = = √2 − √ 1√− 3 + 32 √ 2√− 3+1 √ − The coordinates of the image: [OA OB OC ] = R.[OA OB OC ] (Phan Thi Khanh Van) Linear transformation May 13, 2021 / 27 b) Reflection about the line y = −3x Choose the directional vector: e1 = (1, −3), and the normal vector e2 = (3, 1) to form a basis of R2 −3 −3 f (e2 ) = −e2 = −1 Reflection matrix: − 54 − 53 T = fE E −1 = − 35 The coordinates of the image: [OA OB OC ] = R.[OA OB OC ] f (e1 ) = e1 == (Phan Thi Khanh Van) = Linear transformation − 53 − 17 − 22 − 45 May 13, 2021 / 27 b) Shear transformation S(x, y ) = (x + 3y , y ) Shear matrix: S = The coordinates of the image: S(X ) = S.X = (Phan Thi Khanh Van) Linear transformation 11 10 May 13, 2021 10 / 27 The transformation matrix Let f : U → V be a linear transformation Let E = {e1 , e2 , , en } and F = {f1 , f2 , , fm } be bases of U and V , respectively An m × n matrix whose the i-th column is the coordinate vector of f (ei ) with respect to the basis F is called the transformation matrix of f with respect to E , F Denote: AEF = [f (e1 )]F [f (e2 )]F [f (en )]F [f (x)]F = AEF [x]E Remark: If f : Rn → Rm : [f (ei )]F = F −1 [f (ei )], then, AEF = F −1 [f (e1 )] F −1 [f (e2 )] F −1 [f (en )] = F −1 f (E ) Remark: If f : Rm → Rn , E , F are two corresponding bases Then, AEF = F −1 f (E ) If f : Rn → Rn , E is a basis of R n Then, AE = E −1 f (E ) (Phan Thi Khanh Van) Linear transformation May 13, 2021 13 / 27 Example Let f be a linear transformation f : R3 → R2 : f (x1 , x2 , x3 ) = (2x1 + x2 + x3 , x1 − x2 − x3 ) Find the transformation matrix of f w.r.t E = {(1, 2, 1), (1, 1, 0), (1, 0, 0)}, F = {(1, 3), (2, 4)}   x 1  1 x2 [f (x)] = A[x] = −1 −1 x3 AEF = F −1 fE = F −1 A.E   1 −1 −12 −6 −3 2 1 = 2 0 = 17 −1 −1 2 0 (Phan Thi Khanh Van) Linear transformation May 13, 2021 14 / 27 Theorem Let E and F be bases of U and V , respectively For any linear transformation f : U → V , there exists a unique matrix AEF satisfying: ∀x ∈ U : [f (x)]F = AEF [x]E For any m × n matrix AEF , there exists a unique linear transformation f : U → V such that: ∀x ∈ U : [f (x)]F = AEF [x]E Conclusion: For any linear transformation f , we can find and only matrix A : f (v ) = Av If f is invertible, then f −1 has the matrix A−1 The product of transformations f1 : f1 (v ) = A1 v and f2 : f2 (v ) = A2 v corresponds to A1 A2 This is where the matrix multiplication came from! (Phan Thi Khanh Van) Linear transformation May 13, 2021 15 / 27 Example Let f be a linear transformation f : R3 → R2 The transformation matrix of f w.r.t E = {(1, 1, −1), (2, 1, 0), (1, 0, 0)}, F = {(1, 2), (−2, 3)} is −1 AEF = Find f (x) We have: [f (x)]F = AEF [x]E = AEF E −1 [x] Then [f (x)] = F [f (x)]F = F AEF E −1 [x] −1    x1 2 −1  −2 = 1 0 x2  x3 −1 0   x1 −5 −5x1 + 5x2 [f (x)] = x2  = 18 −32 −21 18x1 − 32x2 − 21x3 x3 (Phan Thi Khanh Van) Linear transformation May 13, 2021 16 / 27 Example Let f be a linear transformation f : R3 → R3 : f (x1 , x2 , x3 ) = (2x1 + x2 + x3 , x1 − x2 − x3 , 4x1 − x2 − x3 ) Find the matrix of f in E = {(1, 1, 1), (1, 1, 0), (1, 0, 0)}, F = {(1, 1, 2), (1, −2, 1), (0, 0, 3)}   1 [f (x)] = 1 −1 −1 [x] = A.[x] ⇒ f (E ) = A.E −1 −1 We have: AEF = F −1 f (E ) = F −1 A.E  −1    1 1 1 = 1 −2 0 1 −1 −1 1 0 = −1 −1 0 (Phan Thi Khanh Van) Linear transformation May 13, 2021 17 / 27 Change of basis matrix and transformation matrix Let f : U → V be a linear transformation; E and E be bases of U; F and F be bases of V Transformation matrix of f w.r.t E , F is AEF Change of basis matrices: PE →E = E −1 E , and QF →F = F −1 F Then, the transformation matrix of f w.r.t E , F is: AE F = QF−1→F AEF PE →E Remark Let f : U → U be a linear transformation; E and E be bases of U The change of basis matrix: PE →E = E −1 E Then, the transformation matrix of f with respect to E : AE = PE−1→E AE PE →E (AE and AE are similar matrices) (Phan Thi Khanh Van) Linear transformation May 13, 2021 18 / 27 Example Given a linear transformation f : R3 → R3 Matrix of f w.r.t  −1  Find the matrix E = {(1, 1, −1), (2, 1, 0), (1, 0, 0)} is AE = 1 −3 −1 representation of f w.r.t the standard basis (standard matrix representation)     12 −24 −17 E =  1 0 ⇒ A = E AE E −1 =  −13 −9  −3 −1 0 (Phan Thi Khanh Van) Linear transformation May 13, 2021 19 / 27 Example Let f be a linear transformation f : R3 → R3 The matrix  of f in −1 E = {(1; 1; 1), (1; 2; −1), (0; 2; 1)} is AE = 1 0 Find the matrix 1 of f w.r.t F = {(1; −1; −1), (1; 1; 1), (0; 0; 1)} −1 AF =F −1 E AE E  F , where:   1 1 E = 1 2, F = −1 0 −1 −1 1 (Phan Thi Khanh Van) Linear transformation May 13, 2021 20 / 27 Kernel and Image of a linear transformation Let f be a linear transformation f : U → V Kernel of the linear transformation f : Kerf = {x ∈ U|f (x) = 0} Image of the linear transformation f : Imf = {y ∈ V |∃x ∈ U : y = f (x)} Theorem Let f be a linear transformation f : U → V Kerf is a subspace of U Imf is a subspace of V dim(Imf ) + dim(Kerf ) = dim(U) Proposition Let f : U → V be a linear transformation If U = span{e1 , e2 , en }, then Imf = span{f (e1 ), f (e2 ) f (en )} (Phan Thi Khanh Van) Linear transformation May 13, 2021 21 / 27 Example Let f be a linear transformation f : R3 → R3 : f (x1 , x2 , x3 ) = (2x1 + x2 + x3 , x1 − x2 − x3 , 4x1 − x2 − x3 ) Find the dimensions, bases of Kerf , Imf Find a basis of Kerf : x ∈ Kerf ⇔ [f (x)]  =0⇔  A[x] =  3 1 −1 −1  −1 −1  ∼  −1 −1  ∼ 1 −1 −1 0 3 General solution: x = (0, −m, m) One basis for Kerf : {(0, −1, 1)}, dim(Kerf ) = Find a basis of Imf : Choose a basis of R3 : {(1, 0, 0), (0, 1, 0), (0, 0, 1)} ⇒ A spanning set of Imf : {f (1, 0, 0), f (0, 1, 0), f (0, 0, 1)} = {(2, 1, 4),  (1, −1, −1),(1, −1,  −1)}  1 1 We have 1 −1 −1 ∼ 0 −3 −3 −1 −1 −3 −3 One basis of Imf : {(2, 1, 4), (1, −1, −1)}, dim(Imf ) = (Phan Thi Khanh Van) Linear transformation May 13, 2021 22 / 27 Example Given f : R3 → R3 : f (1, 1, 1) = (1, 2, 1), f (1, 1, 2) = (2, 1, −1), f (1, 2, 1) = (5, 4, −1) Find the dimensions, bases of Kerf , Imf We have that E = {e1 = (1, 1, 1), e2 = (1, 1, 2), e3 = (1, 2, 1)} is a basis of R3 Then {f (e1 ),  f (e2 ),f (e3 )} is a spanning   set of Imf  5 2  ∼ 0 −3 −6 ∼ 0 −3 −6 −1 −1 −3 −6 0 One basis for Imf : {(1, 2, 1), (2, 1, −1)}, dim(Imf ) = (Phan Thi Khanh Van) Linear transformation May 13, 2021 23 / 27 Find a basis of Kerf : We have: x ∈ Kerf ⇔ [f (x)] = A[x] = ⇔ f (E )E −1 [x] =   −1     1 x1 −4 −1   1 2 x2  = ⇔  ⇔ 2 1 −1 −1 x3 −2 −2   −1 ∼  12 −3  12 −3 −1 ∼ −1 General solution: x = (2m, m, 4m) One basis for Kerf : {(2, 1, 4)}, dim(Ker (f )) = (Phan Thi Khanh Van) Linear transformation May 13, 2021 24 / 27 Exercises The projection f in Oxyz onto the plane P : x − y + 2z = is a linear transformation Find the dimensions, bases of Imf , Kerf The reflection f in Oxyz with respect to the plane x + y + z = is a linear transformation Find the dimensions, bases of Imf , Kerf The rotation f in Oxyz with respect to z-axis by 45o counterclockwise is a linear transformation Find the dimensions, bases of Imf , Kerf (Phan Thi Khanh Van) Linear transformation May 13, 2021 25 / 27 Example Given f : R3 → R3 The matrix of f in the   basis −1  Find the E = {(1, 1, 1), (1, 1, 0), (1, 0, 0)} is AE = 1 −3 −1 dimensions, bases of Kerf , Imf   −8 A = E AE E −1 = 7 −6 2 −4   6x1 − 8x2 + 6x3 Hence, [f (x)] = A.[x] = 7x1 − 6x2 + 2x3  Continue!!! 3x1 − 4x2 + 3x3 (Phan Thi Khanh Van) Linear transformation May 13, 2021 26 / 27 Thank you for your attention! (Phan Thi Khanh Van) Linear transformation May 13, 2021 27 / 27 ... of contents Linear transformation The transformation matrix Change of basis matrix and transformation matrix Kernel and Image of a linear transformation (Phan Thi Khanh Van) Linear transformation. .. (Phan Thi Khanh Van) Linear transformation May 13, 2021 20 / 27 Kernel and Image of a linear transformation Let f be a linear transformation f : U → V Kernel of the linear transformation f : Kerf... −1     −3 1 2  [(2, 3, ? ?4) ]E = 1 −1 0   = − 13 23 0 ? ?4 Or (2, 3, ? ?4) = − 43 (1, 1, 3) − 13 (1, −1, 0) + 23 (2, 0, 0) Consequently, 23 [f (2, 3, ? ?4) ] = − 43 [f (1, 1, 3)] − 13 [f (1,