Transformasi Linier-14

Garis-garis Besar Perkuliahan1. Sistem Persamaan Linier 2. Matriks dan Operasinya 3. Determinan dan Sifat-sifatnya UTS 4. Ruang Vektor 5. Basis dan Dimensi 6. Ruang Hasil Kali Dalam 7. Transformasi Linier UTS 8. Matriks Transformasi Linier 9. Keserupaan 10. Nilai Eigen dan Vektor Eigen 11. Diagonalisasi 1 LINEAR TRANSFORMATIONS Content  6.1 Linear Transformations  6.2 Matrix Transformations  6.3 Kernel and Range  6.4 Transformations and Systems of Linear Equations  6.5 Matrix Representations of Linear Transformations 3 6.1 Linear Transformation . Let u and v be vectors in U and let c be a scalar. A transformation T : U → V is said to be linear if T(u + v) = T(u) + T(v) T(c u) = c T(u) U V 5 .Definition  Let U and V be vector spaces. y1) and (x2. y1) + T(x2.  T((x1.Example 1 Prove that the following transformation T: R2 → R2 is linear.  T is linear. 6 . y1) + (x2. y1) Thus T preserves scalar multiplication. 3x1) + (x2– y2. 3x1) = cT(x1. y2)) = T(x1 + x2. y2) (definition of T) Thus T preserves vector addition. y2) be vectors in R2. 3x1 + 3x2) (definition of T) = (x1 – y1. y) = (x – y. 3x) Solution  Let (x1. 3x2) (vector addition) = T(x1.  T(c(x1. 3cx1) = c(x1 – y1. y1)) = T(cx1. y1 + y2) (vector addition) = (x1 + x2 – y1 – y2. T(x. cy1) = (cx1 – cy1. z) Solution Let (x1. y. z1) + (x2. y1 + y2. y2 . z1) + (x2. z1) + T(x2. z1) + (x2 y2 . z1) + T(x2. y1. y1. z1 + z2) = ((x1 + x2 )( y1 + y2). y2 . z2)) = T(x1 + x2. z) = (xy. y1. z2)) Since vector addition is not preserved. y2 . T(x. y1. z1) and (x2. z2) = (x1 y1 + x2 y2. y2 . z1 + z2) = (x1 y1 + x2 y2 + x1 y2 + x2 y1.Example 2 Show that the following transformation T: R3 → R2 is not linear. z1 + z2) and T(x1. 7 . T((x1. z2))  T(x1. T is not linear. y2 . y1. z2) be vectors in R3. z2)) = (x1y1. in general T((x1. z1 + z2) Thus. T((ax2 + bx + c) + (px2 + qx + r)) = T((a + p)x2 + (b + q)x + (c + r)) By vector addition = (a + p + b + q) x + (c + r) By definition of T = (a + b) x + c + (p + q) x + r =T(ax2 + bx + c) + T(px2 + qx + r)) By definition of T Thus T preserves addition. Show that the following transformation T: P2 → P1 is linear.Example 3 Let Pn be the vector space of real polynomial functions of degree  n. T(ax2 + bx + c) = (a + b) x + c Solution Let ax2 + bx + c and px2 + qx + r be arbitrary elements of P2. 8 . Therefore.Example 3 Let k be a scalar. T is a linear transformation. 9 . T(k(ax2 + bx + c)) = T(kax2 + kbx + kc) By multiplication = (ka + kb) x + kc By definition of T = k((a + b) x + c) =kT(ax2 + bx + c) By definition of T Hence T also preserves scalar multiplication. D(4x3  3x2  2x  1)  12x2  6x  2. D can be interpreted as a mapping of Pn into itself. 10 . For example.Example 4 Let D be the operation of taking the derivative. D maps the element 4 x 3  3x 2  2 x  1 of P3 into 12 x 2  6 x  2 of P3 . Then D( f  g )  Df  Dg D(cf )  cD( f ) The derivation thus preserves addition and scalar multiplication of functions. Let f and g be elements of Pn and c be a scalar. It is a linear transformation. 11 . is linear. Such a linear transformation is called a matrix transformation. and c be a scalar.1 Let A be an m  n matrix. T (x  y )  A(x  y )  Ax  Ay  T ( x)  T ( y ) Addition is preserved. Thus the transformation is linear. Proof Let x and y be vectors in Rn. The transformation T: Rn → Rm. T (cx)  A(cx)  cAx Scalar multiplication is  cT (x) preserved. interpreted as a column matrix. Let x be a vector in Rn. defined by T(x) = Ax.Theorem 6.1. written as column matrices.  1  1 A  0 2  x   5  1  1 3 Solution Since A is a 3  2 matrix. We get  1  1  x  y   T      0 2      2 y  x x   y    1 3  y   x  3 y    Letting x = 5. Example 5 Consider the linear transformation T defined by the following 3  2 matrix A. y = –1 we get  6   5    T      2      2 1   12 . it defines a linear transformation T: R2→R3. and use this result to determine the image of the given vector x. Find the image of an arbitrary vector under T. Example 5 R2 R3  x  x y    2y   y   x 3 y    5 6 1    2    2 T→ 13 . T(x) = T2(T1(x)).Composition of Transformations U V W The composite transformation T = T2 ○ T1 of T1 and T2. 14 .  3)  (12. Solution We get T2  T1 ( x. Determine the image of (2.  x  y) The image of (2. x  y)  (6 x. y) = (2x. –3). –y). 1). x + y) and T2(x.Example 6 Find the T2 T1 of the transformations T1(x.  3) is T2  T1 (2. 15 . y)  T2 (T1 ( x. y) = (3x. y))  T2 (3x. Proof Let U.Theorem 6.1. Let u and v be vectors in U and c be a scalar. If T1 (x)  A1x and T2 (x)  A2 x. then T2 T1 (x)  A2 A1x 16 . V and W be vector spaces and T1: U → V. We use the linearity of T1 and T2 to get T2  T1 (u  v)  T2 (T1 (u  v))  T2 (T1 (u)  T1 ( v))  T2 (T1 (u))  T2 (T1 ( v))  T2  T1 (u)  T2  T1 ( v) T2  T1 (cu)  T2 (T1 (cu))  T2 (cT1 (u))  cT2 (T1 (u))  cT2  T1 (u) Thus T2 ○ T1 is linear. T2: V → W be linear transformations.2 The composite of two linear transformation is itself a linear transformation. 2) (1.0) 17 .3) Scale(2.Non-commutative Composition Scale then Translate: p' = T(S(p)) = (T○S)(p) (5.2) Scale(2.1) (0.0) Translate then Scale: p' = S(T(p)) = (S○T)(p) (8.1) (3.1) (0.4) Translate(3.0) (0.2) Translate(3.1) (6.1) (4.1) (1.2) (2.2) (3. 1  A1   3 0  1 A2   1  2  x   4 4 2 0 4 0 2 Solution T is defined by the product matrix A2A1. We get A2 A1   1  2  3 0  1   5  4  1 4 0 4 2 0  12 0  4 Thus  1  23 T (x)    5  4  1  4     12 0  4 2  4   18 . Find the image of the vector x under T. Let T = T2 ○T1.Example 7 Let T1(x) = A1x and T2(x) = A2x be defined by the following matrices A1 and A2. 2 Matrix Transformation .6. Rotation about the Origin 20 . 21 .  '   T          (a linear transformation) y    y    sin  c os   y  Note tha t  is positive for counterclockwise rotation. negative for a clockwise rotation.Matrix Representation of Rotation x '  r cos(   )  r cos  cos  r sin  sin   x cos  y sin  y '  r sin(   )  r sin  cos  r cos  sin   y cos  x sin   x'    x   cos  sin    x  Thus. Since T is linear. en} for Rn as the set of column matrices. T(v) = T(v1 e1 + … + vn en ) = v1 T(e1) + … + vn T(en) = [ T(e1) … T(en) ] [v1 … vn] t The linear transformation T is then defined by the matrix A = [ T(e1) … T(en) ] A is called the standard matrix of T. e2.Matrix Representation Let T: Rn → Rm be a linear transformation. 22 . Consider the standard basis S = {e1. …. Then for any vector v in Rn can be uniquely written as v = v1e1 + v2e2 + … + vnen. Example 1 2x  y Determine the standard matrix of T       x .      y   3 y   Solution We find the effect of T on the standard basis.  1   2  0  1  T   , T        0    0     1   3 These matrices are the columns of the standard matrix A. A  2 1 0 3 Hence T can be written as a matrix transformation:  x  T      2 1   x   y   0 3  y  23 Dilation and Contraction  x   x  Let T : R  R be defined by T      r   , where r is a scalar. 2 2  y   y  If r > 1, then T moves the points outward from the origin; it is called a dilation of factor r. If 0 < r < 1, then T moves points inward toward the origin; it is called a contraction of factor r. Defined by  r 0 0 r  r > 1, dilation 0 < r < 1, contraction 24 Reflection  x   x  Let T : R  R be defined by T        that maps a point into its 2 2   y   y  mirror image in the x axis. T is called a reflection.  1 0 Defined by   0  1 reflection in x axis 25 Example 2 Determine the matrix that describes a reflection in the x axis. Find the image of the point 4 under this sequence of mappings.   0 1  sin(π / 2 ) cos(  / 2)   0 3 The composite mapping is thus defined by the single product matrix.  . rotation. and dilation are. 1  Solution The matrices that define the reflection. following by a rotation through /2. followed by a dilation of factor 3. respectively.  1 0 cos( / 2)  sin( / 2)   3 0   .  3 0 cos( / 2)  sin( / 2)   1 0  3 0 0 1  1 0      0 3  sin( / 2) cos( / 2)  0 1 0 3  1 0 0 1  0 3   4 0 3  4  3  3 0 The image of the point   is       . 1   3 0   1 12 26 . (d) lines through the origin into lines through the origin. (c) parallel lines into parallel lines. Theorem 6.3 An invertible linear transformation T maps (a) lines into lines. 27 . (b) segments of lines into segments of lines.Transformations Defined by Invertible Matrices An invertible transformation is a mapping T: Rn → Rn defined by T(u) = Au where A is a invertible matrix. It is convenient to use the notation u Au to discuss the images of specific points. Solution The unit square in R2 is the square whose vertices are.Example 3 Consider the transformation T: R2 → R2 defined by A  4 2. R . 2 3 Determine the image of the unit square under this transformation. the points 1  1 0 0 P  . Q . respectively. we get P P Q Q R R O O 1   4 1 6 0  2 0  0                 0   2 1 5   1 3  0  0  28 . O  0 1 1  0 Let us compute the images of these points under the transformation. Multiplying each point by the matrix. A  (4  3)  (2  2)  8  0. The matrix A is invertible. 29 .Example 3 Further. thus line segments are mapped into line segments. We get OP OP PQ PQ QR QR OR OR The square PQRO is transformed into the parallelogram P’Q’R’O.     d   .Example 4 Let T be the orthogonal transformation defined by the following orthogonal matrix A. and distances for the vectors u and v.  1    17.  2 3     2 7  2 d  . Show that T preserve norms. u   .  0 4     2   2     30 . The angle between u and v is equal to the angle between T (u) and T ( v). v    2 2  2 2  0  4 Solution  7   2  2 T (u)  Au    and T ( v)  Av   1    2     2 u  T (u)  2 and v  T ( v)  5. This angle is 53.13.  1 1 2 3 A   1 1 . 31 .Translations A translation is a transformation T: Rn → Rn defined by T(u) = u + v where v is a fixed vector. T will slide this line into another line. 2   y    y  1  Solution The equation y = 2x + 3 describes points on the line of slope 2 and y intercept 3.Example 4 Find the equation of the image of the line y = 2x + 3 under the translation  x  x T          . We want to find the equation of this image line. We get   x    x   2   x   2   x  2   x  T                  y    y   1   2 x  3  1   2 x  4   y  We see that y = 2x for the image point. Thus the equation of the image line is y = 2x. 32 . 3 Kernel and Range .6. 1 Let T: U → V be a linear transformation.Theorem 6. Since 0u = 0U and 0v = 0V and T is linear. a linear transformation maps a zero vector into zero vector.3. Proof Let u be a vector in U and let T(u) = v. Then T(0U) = 0V That is. Let 0V and 0U be the zero vectors of U and V. we get T(0U) = T(0u) = 0T(u) = 0v = 0V 34 . Let 0 be the zero scalar. The kernel is denoted ker(T). The set of vectors in V that are the images of vectors in U is called image (range) of T. The image is denoted im(T). 35 . The set of vectors in U that are mapped into the zero vector of V is called the kernel of T.Definition Let T: U → V be a linear transformation. Theorem 6. (a) The kernel of T is a subspace of U.3. (b) The image of T is a subspace of V. 36 .2 Let T: U → V be a linear transformation. y = 0 Thus ker(T) is the set of all vectors of the form (0. 0) Solution Kernel: ker(T) is the subset that is mapped into (0. im(T) = {(x. z) = (x. 0). z).Example 1 Find the kernel and image of the linear operator T(x. 0). 0. z) = (x. 0). 0) = (0. 37 . if x = 0. ker(T) is the set of all vectors that lie on the z axis. 0. T(x. y. y. 0. Image: The image of T is the set of all vectors of the form (x. We write this as ker(T) = {(0. 0. im(T) is the set of all vectors that lie in the xy- plane. y. y. 0)} Geometrically. z)} Geometrically. y. y. Example 1 Projection T(x. z) = (x. y. y. 0) 38 . span the image of A. namely. T(en).Theorem 6. Express u in terms of the standard basis of Rn. …. the column vectors of A. T(e1).3. 39 . There exist a vector u such that T(u) = v. is spanned by the column vectors of A. u = a1e1 + … + anen Thus v = T(a1e1 + … + anen) = a1T(e1) + … + anT(en) Therefore. defined by T(u) = Au.3 The image of T: Rn → Rm. Proof Let v be a vector in the image. r)} Thus ker(T) is a one-dimensional subspace of R3 with basis (-5. 1). ker(T) = {(-5r.  1 2 3 A  0  1 1 Solution  1 1 4   A is a 3  3 matrix. r. r. 40 . 1.Example 2 Determine the kernel and the image of the transformation defined by the following matrix. Thus A defines a linear operator T: R3→R3 with T(x) = A(x). Kernel:  1 2 3  x1  0 x1  2 x2  3 x3  0 x1  5r 0  1 1  x2   0   x2  x3  0  x2  r  1 1 4   x  0  x1  x2  4 x3  0 x3  r   3    The kernel is the set of vectors of the form (-5r. r). 1. 41 . 1. s + t)} Thus im(T) is a two-dimensional subspace of R3 with basis {(1. (0. t. 0. 0. s(1. 1) Thus the image of T is im(T) = {(s. 1. 1 0 1  1 0 1  1 0 1 1 0 1 2  1 1   0  1  1  0 1 1  0 1 1 3 1 4 0 1 1  0 1 1 0 0 0         The vectors (1. An arbitrary vector in the image will be a linear combination of these vectors.0.1). 1) span the image of T. Example 2 Range: The image is spanned by the column vectors of A. 1) + t(0. 1) and (0.1)}. 3. 42 . implying that dim im(T) = 2. Therefore. dim domain(T) = 3. The domain of T is R3. dim ker(T) = 1. The nonzero row vectors are linearly independent.Theorem 6.4 Let T: U → V be a linear transformation. Thus rank(A) = 2. Then dim ker(T) + dim im(T) = dim domain(T) Example 3 Find the dimensions of the kernel and image of the linear transformation T defined by the matrix 1 0 3    A  0 1 5    Solution 0 0 0 Observe that A is in row echelon form. One-to-one Transformation A transformation T is said to be one-to-one if each element in the image of T corresponds to just one element in the domain of T. one-to-one not one-to-one 43 . This means that T is one-to-one if T(u) = T(v) implies that u = v. assume that the kernel is the zero vector space. The kernel consists of all vectors that are mapped into zero. (2) Conversely.3. u – v = 0 or u = v. Thus T is one-to-one. Since T is one-to-one. However we know that the zero vector must be in the kernel. Therefore.Theorem 6. But the kernel only has the zero vector. Proof (1) Assume that T is one-to-one.5 A linear transformation T is one-to-one if and only if the kernel is the zero vector space. the kernel must consist of a single vector. T(u) – T(v) = 0 ⇒ T(u – v) = 0 Thus u – v is in the kernel. Let u and v be vectors such that T(u) = T(v). Thus the kernel is the zero vector space. 44 . assume that A is invertible.3. Thus ker(T) = 0. defined by T(x) = Ax. This implies that the rank of A is n and that its n column vectors are linearly independent. The rank/nullity theorem implies that dim im(T) = n.6 The transformation T: Rn→Rn.Theorem 6. (2) Conversely. Thus the n columns of A are linearly independent. Proof (1) Let T be one-to-one. and |A|  0. 45 . is one- to-one if and only if A is invertible. Thus T is one- to-one. implying that the rank(A) is n. This means that dim im(T) = n. The rank (nullity) theorem now implies that ker(T) = 0. Matrix B is then invertible. so dim domain(TA) = 4.3. TB is one-to-one. 1 2 5 7  2 0 1     A  0 1 9 8 B  3 4 2      0 0 1 3  0 7 5 Solution (a) Since the rows of A are linearly independent. Therefore.Example 4 Determine whether the linear transformations TA and TB defined by the following matrices are one-to-one.6. (b) It can be shown that |B| = – 9  0. ker(TA)  0 and by Theorem 6. rank(A) = 3. dim ker(TA) = 1. By the rank/nullity theorem.3. TA is not one- to-one. by Theorem 6. 46 .5. Thus. The domain of T is R4. implying that dim im(TA) = 3. ….Theorem 6. If {u1. …. 47 . an. …. T(un)} is linearly independent. Therefore. then {T(u1). T(un)} is linearly independent in V. un} is linearly independent in U.7 Let T: U → V be a one-to-one linear transformation. Proof Consider the equation a1T(u1 ) + + anT(un ) = 0 (1) for scalars a1. a1u1 + + anun = 0 By linear independence of {u1. thus the kernel is the zero vector space. an = 0. Since T is linear. . Returning to equation (1). …. un}. we have a1 = 0. this is equivalent to T(a1u1 + + anun ) = 0 T is one-to-one.3. …. this means that {T(u1). 6.4 Transformations and Systems of Linear Equations . Relationships 49 . The set of solutions is the set of vectors in Rn that are mapped by T into the zero vector.Theorem 6. The set of solutions is the kernel of the transformation and is thus a subspace. 50 . Proof Let T be the linear transformation of Rn into Rm defined by A.4. Ax = 0.1 The set of solutions to a homogeneous system of m linear equations in n variables. is a subspace of Rn. Interpret the set of solutions as a subspace. Sketch the subspace of solutions.Example 1 Solve the following homogeneous system of linear equations. x1  2 x2  3 x3  0  x2  x3  0 Solution x1  x2  4 x3  0 Using Gauss-Jordan elimination we get 1 2 3 0 1 2 3 0 1 2 3 0 1 0 5 0         0 1 1 0  0 1 1 0  0 1 1 0  0 1 1 0          1 1 4 0 0 1 1 0 0 1 1 0 0 0 0 0 Thus x1  5r x1  5x3  0 x1  5x3   x2  r x2  x3  0 x2  x3 x2  r 51 . with basis ( –5.Example 1 The solutions are vectors of the form ( –5r.    1 1 4  52 . r) These vectors form a one- dimensional subspace of R3. This subspace is the kernel of the transformation defined by the matrix of coefficients of the system 1 2 3   0 1 1 . 1). 1. r. Let x1 be a particular solution. 53 .Theorem 6. where z is a vector in the kernel of the transformation T defined by A. Every other solution can be written in the form x = z + x1.4.2 Let Ax = y be a nonhomogeneous system of m linear equations in n variables. Let x be an arbitrary solution.2 Let x1 be a solution. Equating Ax1 and Ax. That is. Thus Ax = y.Proof Theorem 6. We can write x = x1 + z 54 . Then Ax1 = y. x – x1 = z.4. we get Ax1 = Ax  Ax – Ax1 = 0  A(x – x1) = 0  T(x – x1) = 0 Thus x – x1 is in the kernel of T. call it z. Example 2 Solve the following system of linear equations. Sketch the set of solutions. x1  2x2  3x3  11 x2  x3  2 x1  x2  4x3  9 Solution Using Gauss-Jordan elimination. we get 1 2 3 11 1 2 3 11 1 2 3 11 1 0 5 7         0 1 1 2    0 1 1 2    0 1 1 2  0 1 1 2           1 1 4 9 0 1 1 2 0 1 1 2 0 0 0 0 Thus x1  5r  7 x1  5x3  7 x1  5x3  7   x2  r  2 x2  x3  2 x2  x3  2 x2  r The solutions are vectors of the form 55 . 1. 2. 1. 1) + (7. 2. 56 . r) = r(–5. namely the line defined by the vector ( –5. r + 2.Example 2 ( –5r + 7. 1). 0). 0) arbitrary solution element of kernel a particular solution to Ax = y to Ax = y The set of solutions can be represented geometrically by sliding the kernel. in the direction and distance defined by the vector (7. Let T be the linear transformation defined by A. x3. Let x1. x2.Many Systems Consider a number of linear systems. …. Ax = y3. be particular solutions to these systems. …. all having the same matrix of coefficients A. x3. x2. Ax = y1. 57 . Ax = y2. Then the sets of solutions to these systems are ker(T) + x1. ker(T) + x3 These sets are “parallel” sets. each being the ker(T) translated by the amounts x1. ker(T) + x2. Example 3 Analyze the solutions to the following system of equations. x1  2 x2  3 x3  x4  1 2 x1  3 x2  2 x3  x4  4 3 x1  5 x2  5 x3 5 x1  x2  x3  2 x4  3 Solution Solve using Gauss-Jordan elimination. 1 2 3 1 1 1 2 3 1 1 1 0  5  5 5       2 3 2 1 4 0 1 4 3 2 0 1 4 3 2      3 5 5 0 5  0 1 4 3 2 0 0 0 0 0        1 1 1 2 3 0 1 4 3 2 0 0 0 0 0 We get x1  5 x3  5 x4  5 x2  4 x3  3x4  2 Express the leading variables in terms of the free variables. x1  5 x3  5 x4  5 x2  4 x3  3x4  2 58 . s) Separate the parts of this vector as follows. 1. 3. 1). 0). s ) Arbitrary solution to Ax = y  r (5. 0. 1)  (5. 0)  s (5. 0) kernel of mapping a particular solution defined by A to Ax = y Observe that the kernel is a two-dimensional subspace of R4 with basis (5. 4. The set of solutions to the given system is this plane translated in a manner described by the vector (5. 59 . (5. r .Example 3 The arbitrary solution is (5r  5s  5. 0. 0. 0). 4. 0. 4r  3s  2. It is a plane through the origin. 2. 4r  3s  2. 1. 3. r . 2. (5r  5s  5. 6.5 Matrix Representations of Linear Transformations . …. T(un). defining a linear transformation in a basis defines it on the whole domain. un} be a basis for U.1 Let T: U  V be a linear transformation. …. T is defined by its effect on the base vectors. 61 .5. T(un)}. Let {u1. Thus. namely by T(u1). The image of T is spanned by {T(u1).Theorem 6. …. 3). 0) – 2(0. 1. 0) + 3(0. 0. T(1. -2. T(0. express the vector (1. -2. Find T(1. -1) – 2(2. 0) = (8. 0) – 2T(0. –3) 62 . To find T(1. 1) = (3.Example 1 Consider the linear transformation T: R3 → R2 defined as following basis vectors of R3. T(0. 1)) = 1T(1. 3). 0. 0. 0. 1. 1)) = 1(3. 1. 0) + 3T(0. 1). 0. -2. 0) = (3. 3) as a linear combination of the basis vectors and use the linearity of T. 0) Solution Since T is defined on basis vectors for R3. 1) + 3(3. 0) = (2. -2. 3) = T(1(1. it is defined on the whole space. T(1. 0. -1). and T: U → V a linear transformation. 63 .Theorem 6. A is called the matrix A representation of T (or matrix of T) with respect to the bases B and B’. ….2 Let U and V be vector spaces with bases B = {u1. …. vm}. un} and B’ = {v1. respectively. If u is a vector in U with image T(u) having coordinate matrices [u]B and [T(u)]B’. then T  u  B ' =Au B where A = éëéëT ( u 1 )ùû éT ( u )ù ù ë n ûB' û T B' u  T u  The matrix A thus defines a transformation of coordinate vectors of U in the “same way” as T transforms the u B  T  u   B ' vectors of U.5. u2.Example 2 Let T: U → V be a linear transformation. ê ú. and T(u3) are é ù é ù é ù ê 2 3 ú. and ê 1 ú êë -1 úû êë 2 úû êë -4 úû These vectors make up the columns of the matrix of T. T(u1) = 2v1 – v2 T(u2) = 3v1 + 2v2 T(u3) = v1 – 4v2 Find the matrix representation of T with respect to these bases and use this matrix to determine the image of the vector u = 3u1 + 2u2 – u3. u3} and B’ = {v1. T(u2). é ù 2 3 1 A =ê ú êë -1 2 -4 úû 64 . Solution The coordinate matrices of T(u1). v2} of U and V as follows. T is defined relative to bases B = {u1. êë 5 úû 65 . Thus T(u) = 11v1 + 5v2.Example 2 Let us find the image of the vector u = 3u1 + 2u2 – u3 using this matrix. ê -1 ú êë úû é ù é ù 3 ú é 11 ù We get 2 3 1 úê Aéëuùû = ê ê 2 ú=ê ú. B êë -1 2 -4 úûê -1 ú êë 5 úû êë úû é ù T(u) has coordinate vector ê 11 ú. The coordinate matrix of u is é ù ê 3 ú ê ú ê 2 ú. 0). 2) = 3u’1 + 3u’2 é ùé 1 ù é ù The coordinate vectors of T(u1). and T(u3) are thus ê 2 ú. 4). 0) + 0(0. 0) = 2(1. 6) = 3(1. 0) + 4(0. 1. 2) = 1u’1 + 4u’2 T(u3) =T(1. 2) Use this matrix to find the image of the vector u = (2. 3) and u’1 = (1. u3} and {u’1. 2) = 2u’1 + 0u’2 T(u2) =T(0. 0). y. u3 = (1. 2. Find the matrix of T with respect to the bases {u1. 4) = (1. where u1 = (1. 0) + 3(0. 1. u’2 = (0. 3) = (3. u2 = (0. 2. u’2} for R3 and R2.Example 3 Let T: R3 → R2 be the linear transformation defined by T(x. u2.ê ú. andê 3 ú êë 0 úûêë 4 úû êë 3 úû These vectors form the é ù 2 1 3 A =ê ú columns of the matrix of T: êë 0 4 3 úû 66 . 1. 1. z) = (x + y. T(u1) = T(1. Solution We find the effect of T on the basis vectors of R3. 5). 2z). 3. 8) = 1(1. 0) = (2. T(u2). 4) – (1. 0) + 5(0.Example 3 Let us now use A to find the images of the vectors u = (2. 0) + (0. 5) = (2 + 3. 10) 67 . 2z). 5) = 3(1. z) = (x + y. 2. 1. We can check this result directly using the definition T(x. 1. 3) = 3u1 + 2u2 + (– 1)u3  3   The coordinate matrix of u is thus  2 . y. this gives T(u) = T(2. 3. 5). We determine the coordinate vector of u. T(u) = 5u’1 + 5u’2 = 5(1. It can be shown that u = (2. 3. 5). 3. 10). 3. 2  5) = (5. 2) = (5. For u = (2.   The coordinate matrix of T(u) is  3   1  2 1 3    5     2     0 4 3    5   1 Therefore. and T(1) are thus ê 1 ú. where u1 = x2. u3} and {u’1. u’2} of P2 and P1. T(x). defined by T(ax2 + bx + c) = (a + b)x – c. u’2 = 1 Use this matrix to find the image of u = 3x2 + 2x – 1 . u2 = x. u3 = 1 and u’1 = x. and ê 0 ú êë 0 úû êë 0 úû êë -1 úû The matrix of T is thus é ù 1 1 0 A =ê ú êë 0 0 -1 úû 68 .Example 4 Consider the linear transformation T: P2 → P1. Find the matrix of T with respect to the basis {u1. T(u1) = T(x2) = x = 1x + 0(1) = 1u’1 + 0u’2 T(u2) = T(x) = x = 1x + 0(1) = 1u’1 + 0u’2 T(u3) = T(1) = –1 = 0x + (-1)(1) = 0u’1 + (–1)u’2 é ù é ù é ù The coordinate vectors of T(x2). u2. ê 1 ú. Solution Consider the effect of T on each basis vectors of P2. The coordinate matrix of u relative to the basis {x2.Example 4 Let us now use T to find the image of u = 3x2 + 2x – 1. x. 69 . 1} is é ù ê 3 ú ê 2 ú ê -1 ú êë úû We get é ù é ùê 3 ú é 5 ù ê 1 1 0 úê 2 ú = ê ú êë 0 0 -1 úûê -1 ú êë 1 úû êë úû Therefore T(u) = 5u’1 + 1u’2 = 5x + 1. Example 5 Let D  d be the operation of taking the derivative. Find the dx matrix of D with respect to the basis {x2. Solution We examine the effect of D on the basis vectors. D(x 2 )  2x  0x 2  2x  1(1) D(x )  1  0x 2  0x  0(1) D(1)  0  0x 2  0x  0(1) The matrix of D is thus é ù ê 0 0 0 ú A =ê 2 0 0 ú ê 0 1 0 ú êë úû 70 . 1} of P2. x. 3 Let U be a vector space with B and B’. we know that a = Pa’ and Aa = P(A’a’) The second equation may be written P–1Aa = A’a’ Substituting a = Pa’ into the equation gives P–1APa’ = A’a’ The effect of the matrices P–1AP and A’ as transformation on an arbitrary coordinate vector a’ is the same. having matrix A with respect to the first basis and A’ with respect to the second basis. Let P be the transition matrix from B’ to B. then A’ = P–1AP Proof Consider a vector u in U. 71 .Theorem 6. Thus these matrices are equal. Since P is the transition matrix from B’ to B.5. If T is a linear transformation on U. Let its coordinate matrices relative to B and B’ be a and a’. The coordinate matrices of T(u) are Aa and A’a’. Solution The effect of T on the standard basis vectors is T(1. -1) = 1(1. 0) = (2. 1) = 0(1. 0) + 3(0. 1)} on R2. y) = (2x. 1) = 2(1. 1) The transition matrix is é ù -2 1 ú P =ê êë 3 -1 úû Therefore -1 é ù é ùé ù é ùé ùé ù é ù -1 -2 1 2 0 -2 1 1 1 2 0 -2 1 ú=ê -3 2 ú A¢ = P AP = ê ú ê úê ú=ê úê úê êë 3 -1 úû êë 1 1 úûêë 3 -1 úû êë 3 2 úûêë 1 1 úûêë 3 -1 úû êë -10 6 úû 72 . -1)}. 1) (1. 3) = -2(1.Example 6 Consider the linear transformation T(x. x + y) on R2. 0) + 1(0. (0. 1) = (0. 0). Find the matrix of T with respect to the basis B = {(1. 0) – 1(0. (-2. Determine the matrix A’ with respect to B’ = {(-2. 3). 1) T(0. (1. 0) + 1(0. 1) é ù 2 0 The matrix of T relative to the standard basis is A = ê ú êë 1 1 úû Write the vectors of B’ in terms of those of B.