la08-lt - LINEAR ALGEBRA W W L CHEN c W W L Chen 1997 2005...

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Unformatted text preview: LINEAR ALGEBRA W W L CHEN c W W L Chen, 1997, 2005. This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 8 LINEAR TRANSFORMATIONS 8.1. Euclidean Linear Transformations By a transformation from Rn into Rm , we mean a function of the type T : Rn → Rm , with domain Rn and codomain Rm . For every vector x ∈ Rn , the vector T (x) ∈ Rm is called the image of x under the transformation T , and the set R(T ) = {T (x) : x ∈ Rn }, of all images under T , is called the range of the transformation T . Remark. For our convenience later, we have chosen to use R(T ) instead of the usual T (Rn ) to denote the range of the transformation T . For every x = (x1 , . . . , xn ) ∈ Rn , we can write T (x) = T (x1 , . . . , xn ) = (y1 , . . . , ym ). Here, for every i = 1, . . . , m, we have (1) yi = Ti (x1 , . . . , xn ), where Ti : Rn → R is a real valued function. Definition. A transformation T : Rn → Rm is called a linear transformation if there exists a real matrix a11 . . A= . am1 Chapter 8 : Linear Transformations ... a1n . . . . . . amn page 1 of 34 c Linear Algebra W W L Chen, 1997, 2005 such that for every x = (x1 , . . . , xn ) ∈ Rn , we have T (x1 , . . . , xn ) = (y1 , . . . , ym ), where y1 = a11 x1 + . . . + a1n xn , . . . ym = am1 x1 + . . . + amn xn , or, in matrix notation, y1 a11 . . . . = . . am1 ym (2) ... x1 a1n . . . . . . . ... amn xn The matrix A is called the standard matrix for the linear transformation T . Remarks. (1) In other words, a transformation T : Rn → Rm is linear if the equation (1) for every i = 1, . . . , m is linear. (2) If we write x ∈ Rn and y ∈ Rm as column matrices, then (2) can be written in the form y = Ax, and so the linear transformation T can be interpreted as multiplication of x ∈ Rn by the standard matrix A. Definition. A linear transformation T : Rn → Rm is said to be a linear operator if n = m. In this case, we say that T is a linear operator on Rn . Example 8.1.1. The linear transformation T : R5 → R3 , defined by the equations y1 = 2x1 + 3x2 + 5x3 + 7x4 − 9x5 , 3x2 + 4x3 + 2x5 , y2 = y3 = x1 + 3x3 − 2x4 , can be expressed in matrix form as y1 2 y2 = 0 1 y3 3 3 0 5 4 3 7 0 −2 x1 −9 x2 2 x3 . 0 x4 x5 If (x1 , x2 , x3 , x4 , x5 ) = (1, 0, 1, 0, 1), then y1 235 7 y2 = 0 3 4 0 1 0 3 −2 y3 1 −9 0 −2 2 1 = 6 , 0 0 4 1 so that T (1, 0, 1, 0, 1) = (−2, 6, 4). Example 8.1.2. Suppose that A is the zero m × n matrix. The linear transformation T : Rn → Rm , where T (x) = Ax for every x ∈ Rn , is the zero transformation from Rn into Rm . Clearly T (x) = 0 for every x ∈ Rn . Example 8.1.3. Suppose that I is the identity n × n matrix. The linear operator T : Rn → Rn , where T (x) = I x for every x ∈ Rn , is the identity operator on Rn . Clearly T (x) = x for every x ∈ Rn . Chapter 8 : Linear Transformations page 2 of 34 c Linear Algebra W W L Chen, 1997, 2005 PROPOSITION 8A. Suppose that T : Rn → Rm is a linear transformation, and that {e1 , . . . , en } is the standard basis for Rn . Then the standard matrix for T is given by A = ( T (e1 ) ... T (en ) ) , where T (ej ) is a column matrix for every j = 1, . . . , n. Proof. This follows immediately from (2). 8.2. Linear Operators on R2 In this section, we consider the special case when n = m = 2, and study linear operators on R2 . For every x ∈ R2 , we shall write x = (x1 , x2 ). Example 8.2.1. Consider reflection across the x2 -axis, so that T (x1 , x2 ) = (−x1 , x2 ). Clearly we have T (e1 ) = −1 0 and 0 1 T (e2 ) = , and so it follows from Proposition 8A that the standard matrix is given by A= −1 0 0 1 . It is not difficult to see that the standard matrices for reflection across the x1 -axis and across the line x1 = x2 are given respectively by A= 10 0 −1 and A= 0 1 1 0 . Also, the standard matrix for reflection across the origin is given by A= −1 0 0 −1 . We give a summary in the table below: Linear operator Equations Standard matrix Reflection across x2 -axis y1 = −x1 y2 = x2 −1 0 Reflection across x1 -axis y1 = x1 y2 = −x2 1 0 0 −1 Reflection across x1 = x2 y1 = x2 y2 = x1 0 1 1 0 Reflection across origin y1 = −x1 y2 = −x2 −1 0 0 1 0 −1 Example 8.2.2. For orthogonal projection onto the x1 -axis, we have T (x1 , x2 ) = (x1 , 0), with standard matrix A= Chapter 8 : Linear Transformations 1 0 0 0 . page 3 of 34 c Linear Algebra W W L Chen, 1997, 2005 Similarly, the standard matrix for orthogonal projection onto the x2 -axis is given by 0 0 A= 0 1 . We give a summary in the table below: Linear operator Equations Standard matrix Orthogonal projection onto x1 -axis y1 = x1 y2 = 0 1 0 0 0 Orthogonal projection onto x2 -axis y1 = 0 y2 = x2 0 0 0 1 Example 8.2.3. For anticlockwise rotation by an angle θ, we have T (x1 , x2 ) = (y1 , y2 ), where y1 + iy2 = (x1 + ix2 )(cos θ + i sin θ), and so y1 y2 = − sin θ cos θ cos θ sin θ x1 x2 . It follows that the standard matrix is given by A= − sin θ cos θ cos θ sin θ . We give a summary in the table below: Linear operator Equations Standard matrix y1 = x1 cos θ − x2 sin θ y2 = x1 sin θ + x2 cos θ Anticlockwise rotation by angle θ cos θ sin θ − sin θ cos θ Example 8.2.4. For contraction or dilation by a non-negative scalar k , we have T (x1 , x2 ) = (kx1 , kx2 ), with standard matrix A= k 0 0 k . The operator is called a contraction if 0 < k < 1 and a dilation if k > 1, and can be extended to negative values of k by noting that for k < 0, we have k 0 0 k = −1 0 0 −1 −k 0 0 −k . This describes contraction or dilation by non-negative scalar −k followed by reflection across the origin. We give a summary in the table below: Linear operator Contraction or dilation by factor k Chapter 8 : Linear Transformations Equations y1 = kx1 y2 = kx2 Standard matrix k 0 0 k page 4 of 34 c Linear Algebra W W L Chen, 1997, 2005 Example 8.2.5. For expansion or compression in the x1 -direction by a positive factor k , we have T (x1 , x2 ) = (kx1 , x2 ), with standard matrix k 0 A= 0 1 . This can be extended to negative values of k by noting that for k < 0, we have k 0 0 1 = −1 0 −k 0 0 1 0 1 . This describes expansion or compression in the x1 -direction by positive factor −k followed by reflection across the x2 -axis. Similarly, for expansion or compression in the x2 -direction by a non-zero factor k , we have the standard matrix 1 0 A= 0 k . We give a summary in the table below: Linear operator Equations Standard matrix Expansion or compression in x1 -direction y1 = kx1 y2 = x2 k 0 0 1 Expansion or compression in x2 -direction y1 = x1 y2 = kx2 1 0 0 k Example 8.2.6. For shears in the x1 -direction with factor k , we have T (x1 , x2 ) = (x1 + kx2 , x2 ), with standard matrix 1 0 A= k 1 . For the case k = 1, we have the following. • • • / • ~ ~ ~ ~ ~ ~ ~ • • T (k=1) • ~ ~ ~ ~ ~ ~ ~ • For the case k = −1, we have the following. • • • • Chapter 8 : Linear Transformations T (k=−1) / @ •@ @ @ @ @ @ • @ •@ @ @ @ @ @ • page 5 of 34 c Linear Algebra W W L Chen, 1997, 2005 Similarly, for shears in the x2 -direction with factor k , we have standard matrix 1 k A= 0 1 . We give a summary in the table below: Linear operator Equations Standard matrix Shear in x1 -direction y1 = x1 + kx2 y2 = x2 1 0 k 1 Shear in x2 -direction y1 = x1 y2 = kx1 + x2 1 k 0 1 Example 8.2.7. Consider a linear operator T : R2 → R2 which consists of a reflection across the x2 -axis, followed by a shear in the x1 -direction with factor 3 and then reflection across the x1 -axis. To find the standard matrix, consider the effect of T on a standard basis {e1 , e2 } of R2 . Note that e1 = 1 0 → −1 0 → −1 0 → −1 0 = T (e1 ), e2 = 0 1 → 0 1 → 3 1 → 3 −1 = T (e2 ), so it follows from Proposition 8A that the standard matrix for T is A= −1 0 3 −1 . Let us summarize the above and consider a few special cases. We have the following table of invertible linear operators with k = 0. Clearly, if A is the standard matrix for an invertible linear operator T , then the inverse matrix A−1 is the standard matrix for the inverse linear operator T −1 . Linear operator T Standard matrix A Inverse matrix A−1 Reflection across line x1 =x2 0 1 1 0 0 1 1 0 Expansion or compression in x1 −direction k 0 0 1 k −1 0 Expansion or compression in x2 −direction 1 0 0 k Shear in x1 −direction 1 0 Shear in x2 −direction 1 k Linear operator T −1 Reflection across line x1 =x2 0 1 Expansion or compression in x1 −direction 1 0 0 k −1 Expansion or compression in x2 −direction k 1 1 0 −k 1 Shear in x1 −direction 0 1 1 −k 0 1 Shear in x2 −direction Next, let us consider the question of elementary row operations on 2 × 2 matrices. It is not difficult to see that an elementary row operation performed on a 2 × 2 matrix A has the effect of multiplying the Chapter 8 : Linear Transformations page 6 of 34 c Linear Algebra W W L Chen, 1997, 2005 matrix A by some elementary matrix E to give the product EA. We have the following table. Elementary row operation Elementary matrix E Interchanging the two rows 0 1 1 0 Multiplying row 1 by non-zero factor k k 0 0 1 Multiplying row 2 by non-zero factor k 1 0 0 k Adding k times row 2 to row 1 1 0 k 1 Adding k times row 1 to row 2 1 k 0 1 Now, we know that any invertible matrix A can be reduced to the identity matrix by a finite number of elementary row operations. In other words, there exist a finite number of elementary matrices E1 , . . . , Es of the types above with various non-zero values of k such that Es . . . E1 A = I, so that − − A = E1 1 . . . Es 1 . We have proved the following result. PROPOSITION 8B. Suppose that the linear operator T : R2 → R2 has standard matrix A, where A is invertible. Then T is the product of a succession of finitely many reflections, expansions, compressions and shears. In fact, we can prove the following result concerning images of straight lines. PROPOSITION 8C. Suppose that the linear operator T : R2 → R2 has standard matrix A, where A is invertible. Then (a) the image under T of a straight line is a straight line; (b) the image under T of a straight line through the origin is a straight line through the origin; and (c) the images under T of parallel straight lines are parallel straight lines. Proof. Suppose that T (x1 , x2 ) = (y1 , y2 ). Since A is invertible, we have x = A−1 y, where x= x1 x2 and y= y1 y2 . The equation of a straight line is given by αx1 + βx2 = γ or, in matrix form, by (α β) x1 x2 = (γ ). Hence (α Chapter 8 : Linear Transformations β ) A−1 y1 y2 = (γ ). page 7 of 34 c Linear Algebra W W L Chen, 1997, 2005 Let (α β ) = (α β ) A−1 . Then (α β) y1 y2 = (γ ). In other words, the image under T of the straight line αx1 + βx2 = γ is α y1 + β y2 = γ , clearly another straight line. This proves (a). To prove (b), note that straight lines through the origin correspond to γ = 0. To prove (c), note that parallel straight lines correspond to different values of γ for the same values of α and β . 8.3. Elementary Properties of Euclidean Linear Transformations In this section, we establish a number of simple properties of euclidean linear transformations. PROPOSITION 8D. Suppose that T1 : Rn → Rm and T2 : Rm → Rk are linear transformations. Then T = T2 ◦ T1 : Rn → Rk is also a linear transformation. Proof. Since T1 and T2 are linear transformations, they have standard matrices A1 and A2 respectively. In other words, we have T1 (x) = A1 x for every x ∈ Rn and T2 (y) = A2 y for every y ∈ Rm . It follows that T (x) = T2 (T1 (x)) = A2 A1 x for every x ∈ Rn , so that T has standard matrix A2 A1 . Example 8.3.1. Suppose that T1 : R2 → R2 is anticlockwise rotation by π/2 and T2 : R2 → R2 is orthogonal projection onto the x1 -axis. Then the respective standard matrices are A1 = −1 0 0 1 and A2 = 1 0 0 0 . It follows that the standard matrices for T2 ◦ T1 and T1 ◦ T2 are respectively A2 A1 = −1 0 0 0 and A 1 A2 = 0 1 0 0 . Hence T2 ◦ T1 and T1 ◦ T2 are not equal. Example 8.3.2. Suppose that T1 : R2 → R2 is anticlockwise rotation by θ and T2 : R2 → R2 is anticlockwise rotation by φ. Then the respective standard matrices are A1 = cos θ sin θ − sin θ cos θ and A2 = cos φ − sin φ sin φ cos φ . It follows that the standard matrix for T2 ◦ T1 is A2 A1 = cos φ cos θ − sin φ sin θ sin φ cos θ + cos φ sin θ − cos φ sin θ − sin φ cos θ cos φ cos θ − sin φ sin θ = cos(φ + θ) − sin(φ + θ) sin(φ + θ) cos(φ + θ) . Hence T2 ◦ T1 is anticlockwise rotation by φ + θ. Example 8.3.3. The reader should check that in R2 , reflection across the x1 -axis followed by reflection across the x2 -axis gives reflection across the origin. Linear transformations that map distinct vectors to distinct vectors are of special importance. Chapter 8 : Linear Transformations page 8 of 34 c Linear Algebra W W L Chen, 1997, 2005 Definition. A linear transformation T : Rn → Rm is said to be one-to-one if for every x , x ∈ Rn , we have x = x whenever T (x ) = T (x ). Example 8.3.4. If we consider linear operators T : R2 → R2 , then T is one-to-one precisely when the standard matrix A is invertible. To see this, suppose first of all that A is invertible. If T (x ) = T (x ), then Ax = Ax . Multiplying on the left by A−1 , we obtain x = x . Suppose next that A is not invertible. Then there exists x ∈ R2 such that x = 0 and Ax = 0. On the other hand, we clearly have A0 = 0. It follows that T (x) = T (0), so that T is not one-to-one. PROPOSITION 8E. Suppose that the linear operator T : Rn → Rn has standard matrix A. Then the following statements are equivalent: (a) The matrix A is invertible. (b) The linear operator T is one-to-one. (c) The range of T is Rn ; in other words, R(T ) = Rn . Proof. ((a)⇒(b)) Suppose that T (x ) = T (x ). Then Ax = Ax . Multiplying on the left by A−1 gives x =x . ((b)⇒(a)) Suppose that T is one-to-one. Then the system Ax = 0 has unique solution x = 0 in Rn . It follows that A can be reduced by elementary row operations to the identity matrix I , and is therefore invertible. ((a)⇒(c)) For any y ∈ Rn , clearly x = A−1 y satisfies Ax = y, so that T (x) = y. ((c)⇒(a)) Suppose that {e1 , . . . , en } is the standard basis for Rn . Let x1 , . . . , xn ∈ Rn be chosen to satisfy T (xj ) = ej , so that Axj = ej , for every j = 1, . . . , n. Write C = ( x1 . . . xn ) . Then AC = I , so that A is invertible. Definition. Suppose that the linear operator T : Rn → Rn has standard matrix A, where A is invertible. Then the linear operator T −1 : Rn → Rn , defined by T −1 (x) = A−1 x for every x ∈ Rn , is called the inverse of the linear operator T . Remark. Clearly T −1 (T (x)) = x and T (T −1 (x)) = x for every x ∈ Rn . Example 8.3.5. Consider the linear operator T : R2 → R2 , defined by T (x) = Ax for every x ∈ R2 , where A= 1 1 1 2 . Clearly A is invertible, and A−1 = 2 −1 −1 1 . Hence the inverse linear operator is T −1 : R2 → R2 , defined by T −1 (x) = A−1 x for every x ∈ R2 . Example 8.3.6. Suppose that T : R2 → R2 is anticlockwise rotation by angle θ. The reader should check that T −1 : R2 → R2 is anticlockwise rotation by angle 2π − θ. Next, we study the linearity properties of euclidean linear transformations which we shall use later to discuss linear transformations in arbitrary real vector spaces. Chapter 8 : Linear Transformations page 9 of 34 c Linear Algebra W W L Chen, 1997, 2005 PROPOSITION 8F. A transformation T : Rn → Rm is linear if and only if the following two conditions are satisfied: (a) For every u, v ∈ Rn , we have T (u + v) = T (u) + T (v). (b) For every u ∈ Rn and c ∈ R, we have T (cu) = cT (u). Proof. Suppose first of all that T : Rn → Rm is a linear transformation. Let A be the standard matrix for T . Then for every u, v ∈ Rn and c ∈ R, we have T (u + v) = A(u + v) = Au + Av = T (u) + T (v) and T (cu) = A(cu) = c(Au) = cT (u). Suppose now that (a) and (b) hold. To show that T is linear, we need to find a matrix A such that T (x) = Ax for every x ∈ Rn . Suppose that {e1 , . . . , en } is the standard basis for Rn . As suggested by Proposition 8A, we write A = ( T (e1 ) . . . T (en ) ) , where T (ej ) is a column matrix for every j = 1, . . . , n. For any vector x1 . x= . . xn in Rn , we have x1 . Ax = ( T (e1 ) . . . T (en ) ) . = x1 T (e1 ) + . . . + xn T (en ). . xn Using (b) on each summand and then using (a) inductively, we obtain Ax = T (x1 e1 ) + . . . + T (xn en ) = T (x1 e1 + . . . + xn en ) = T (x) as required. To conclude our study of euclidean linear transformations, we briefly mention the problem of eigenvalues and eigenvectors of euclidean linear operators. Definition. Suppose that T : Rn → Rn is a linear operator. Then any real number λ ∈ R is called an eigenvalue of T if there exists a non-zero vector x ∈ Rn such that T (x) = λx. This non-zero vector x ∈ Rn is called an eigenvector of T corresponding to the eigenvalue λ. Remark. Note that the equation T (x) = λx is equivalent to the equation Ax = λx. It follows that there is no distinction between eigenvalues and eigenvectors of T and those of the standard matrix A. We therefore do not need to discuss this problem any further. 8.4. General Linear Transformations Suppose that V and W are real vector spaces. To define a linear transformation from V into W , we are motivated by Proposition 8F which describes the linearity properties of euclidean linear transformations. Chapter 8 : Linear Transformations page 10 of 34 c Linear Algebra W W L Chen, 1997, 2005 By a transformation from V into W , we mean a function of the type T : V → W , with domain V and codomain W . For every vector u ∈ V , the vector T (u) ∈ W is called the image of u under the transformation T . Definition. A transformation T : V → W from a real vector space V into a real vector space W is called a linear transformation if the following two conditions are satisfied: (LT1) For every u, v ∈ V , we have T (u + v) = T (u) + T (v). (LT2) For every u ∈ V and c ∈ R, we have T (cu) = cT (u). Definition. A linear transformation T : V → V from a real vector space V into itself is called a linear operator on V . Example 8.4.1. Suppose that V and W are two real vector spaces. The transformation T : V → W , where T (u) = 0 for every u ∈ V , is clearly linear, and is called the zero transformation from V to W . Example 8.4.2. Suppose that V is a real vector space. The transformation I : V → V , where I (u) = u for every u ∈ V , is clearly linear, and is called the identity operator on V . Example 8.4.3. Suppose that V is a real vector space, and that k ∈ R is fixed. The transformation T : V → V , where T (u) = k u for every u ∈ V , is clearly linear. This operator is called a dilation if k > 1 and a contraction if 0 < k < 1. Example 8.4.4. Suppose that V is a finite dimensional vector space, with basis {w1 , . . . , wn }. Define a transformation T : V → Rn as follows. For every u ∈ V , there exists a unique vector (β1 , . . . , βn ) ∈ Rn such that u = β1 w1 + . . . + βn wn . We let T (u) = (β1 , . . . , βn ). In other words, the transformation T gives the coordinates of any vector u ∈ V with respect to the given basis {w1 , . . . , wn }. Suppose now that v = γ1 w1 + . . . + γn wn is another vector in V . Then u + v = (β1 + γ1 )w1 + . . . + (βn + γn )wn , so that T (u + v) = (β1 + γ1 , . . . , βn + γn ) = (β1 , . . . , βn ) + (γ1 , . . . , γn ) = T (u) + T (v). Also, if c ∈ R, then cu = cβ1 w1 + . . . + cβn wn , so that T (cu) = (cβ1 , . . . , cβn ) = c(β1 , . . . , βn ) = cT (u). Hence T is a linear transformation. We shall return to this in greater detail in the next section. Example 8.4.5. Suppose that Pn denotes the vector space of all polynomials with real coefficients and degree at most n. Define a transformation T : Pn → Pn as follows. For every polynomial p = p0 + p1 x + . . . + pn xn in Pn , we let T (p) = pn + pn−1 x + . . . + p0 xn . Suppose now that q = q0 + q1 x + . . . + qn xn is another polynomial in Pn . Then p + q = (p0 + q0 ) + (p1 + q1 )x + . . . + (pn + qn )xn , so that T (p + q ) = (pn + qn ) + (pn−1 + qn−1 )x + . . . + (p0 + q0 )xn = (pn + pn−1 x + . . . + p0 xn ) + (qn + qn−1 x + . . . + q0 xn ) = T (p) + T (q ). Chapter 8 : Linear Transformations page 11 of 34 c Linear Algebra W W L Chen, 1997, 2005 Also, for any c ∈ R, we have cp = cp0 + cp1 x + . . . + cpn xn , so that T (cp) = cpn + cpn−1 x + . . . + cp0 xn = c(pn + pn−1 x + . . . + p0 xn ) = cT (p). Hence T is a linear transformation. Example 8.4.6. Let V denote the vector space of all real valued functions differentiable everywhere in R, and let W denote the vector space of all real valued functions defined on R. Consider the transformation T : V → W , where T (f ) = f for every f ∈ V . It is easy to check from properties of derivatives that T is a linear transformation. Example 8.4.7. Let V denote the vector space of all real valued functions that are Riemann integrable over the interval [0, 1]. Consider the transformation T : V → R, where 1 T (f ) = f (x) dx 0 for every f ∈ V . It is easy to check from properties of the Riemann integral that T is a linear transformation. Consider a linear transformation T : V → W from a finite dimensional real vector space V into a real vector space W . Suppose that {v1 , . . . , vn } is a basis of V . Then every u ∈ V can be written uniquely in the form u = β1 v1 + . . . + βn vn , where β1 , . . . , βn ∈ R. It follows that T (u) = T (β1 v1 + . . . + βn vn ) = T (β1 v1 ) + . . . + T (βn vn ) = β1 T (v1 ) + . . . + βn T (vn ). We have therefore proved the following generalization of Proposition 8A. PROPOSITION 8G. Suppose that T : V → W is a linear transformation from a finite dimensional real vector space V into a real vector space W . Suppose further that {v1 , . . . , vn } is a basis of V . Then T is completely determined by T (v1 ), . . . , T (vn ). Example 8.4.8. Consider a linear transformation T : P2 → R, where T (1) = 1, T (x) = 2 and T (x2 ) = 3. Since {1, x, x2 } is a basis of P2 , this linear transformation is completely determined. In particular, we have, for example, T (5 − 3x + 2x2 ) = 5T (1) − 3T (x) + 2T (x2 ) = 5. Example 8.4.9. Consider a linear transformation T : R4 → R, where T (1, 0, 0, 0) = 1, T (1, 1, 0, 0) = 2, T (1, 1, 1, 0) = 3 and T (1, 1, 1, 1) = 4. Since {(1, 0, 0, 0), (1, 1, 0, 0), (1, 1, 1, 0), (1, 1, 1, 1)} is a basis of R4 , this linear transformation is completely determined. In particular, we have, for example, T (6, 4, 3, 1) = T (2(1, 0, 0, 0) + (1, 1, 0, 0) + 2(1, 1, 1, 0) + (1, 1, 1, 1)) = 2T (1, 0, 0, 0) + T (1, 1, 0, 0) + 2T (1, 1, 1, 0) + T (1, 1, 1, 1) = 14. We also have the following generalization of Proposition 8D. PROPOSITION 8H. Suppose that V, W, U are real vector spaces. Suppose further that T1 : V → W and T2 : W → U are linear transformations. Then T = T2 ◦ T1 : V → U is also a linear transformation. Proof. Suppose that u, v ∈ V . Then T (u + v) = T2 (T1 (u + v)) = T2 (T1 (u) + T1 (v)) = T2 (T1 (u)) + T2 (T1 (v)) = T (u) + T (v). Also, if c ∈ R, then T (cu) = T2 (T1 (cu)) = T2 (cT1 (u)) = cT2 (T1 (u)) = cT (u). Hence T is a linear transformation. Chapter 8 : Linear Transformations page 12 of 34 c Linear Algebra W W L Chen, 1997, 2005 8.5. Change of Basis Suppose that V is a real vector space, with basis B = {u1 , . . . , un }. Then every vector u ∈ V can be written uniquely as a linear combination (3) u = β1 u1 + . . . + βn un , where β1 , . . . , βn ∈ R. It follows that the vector u can be identified with the vector (β1 , . . . , βn ) ∈ Rn . Definition. Suppose that u ∈ V and (3) holds. Then the matrix β1 . [u]B = . . βn is called the coordinate matrix of u relative to the basis B = {u1 , . . . , un }. Example 8.5.1. The vectors u1 = (1, 2, 1, 0), u3 = (2, −10, 0, 0), u2 = (3, 3, 3, 0), u4 = (−2, 1, −6, 2) are linearly independent in R4 , and so B = {u1 , u2 , u3 , u4 } is a basis of R4 . It follows that for any u = (x, y, z, w) ∈ R4 , we can write u = β1 u1 + β2 u2 + β3 u3 + β4 u4 . In matrix notation, this becomes x 1 y 2 = z 1 w 0 3 3 3 0 2 −10 0 0 −2 β1 1 β2 , −6 β3 2 β4 so that 1 β1 β2 2 [u]B = = 1 β3 0 β4 3 3 3 0 2 −10 0 0 −1 x −2 1 y . −6 z 2 w Remark. Consider a function φ : V → Rn , where φ(u) = [u]B for every u ∈ V . It is not difficult to see that this function gives rise to a one-to-one correspondence between the elements of V and the elements of Rn . Furthermore, note that [u + v]B = [u]B + [v]B and [cu]B = c[u]B , so that φ(u + v) = φ(u) + φ(v) and φ(cu) = cφ(u) for every u, v ∈ V and c ∈ R. Thus φ is a linear transformation, and preserves much of the structure of V . We also say that V is isomorphic to Rn . In practice, once we have made this identification between vectors and their coordinate matrices, then we can basically forget about the basis B and imagine that we are working in Rn with the standard basis. Clearly, if we change from one basis B = {u1 , . . . , un } to another basis C = {v1 , . . . , vn } of V , then we also need to find a way of calculating [u]C in terms of [u]B for every vector u ∈ V . To do this, note that each of the vectors v1 , . . . , vn can be written uniquely as a linear combination of the vectors u1 , . . . , un . Suppose that for i = 1, . . . , n, we have vi = a1i u1 + . . . + ani un , Chapter 8 : Linear Transformations where a1i , . . . , ani ∈ R, page 13 of 34 c Linear Algebra W W L Chen, 1997, 2005 so that a1i . [vi ]B = . . . ani For every u ∈ V , we can write where β1 , . . . , βn , γ1 , . . . , γn ∈ R, u = β1 u1 + . . . + βn un = γ1 v1 + . . . + γn vn , so that β1 . [u]B = . . and βn γ1 . [u]C = . . . γn Clearly u = γ1 v1 + . . . + γn vn = γ1 (a11 u1 + . . . + an1 un ) + . . . + γn (a1n u1 + . . . + ann un ) = (γ1 a11 + . . . + γn a1n )u1 + . . . + (γ1 an1 + . . . + γn ann )un = β1 u1 + . . . + βn un . Hence β1 = γ1 a11 + . . . + γn a1n , . . . βn = γ1 an1 + . . . + γn ann . Written in matrix notation, we have β1 a11 . . . . = . . βn an1 ... γ1 a1n . . . . . . . . . . ann γn We have proved the following result. PROPOSITION 8J. Suppose that B = {u1 , . . . , un } and C = {v1 , . . . , vn } are two bases of a real vector space V . Then for every u ∈ V , we have [u]B = P [u]C , where the columns of the matrix P = ( [v1 ]B ... [vn ]B ) are precisely the coordinate matrices of the elements of C relative to the basis B . Remark. Strictly speaking, Proposition 8J gives [u]B in terms of [u]C . However, note that the matrix P is invertible (why?), so that [u]C = P −1 [u]B . Definition. The matrix P in Proposition 8J is sometimes called the transition matrix from the basis C to the basis B . Chapter 8 : Linear Transformations page 14 of 34 c Linear Algebra W W L Chen, 1997, 2005 Example 8.5.2. We know that with u1 = (1, 2, 1, 0), u3 = (2, −10, 0, 0), u2 = (3, 3, 3, 0), u4 = (−2, 1, −6, 2), and with v2 = (1, −1, 1, 0), v1 = (1, 2, 1, 0), v3 = (1, 0, −1, 0), v4 = (0, 0, 0, 2), both B = {u1 , u2 , u3 , u4 } and C = {v1 , v2 , v3 , v4 } are bases of R4 . It is easy to check that v1 = u1 , v2 = −2u1 + u2 , v3 = 11u1 − 4u2 + u3 , v4 = −27u1 + 11u2 − 2u3 + u4 , so that P = ( [v1 ]B [v2 ]B [v3 ]B 1 0 [v4 ]B ) = 0 0 −2 1 0 0 11 −4 1 0 −27 11 . −2 1 −3 4 1 0 −1 −3 . 2 1 Hence [u]B = P [u]C for every u ∈ R4 . It is also easy to check that u1 = v1 , u2 = 2v1 + v2 , u3 = −3v1 + 4v2 + v3 , u4 = −v1 − 3v2 + 2v3 + v4 , so that Q = ( [u1 ]C [u2 ]C [u3 ]C 12 01 [u4 ]C ) = 00 00 Hence [u]C = Q[u]B for every u ∈ R4 . Note that P Q = I . Now let u = (6, −1, 2, 2). We can check that u = v1 + 3v2 + 2v3 + v4 , so that 1 3 [u]C = . 2 1 Then 1 0 [u]B = 0 0 −2 1 0 0 11 −4 1 0 −27 1 −10 11 3 6 = . −2 2 0 1 1 1 Check that u = −10u1 + 6u2 + u4 . Chapter 8 : Linear Transformations page 15 of 34 c Linear Algebra W W L Chen, 1997, 2005 Example 8.5.3. Consider the vector space P2 . It is not too difficult to check that u2 = 1 + x2 , u1 = 1 + x, u3 = x + x2 form a basis of P2 . Let u = 1 + 4x − x2 . Then u = β1 u1 + β2 u2 + β3 u3 , where 1 + 4x − x2 = β1 (1 + x) + β2 (1 + x2 ) + β3 (x + x2 ) = (β1 + β2 ) + (β1 + β3 )x + (β2 + β3 )x2 , so that β1 + β2 = 1, β1 + β3 = 4 and β2 + β3 = −1. Hence (β1 , β2 , β3 ) = (3, −2, 1). If we write B = {u1 , u2 , u3 }, then 3 [u]B = −2 . 1 On the other hand, it is also not too difficult to check that v1 = 1, v2 = 1 + x, v3 = 1 + x + x2 form a basis of P2 . Also u = γ1 v1 + γ2 v2 + γ3 v3 , where 1 + 4x − x2 = γ1 + γ2 (1 + x) + γ3 (1 + x + x2 ) = (γ1 + γ2 + γ3 ) + (γ2 + γ3 )x + γ3 x2 , so that γ1 + γ2 + γ3 = 1, γ2 + γ3 = 4 and γ3 = −1. Hence (γ1 , γ2 , γ3 ) = (−3, 5, −1). If we write C = {v1 , v2 , v3 }, then −3 [u]C = 5 . −1 Next, note that v1 = 1 u1 + 1 u2 − 1 u3 , 2 2 2 v2 = u1 , v3 = 1 u1 + 1 u2 + 1 u3 . 2 2 2 Hence P = ( [v1 ]B [v2 ]B 1/2 [v3 ]B ) = 1/2 −1/2 To verify that [u]B = P [u]C , note that 3 1/2 −2 = 1/2 1 −1/2 1 0 0 1 1/2 0 1/2 . 0 1/2 1/2 −3 1/2 5 . 1/2 −1 8.6. Kernel and Range Consider first of all a euclidean linear transformation T : Rn → Rm . Suppose that A is the standard matrix for T . Then the range of the transformation T is given by R(T ) = {T (x) : x ∈ Rn } = {Ax : x ∈ Rn }. Chapter 8 : Linear Transformations page 16 of 34 c Linear Algebra W W L Chen, 1997, 2005 It follows that R(T ) is the set of all linear combinations of the columns of the matrix A, and is therefore the column space of A. On the other hand, the set {x ∈ Rn : Ax = 0} is the nullspace of A. Recall that the sum of the dimension of the nullspace of A and dimension of the column space of A is equal to the number of columns of A. This is known as the Rank-nullity theorem. The purpose of this section is to extend this result to the setting of linear transformations. To do this, we need the following generalization of the idea of the nullspace and the column space. Definition. Suppose that T : V → W is a linear transformation from a real vector space V into a real vector space W . Then the set ker(T ) = {u ∈ V : T (u) = 0} is called the kernel of T , and the set R(T ) = {T (u) : u ∈ V } is called the range of T . Example 8.6.1. For a euclidean linear transformation T with standard matrix A, we have shown that ker(T ) is the nullspace of A, while R(T ) is the column space of A. Example 8.6.2. Suppose that T : V → W is the zero transformation. Clearly we have ker(T ) = V and R(T ) = {0}. Example 8.6.3. Suppose that T : V → V is the identity operator on V . Clearly we have ker(T ) = {0} and R(T ) = V . Example 8.6.4. Suppose that T : R2 → R2 is orthogonal projection onto the x1 -axis. Then ker(T ) is the x2 -axis, while R(T ) is the x1 -axis. Example 8.6.5. Suppose that T : Rn → Rn is one-to-one. Then ker(T ) = {0} and R(T ) = Rn , in view of Proposition 8E. Example 8.6.6. Consider the linear transformation T : V → W , where V denotes the vector space of all real valued functions differentiable everywhere in R, where W denotes the space of all real valued functions defined in R, and where T (f ) = f for every f ∈ V . Then ker(T ) is the set of all differentiable functions with derivative 0, and so is the set of all constant functions in R. Example 8.6.7. Consider the linear transformation T : V → R, where V denotes the vector space of all real valued functions Riemann integrable over the interval [0, 1], and where 1 T (f ) = f (x) dx 0 for every f ∈ V . Then ker(T ) is the set of all Riemann integrable functions in [0, 1] with zero mean, while R(T ) = R. PROPOSITION 8K. Suppose that T : V → W is a linear transformation from a real vector space V into a real vector space W . Then ker(T ) is a subspace of V , while R(T ) is a subspace of W . Chapter 8 : Linear Transformations page 17 of 34 c Linear Algebra W W L Chen, 1997, 2005 Proof. Since T (0) = 0, it follows that 0 ∈ ker(T ) ⊆ V and 0 ∈ R(T ) ⊆ W . For any u, v ∈ ker(T ), we have T (u + v) = T (u) + T (v) = 0 + 0 = 0, so that u + v ∈ ker(T ). Suppose further that c ∈ R. Then T (cu) = cT (u) = c0 = 0, so that cu ∈ ker(T ). Hence ker(T ) is a subspace of V . Suppose next that w, z ∈ R(T ). Then there exist u, v ∈ V such that T (u) = w and T (v) = z. Hence T (u + v) = T (u) + T (v) = w + z, so that w + z ∈ R(T ). Suppose further that c ∈ R. Then T (cu) = cT (u) = cw, so that cw ∈ R(T ). Hence R(T ) is a subspace of W . To complete this section, we prove the following generalization of the Rank-nullity theorem. PROPOSITION 8L. Suppose that T : V → W is a linear transformation from an n-dimensional real vector space V into a real vector space W . Then dim ker(T ) + dim R(T ) = n. Proof. Suppose first of all that dim ker(T ) = n. Then ker(T ) = V , and so R(T ) = {0}, and the result follows immediately. Suppose next that dim ker(T ) = 0, so that ker(T ) = {0}. If {v1 , . . . , vn } is a basis of V , then it follows that T (v1 ), . . . , T (vn ) are linearly independent in W , for otherwise there exist c1 , . . . , cn ∈ R, not all zero, such that c1 T (v1 ) + . . . + cn T (vn ) = 0, so that T (c1 v1 + . . . + cn vn ) = 0, a contradiction since c1 v1 + . . . + cn vn = 0. On the other hand, elements of R(T ) are linear combinations of T (v1 ), . . . , T (vn ). Hence dim R(T ) = n, and the result again follows immediately. We may therefore assume that dim ker(T ) = r, where 1 ≤ r < n. Let {v1 , . . . , vr } be a basis of ker(T ). This basis can be extended to a basis {v1 , . . . , vr , vr+1 , . . . , vn } of V . It suffices to show that {T (vr+1 ), . . . , T (vn )} (4) is a basis of R(T ). Suppose that u ∈ V . Then there exist β1 , . . . , βn ∈ R such that u = β1 v1 + . . . + βr vr + βr+1 vr+1 + . . . + βn vn , so that T (u) = β1 T (v1 ) + . . . + βr T (vr ) + βr+1 T (vr+1 ) + . . . + βn T (vn ) = βr+1 T (vr+1 ) + . . . + βn T (vn ). It follows that (4) spans R(T ). It remains to prove that its elements are linearly independent. Suppose that cr+1 , . . . , cn ∈ R and (5) Chapter 8 : Linear Transformations cr+1 T (vr+1 ) + . . . + cn T (vn ) = 0. page 18 of 34 c Linear Algebra W W L Chen, 1997, 2005 We need to show that (6) cr+1 = . . . = cn = 0. By linearity, it follows from (5) that T (cr+1 vr+1 + . . . + cn vn ) = 0, so that cr+1 vr+1 + . . . + cn vn ∈ ker(T ). Hence there exist c1 , . . . , cr ∈ R such that cr+1 vr+1 + . . . + cn vn = c1 v1 + . . . + cr vr , so that c1 v1 + . . . + cr vr − cr+1 vr+1 − . . . − cn vn = 0. Since {v1 , . . . , vn } is a basis of V , it follows that c1 = . . . = cr = cr+1 = . . . = cn = 0, so that (6) holds. This completes the proof. Remark. We sometimes say that dim R(T ) and dim ker(T ) are respectively the rank and the nullity of the linear transformation T . 8.7. Inverse Linear Transformations In this section, we generalize some of the ideas first discussed in Section 8.3. Definition. A linear transformation T : V → W from a real vector space V into a real vector space W is said to be one-to-one if for every u , u ∈ V , we have u = u whenever T (u ) = T (u ). The result below follows immediately from our definition. PROPOSITION 8M. Suppose that T : V → W is a linear transformation from a real vector space V into a real vector space W . Then T is one-to-one if and only if ker(T ) = {0}. Proof. (⇒) Clearly 0 ∈ ker(T ). Suppose that ker(T ) = {0}. Then there exists a non-zero v ∈ ker(T ). It follows that T (v) = T (0), and so T is not one-to-one. (⇐) Suppose that ker(T ) = {0}. Given any u , u ∈ V , we have T (u ) − T (u ) = T (u − u ) = 0 if and only if u − u = 0; in other words, if and only if u = u . We have the following generalization of Proposition 8E. PROPOSITION 8N. Suppose that T : V → V is a linear operator on a finite-dimensional real vector space V . Then the following statements are equivalent: (a) The linear operator T is one-to-one. (b) We have ker(T ) = {0}. (c) The range of T is V ; in other words, R(T ) = V . Proof. The equivalence of (a) and (b) is established by Proposition 8M. The equivalence of (b) and (c) follows from Proposition 8L. Chapter 8 : Linear Transformations page 19 of 34 c Linear Algebra W W L Chen, 1997, 2005 Suppose that T : V → W is a one-to-one linear transformation from a real vector space V into a real vector space W . Then for every w ∈ R(T ), there exists exactly one u ∈ V such that T (u) = w. We can therefore define a transformation T −1 : R(T ) → V by writing T −1 (w) = u, where u ∈ V is the unique vector satisfying T (u) = w. PROPOSITION 8P. Suppose that T : V → W is a one-to-one linear transformation from a real vector space V into a real vector space W . Then T −1 : R(T ) → V is a linear transformation. Proof. Suppose that w, z ∈ R(T ). Then there exist u, v ∈ V such that T −1 (w) = u and T −1 (z) = v. It follows that T (u) = w and T (v) = z, so that T (u + v) = T (u) + T (v) = w + z, whence T −1 (w + z) = u + v = T −1 (w) + T −1 (z). Suppose further that c ∈ R. Then T (cu) = cw, so that T −1 (cw) = cu = cT −1 (w). This completes the proof. We also have the following result concerning compositions of linear transformations and which requires no further proof, in view of our knowledge concerning inverse functions. PROPOSITION 8Q. Suppose that V, W, U are real vector spaces, and that T1 : V → W and T2 : W → U are one-to-one linear transformations. Then (a) the linear transformation T2 ◦ T1 : V → U is one-to-one; and − − (b) (T2 ◦ T1 )−1 = T1 1 ◦ T2 1 . 8.8. Matrices of General Linear Transformations Suppose that T : V → W is a linear transformation from a real vector space V to a real vector space W . Suppose further that the vector spaces V and W are finite dimensional, with dim V = n and dim W = m. We shall show that if we make use of a basis B of V and a basis C of W , then it is possible to describe T indirectly in terms of some matrix A. The main idea is to make use of coordinate matrices relative to the bases B and C . Let us recall some discussion in Section 8.5. Suppose that B = {v1 , . . . , vn } is a basis of V . Then every vector v ∈ V can be written uniquely as a linear combination (7) where β1 , . . . , βn ∈ R. v = β1 v1 + . . . + βn vn , The matrix (8) β1 . [v]B = . . βn is the coordinate matrix of v relative to the basis B . Consider now a transformation φ : V → Rn , where φ(v) = [v]B for every v ∈ V . The proof of the following result is straightforward. PROPOSITION 8R. Suppose that the real vector space V has basis B = {v1 , . . . , vn }. Then the transformation φ : V → Rn , where φ(v) = [v]B satisfies (7) and (8) for every v ∈ V , is a oneto-one linear transformation, with range R(φ) = Rn . Furthermore, the inverse linear transformation φ−1 : Rn → V is also one-to-one, with range R(φ−1 ) = V . Chapter 8 : Linear Transformations page 20 of 34 c Linear Algebra W W L Chen, 1997, 2005 Suppose next that C = {w1 , . . . , wm } is a basis of W . Then we can define a linear transformation ψ : W → Rm , where ψ (w) = [w]C for every w ∈ W , in a similar way. We now have the following diagram of linear transformations. T V O φ−1 /W O ψ −1 φ Rn ψ Rm Clearly the composition S = ψ ◦ T ◦ φ−1 : Rn → Rm is a euclidean linear transformation, and can therefore be described in terms of a standard matrix A. Our task is to determine this matrix A in terms of T and the bases B and C . We know from Proposition 8A that A = ( S (e1 ) . . . S (en ) ) , where {e1 , . . . , en } is the standard basis for Rn . For every j = 1, . . . , n, we have S (ej ) = (ψ ◦ T ◦ φ−1 )(ej ) = ψ (T (φ−1 (ej ))) = ψ (T (vj )) = [T (vj )]C . It follows that A = ( [T (v1 )]C (9) . . . [T (vn )]C ) . Definition. The matrix A given by (9) is called the matrix for the linear transformation T with respect to the bases B and C . We now have the following diagram of linear transformations. T V O φ−1 /W O ψ −1 φ S Rn ψ / Rm Hence we can write T as the composition T = ψ −1 ◦ S ◦ φ : V → W. For every v ∈ V , we have the following: v φ Chapter 8 : Linear Transformations / [v]B S / A[v]B ψ −1 / ψ −1 (A[v]B ) page 21 of 34 c Linear Algebra W W L Chen, 1997, 2005 More precisely, if v = β1 v1 + . . . + βn vn , then β1 . [v]B = . . and βn β1 γ1 . . A[v]B = A . = . , . . βn γm say, and so T (v) = ψ −1 (A[v]B ) = γ1 w1 + . . . + γm wm . We have proved the following result. PROPOSITION 8S. Suppose that T : V → W is a linear transformation from a real vector space V into a real vector space W . Suppose further that V and W are finite dimensional, with bases B and C respectively, and that A is the matrix for the linear transformation T with respect to the bases B and C . Then for every v ∈ V , we have T (v) = w, where w ∈ W is the unique vector satisfying [w]C = A[v]B . Remark. In the special case when V = W , the linear transformation T : V → W is a linear operator on T . Of course, we may choose a basis B for the domain V of T and a basis C for the codomain V of T . In the case when T is the identity linear operator, we often choose B = C since this represents a change of basis. In the case when T is not the identity operator, we often choose B = C for the sake of convenience; we then say that A is the matrix for the linear operator T with respect to the basis B . Example 8.8.1. Consider an operator T : P3 → P3 on the real vector space P3 of all polynomials with real coefficients and degree at most 3, where for every polynomial p(x) in P3 , we have T (p(x)) = xp (x), the product of x with the formal derivative p (x) of p(x). The reader is invited to check that T is a linear operator. Now consider the basis B = {1, x, x2 , x3 } of P3 . The matrix for T with respect to B is given by A = ( [T (1)]B [T (x)]B [T (x2 )]B [T (x3 )]B ) = ( [0]B [x]B [2x2 ]B 0 0 3 [3x ]B ) = 0 0 0 1 0 0 0 0 2 0 0 0 . 0 3 Suppose that p(x) = 1 + 2x + 4x2 + 3x3 . Then 1 2 [p(x)]B = 4 3 and 0 0 A[p(x)]B = 0 0 0 1 0 0 0 0 2 0 0 1 0 02 2 = , 0 4 8 3 3 9 so that T (p(x)) = 2x + 8x2 + 9x3 . This can be easily verified by noting that T (p(x)) = xp (x) = x(2 + 8x + 9x2 ) = 2x + 8x2 + 9x3 . In general, if p(x) = p0 + p1 x + p2 x2 + p3 x3 , then p0 p [p(x)]B = 1 p2 p3 and 0 0 A[p(x)]B = 0 0 0 1 0 0 0 0 2 0 0 0 p0 0 p1 p1 , = 0 p2 2p2 3 p3 3p3 so that T (p(x)) = p1 x + 2p2 x2 + 3p3 x3 . Observe that T (p(x)) = xp (x) = x(p1 + 2p2 x + 3p3 x2 ) = p1 x + 2p2 x2 + 3p3 x3 , verifying our result. Chapter 8 : Linear Transformations page 22 of 34 c Linear Algebra W W L Chen, 1997, 2005 Example 8.8.2. Consider the linear operator T : R2 → R2 , given by T (x1 , x2 ) = (2x1 + x2 , x1 + 3x2 ) for every (x1 , x2 ) ∈ R2 . Consider also the basis B = {(1, 0), (1, 1)} of R2 . Then the matrix for T with respect to B is given by A = ( [T (1, 0)]B [T (1, 1)]B ) = ( [(2, 1)]B −1 4 1 1 [(3, 4)]B ) = . Suppose that (x1 , x2 ) = (3, 2). Then 1 2 [(3, 2)]B = and A[(3, 2)]B = 1 1 −1 4 1 2 = −1 9 , so that T (3, 2) = −(1, 0) + 9(1, 1) = (8, 9). This can be easily verified directly. In general, we have [(x1 , x2 )]B = x1 − x2 x2 and A[(x1 , x2 )]B = 1 1 −1 4 x1 − x2 x2 = x1 − 2x2 x1 + 3x2 , so that T (x1 , x2 ) = (x1 − 2x2 )(1, 0) + (x1 + 3x2 )(1, 1) = (2x1 + x2 , x1 + 3x2 ). Example 8.8.3. Suppose that T : Rn → Rm is a linear transformation. Suppose further that B and C are the standard bases for Rn and Rm respectively. Then the matrix for T with respect to B and C is given by A = ( [T (e1 )]C . . . [T (en )]C ) = ( T (e1 ) . . . T (en ) ) , so it follows from Proposition 8A that A is simply the standard matrix for T . Suppose now that T1 : V → W and T2 : W → U are linear transformations, where the real vector spaces V, W, U are finite dimensional, with respective bases B = {v1 , . . . , vn }, C = {w1 , . . . , wm } and D = {u1 , . . . , uk }. We then have the following diagram of linear transformations. T1 V O φ−1 ψ −1 φ Rn /W O S1 T2 η −1 ψ / Rm /U O S2 η / Rk Here η : U → Rk , where η (u) = [u]D for every u ∈ U , is a linear transformation, and S1 = ψ ◦ T1 ◦ φ−1 : Rn → Rm and S2 = η ◦ T2 ◦ ψ −1 : Rm → Rk are euclidean linear transformations. Suppose that A1 and A2 are respectively the standard matrices for S1 and S2 , so that they are respectively the matrix for T1 with respect to B and C and the matrix for T2 with respect to C and D. Clearly S2 ◦ S1 = η ◦ T2 ◦ T1 ◦ φ−1 : Rn → Rk . It follows that A2 A1 is the standard matrix for S2 ◦ S1 , and so is the matrix for T2 ◦ T1 with respect to the bases B and D. To summarize, we have the following result. Chapter 8 : Linear Transformations page 23 of 34 c Linear Algebra W W L Chen, 1997, 2005 PROPOSITION 8T. Suppose that T1 : V → W and T2 : W → U are linear transformations, where the real vector spaces V, W, U are finite dimensional, with bases B , C , D respectively. Suppose further that A1 is the matrix for the linear transformation T1 with respect to the bases B and C , and that A2 is the matrix for the linear transformation T2 with respect to the bases C and D. Then A2 A1 is the matrix for the linear transformation T2 ◦ T1 with respect to the bases B and D. Example 8.8.4. Consider the linear operator T1 : P3 → P3 , where for every polynomial p(x) in P3 , we have T1 (p(x)) = xp (x). We have already shown that the matrix for T1 with respect to the basis B = {1, x, x2 , x3 } of P3 is given by 0 0 A1 = 0 0 0 1 0 0 0 0 2 0 0 0 . 0 3 Consider next the linear operator T2 : P3 → P3 , where for every polynomial q (x) = q0 + q1 x + q2 x2 + q3 x3 in P3 , we have T2 (q (x)) = q (1 + x) = q0 + q1 (1 + x) + q2 (1 + x)2 + q3 (1 + x)3 . We have T2 (1) = 1, T2 (x) = 1 + x, T2 (x2 ) = 1 + 2x + x2 and T2 (x3 ) = 1 + 3x + 3x2 + x3 , so that the matrix for T2 with respect to B is given by A2 = ( [T2 (1)]B [T2 (x)]B 1 0 [T2 (x3 )]B ) = 0 0 [T2 (x2 )]B 1 1 0 0 1 2 1 0 1 3 . 3 1 Consider now the composition T = T2 ◦ T1 : P3 → P3 . Let A denote the matrix for T with respect to B . By Proposition 8T, we have 1 0 A = A2 A1 = 0 0 1 1 0 0 1 2 1 0 1 0 30 3 0 1 0 0 1 0 0 0 0 0 0 = 0 0 3 0 0 0 2 0 1 1 0 0 2 4 2 0 3 9 . 9 3 Suppose that p(x) = p0 + p1 x + p2 x2 + p3 x3 . Then p0 p [p(x)]B = 1 p2 p3 and 0 0 A[p(x)]B = 0 0 1 1 0 0 2 4 2 0 p1 + 2p2 + 3p3 3 p0 9 p1 p1 + 4p2 + 9p3 , = 9 p2 2p2 + 9p3 3 p3 3p3 so that T (p(x)) = (p1 + 2p2 + 3p3 ) + (p1 + 4p2 + 9p3 )x + (2p2 + 9p3 )x2 + 3p3 x3 . We can check this directly by noting that T (p(x)) = T2 (T1 (p(x))) = T2 (p1 x + 2p2 x2 + 3p3 x3 ) = p1 (1 + x) + 2p2 (1 + x)2 + 3p3 (1 + x)3 = (p1 + 2p2 + 3p3 ) + (p1 + 4p2 + 9p3 )x + (2p2 + 9p3 )x2 + 3p3 x3 . Example 8.8.5. Consider the linear operator T : R2 → R2 , given by T (x1 , x2 ) = (2x1 + x2 , x1 + 3x2 ) for every (x1 , x2 ) ∈ R2 . We have already shown that the matrix for T with respect to the basis B = {(1, 0), (1, 1)} of R2 is given by A= Chapter 8 : Linear Transformations 1 1 −1 4 . page 24 of 34 c Linear Algebra W W L Chen, 1997, 2005 Consider the linear operator T 2 : R2 → R2 . By Proposition 8T, the matrix for T 2 with respect to B is given by A2 = 1 1 −1 4 −1 4 1 1 0 5 = −5 15 . Suppose that (x1 , x2 ) ∈ R2 . Then [(x1 , x2 )]B = x1 − x2 x2 and −5 15 0 5 A2 [(x1 , x2 )]B = x1 − x2 x2 = −5x2 5x1 + 10x2 , so that T (x1 , x2 ) = −5x2 (1, 0) + (5x1 + 10x2 )(1, 1) = (5x1 + 5x2 , 5x1 + 10x2 ). The reader is invited to check this directly. A simple consequence of Propositions 8N and 8T is the following result concerning inverse linear transformations. PROPOSITION 8U. Suppose that T : V → V is a linear operator on a finite dimensional real vector space V with basis B . Suppose further that A is the matrix for the linear operator T with respect to the basis B . Then T is one-to-one if and only if A is invertible. Furthermore, if T is one-to-one, then A−1 is the matrix for the inverse linear operator T −1 : V → V with respect to the basis B . Proof. Simply note that T is one-to-one if and only if the system Ax = 0 has only the trivial solution x = 0. The last assertion follows easily from Proposition 8T, since if A denotes the matrix for the inverse linear operator T −1 with respect to B , then we must have A A = I , the matrix for the identity operator T −1 ◦ T with respect to B . Example 8.8.6. Consider the linear operator T : P3 → P3 , where for every q (x) = q0 + q1 x + q2 x2 + q3 x3 in P3 , we have T (q (x)) = q (1 + x) = q0 + q1 (1 + x) + q2 (1 + x)2 + q3 (1 + x)3 . We have already shown that the matrix for T with respect to the basis B = {1, x, x2 , x3 } is given by 1 0 A= 0 0 1 1 0 0 1 2 1 0 1 3 . 3 1 This matrix is invertible, so it follows that T is one-to-one. 1 −1 1 0 1 −2 −1 A = 00 1 00 0 Furthermore, it can be checked that −1 3 . −3 1 Suppose that p(x) = p0 + p1 x + p2 x2 + p3 x3 . Then p0 p [p(x)]B = 1 p2 p3 and 1 0 A−1 [p(x)]B = 0 0 −1 1 0 0 1 −2 1 0 p 0 − p1 + p 2 − p3 −1 p0 3 p1 p1 − 2p2 + 3p3 , = p2 p2 − 3p3 −3 1 p3 p3 so that T −1 (p(x)) = (p0 − p1 + p2 − p3 ) + (p1 − 2p2 + 3p3 )x + (p2 − 3p3 )x2 + p3 x3 = p0 + p1 (x − 1) + p2 (x2 − 2x + 1) + p3 (x3 − 3x2 + 3x − 1) = p0 + p1 (x − 1) + p2 (x − 1)2 + p3 (x − 1)3 = p(x − 1). Chapter 8 : Linear Transformations page 25 of 34 c Linear Algebra W W L Chen, 1997, 2005 8.9. Change of Basis Suppose that V is a finite dimensional real vector space, with one basis B = {v1 , . . . , vn } and another basis B = {u1 , . . . , un }. Suppose that T : V → V is a linear operator on V . Let A denote the matrix for T with respect to the basis B , and let A denote the matrix for T with respect to the basis B . If v ∈ V and T (v) = w, then (10) [w]B = A[v]B and (11) [w]B = A [v]B . We wish to find the relationship between A and A. Recall Proposition 8J, that if P = ( [u1 ]B . . . [un ]B ) denotes the transition matrix from the basis B to the basis B , then (12) [v]B = P [v]B and [w]B = P [w]B . Note that the matrix P can also be interpreted as the matrix for the identity operator I : V → V with respect to the bases B and B . It is easy to see that the matrix P is invertible, and P −1 = ( [v1 ]B . . . [vn ]B ) denotes the transition matrix from the basis B to the basis B , and can also be interpreted as the matrix for the identity operator I : V → V with respect to the bases B and B . Combining (10) and (12), we conclude that [w]B = P −1 [w]B = P −1 A[v]B = P −1 AP [v]B . Comparing this with (11), we conclude that P −1 AP = A . (13) This implies that A = P AP −1 . (14) Remark. We can use the notation A = [T ]B and A = [T ]B to denote that A and A are the matrices for T with respect to the basis B and with respect to the basis B respectively. We can also write P = [I ]B,B to denote that P is the transition matrix from the basis B to the basis B , so that P −1 = [I ]B ,B . Chapter 8 : Linear Transformations page 26 of 34 c Linear Algebra W W L Chen, 1997, 2005 Then (13) and (14) become respectively [I ]B ,B [T ]B [I ]B,B = [T ]B and [I ]B,B [T ]B [I ]B ,B = [T ]B . We have proved the following result. PROPOSITION 8V. Suppose that T : V → V is a linear operator on a finite dimensional real vector space V , with bases B = {v1 , . . . , vn } and B = {u1 , . . . , un }. Suppose further that A and A are the matrices for T with respect to the basis B and with respect to the basis B respectively. Then P −1 AP = A A = P AP −1 , and where P = ( [u1 ]B ... [un ]B ) denotes the transition matrix from the basis B to the basis B . Remarks. (1) We have the following picture. v T m6 v mmm Immm m mmm mmm / w RR RRR RRI RRR RRR R( T /w _______________________ [v]B O OO OO OO P OOO O' A [v]B A / [w]B 7 oo oo oo 1 o oo P − oo / [w]B (2) The idea can be extended to the case of linear transformations T : V → W from a finite dimensional real vector space into another, with a change of basis in V and a change of basis in W . Example 8.9.1. Consider the vector space P3 of all polynomials with real coefficients and degree at most 3, with bases B = {1, x, x2 , x3 } and B = {1, 1 + x, 1 + x + x2 , 1 + x + x2 + x3 }. Consider also the linear operator T : P3 → P3 , where for every polynomial p(x) = p0 + p1 x + p2 x2 + p3 x3 , we have T (p(x)) = (p0 + p1 ) + (p1 + p2 )x + (p2 + p3 )x2 + (p0 + p3 )x3 . Let A denote the matrix for T with respect to the basis B . Then T (1) = 1 + x3 , T (x) = 1 + x, T (x2 ) = x + x2 and T (x3 ) = x2 + x3 , and so A = ( [T (1)]B [T (x)]B [T (x2 )]B 1 0 [T (x3 )]B ) = 0 1 1 1 0 0 0 1 1 0 0 0 . 1 1 Next, note that the transition matrix from the basis B to the basis B is given by P = ( [1]B [1 + x]B Chapter 8 : Linear Transformations [1 + x + x2 ]B 1 0 [1 + x + x2 + x3 ]B ) = 0 0 1 1 0 0 1 1 1 0 1 1 . 1 1 page 27 of 34 c Linear Algebra W W L Chen, 1997, 2005 It can be checked that P −1 1 0 = 0 0 −1 1 0 0 0 −1 1 0 0 0 , −1 1 and so 1 0 −1 A = P AP = 0 0 −1 1 0 0 0 −1 1 0 0 1 0 0 −1 0 1 1 1 1 0 0 0 1 1 0 0 1 00 1 0 1 0 1 1 0 0 1 1 1 0 1 1 1 0 = 1 −1 1 1 1 1 −1 1 0 1 0 1 0 0 0 2 is the matrix for T with respect to the basis B . It follows that T (1) = 1 − (1 + x + x2 ) + (1 + x + x2 + x3 ) = 1 + x3 , T (1 + x) = 1 + (1 + x) − (1 + x + x2 ) + (1 + x + x2 + x3 ) = 2 + x + x3 , T (1 + x + x2 ) = (1 + x) + (1 + x + x2 + x3 ) = 2 + 2x + x2 + x3 , T (1 + x + x2 + x3 ) = 2(1 + x + x2 + x3 ) = 2 + 2x + 2x2 + 2x3 . These can be verified directly. 8.10. Eigenvalues and Eigenvectors Definition. Suppose that T : V → V is a linear operator on a finite dimensional real vector space V . Then any real number λ ∈ R is called an eigenvalue of T if there exists a non-zero vector v ∈ V such that T (v) = λv. This non-zero vector v ∈ V is called an eigenvector of T corresponding to the eigenvalue λ. The purpose of this section is to show that the problem of eigenvalues and eigenvectors of the linear operator T can be reduced to the problem of eigenvalues and eigenvectors of the matrix for T with respect to any basis B of V . The starting point of our argument is the following theorem, the proof of which is left as an exercise. PROPOSITION 8W. Suppose that T : V → V is a linear operator on a finite dimensional real vector space V , with bases B and B . Suppose further that A and A are the matrices for T with respect to the basis B and with respect to the basis B respectively. Then (a) det A = det A ; (b) A and A have the same rank; (c) A and A have the same characteristic polynomial; (d) A and A have the same eigenvalues; and (e) the dimension of the eigenspace of A corresponding to an eigenvalue λ is equal to the dimension of the eigenspace of A corresponding to λ. We also state without proof the following result. PROPOSITION 8X. Suppose that T : V → V is a linear operator on a finite dimensional real vector space V . Suppose further that A is the matrix for T with respect to a basis B of V . Then (a) the eigenvalues of T are precisely the eigenvalues of A; and (b) a vector u ∈ V is an eigenvector of T corresponding to an eigenvalue λ if and only if the coordinate matrix [u]B is an eigenvector of A corresponding to the eigenvalue λ. Chapter 8 : Linear Transformations page 28 of 34 c Linear Algebra W W L Chen, 1997, 2005 Suppose now that A is the matrix for a linear operator T : V → V on a finite dimensional real vector space V with respect to a basis B = {v1 , . . . , vn }. If A can be diagonalized, then there exists an invertible matrix P such that P −1 AP = D is a diagonal matrix. Furthermore, the columns of P are eigenvectors of A, and so are the coordinate matrices of eigenvectors of T with respect to the basis B . In other words, P = ( [u1 ]B . . . [un ]B ) , where B = {u1 , . . . , un } is a basis of V consiting of eigenvectors of T . Furthermore, P is the transition matrix from the basis B to the basis B . It follows that the matrix for T with respect to the basis B is given by λ1 D= .. , . λn where λ1 , . . . , λn are the eigenvalues of T . Example 8.10.1. Consider the vector space P2 of all polynomials with real coefficients and degree at most 2, with basis B = {1, x, x2 }. Consider also the linear operator T : P2 → P2 , where for every polynomial p(x) = p0 + p1 x + p2 x2 , we have T (p(x)) = (5p0 − 2p1 ) + (6p1 + 2p2 − 2p0 )x + (2p1 + 7p2 )x2 . Then T (1) = 5 − 2x, T (x) = −2 + 6x + 2x2 and T (x2 ) = 2x + 7x2 , so that the matrix for T with respect to the basis B is given by A = ( [T (1)]B [T (x)]B 5 [T (x2 )]B ) = −2 0 −2 6 2 0 2. 7 It is a simple exercise to show that the matrix A has eigenvalues 3, 6, 9, with corresponding eigenvectors 2 x1 = 2 , −1 2 x2 = −1 , 2 −1 x3 = 2 , 2 so that writing 2 P = 2 −1 −1 2 , 2 2 −1 2 we have 3 P −1 AP = 0 0 0 6 0 0 0. 9 Now let B = {p1 (x), p2 (x), p3 (x)}, where 2 [p1 (x)]B = 2 , −1 Chapter 8 : Linear Transformations 2 [p2 (x)]B = −1 , 2 −1 [p3 (x)]B = 2 . 2 page 29 of 34 c Linear Algebra W W L Chen, 1997, 2005 Then P is the transition matrix from the basis B to the basis B , and D is the matrix for T with respect to the basis B . Clearly p1 (x) = 2 + 2x − x2 , p2 (x) = 2 − x + 2x2 and p3 (x) = −1 + 2x + 2x2 . Note now that T (p1 (x)) = T (2 + 2x − x2 ) = 6 + 6x − 3x2 = 3p1 (x), T (p2 (x)) = T (2 − x + 2x2 ) = 12 − 6x + 12x2 = 6p2 (x), T (p3 (x)) = T (−1 + 2x + 2x2 ) = −9 + 18x + 18x2 = 9p3 (x). Problems for Chapter 8 1. Consider the transformation T : R3 → R4 , given by T (x1 , x2 , x3 ) = (x1 + x2 + x3 , x2 + x3 , 3x1 + x2 , 2x2 + x3 ) for every (x1 , x2 , x3 ) ∈ R3 . a) Find the standard matrix A for T . b) By reducing A to row echelon form, determine the dimension of the kernel of T and the dimension of the range of T . 2. Consider a linear operator T : R3 → R3 with standard matrix 123 A = 2 1 3. 132 Let {e1 , e2 , e3 } denote the standard basis for R3 . a) Find T (ej ) for every j = 1, 2, 3. b) Find T (2e1 + 5e2 + 3e3 ). c) Is T invertible? Justify your assertion. 3. Consider the linear operator T : R2 → R2 with standard matrix A= 1 0 1 1 . a) Find the image under T of the line x1 + 2x2 = 3. b) Find the image under T of the circle x2 + x2 = 1. 1 2 4. For each of the following, determine whether the given transformation is linear: a) T : V → R, where V is a real inner product space and T (u) = u . b) T : M2,2 (R) → M2,3 (R), where B ∈ M2,3 (R) is fixed and T (A) = AB . c) T : M3,4 (R) → M4,3 (R), where T (A) = At . d) T : P2 → P2 , where T (p0 + p1 x + p2 x2 ) = p0 + p1 (2 + x) + p2 (2 + x)2 . e) T : P2 → P2 , where T (p0 + p1 x + p2 x2 ) = p0 + p1 x + (p2 + 1)x2 . 5. Suppose that T : R3 → R3 is a linear transformation satisfying the conditions T (1, 0, 0) = (2, 4, 1), T (1, 1, 0) = (3, 0, 2) and T (1, 1, 1) = (1, 4, 6). a) Evaluate T (5, 3, 2). b) Find T (x1 , x2 , x3 ) for every (x1 , x2 , x3 ) ∈ R3 . 6. Suppose that T : R3 → R3 is orthogonal projection onto the x1 x2 -plane. a) Find the standard matrix A for T . b) Find A2 . c) Show that T ◦ T = T . Chapter 8 : Linear Transformations page 30 of 34 c Linear Algebra W W L Chen, 1997, 2005 7. Consider the bases B = {u1 , u2 , u3 } and C = {v1 , v2 , v3 } of R3 , where u1 = (2, 1, 1), u2 = (2, −1, 1), u3 = (1, 2, 1), v1 = (3, 1, −5), v2 = (1, 1, −3) and v3 = (−1, 0, 2). a) Find the transition matrix from the basis C to the basis B . b) Find the transition matrix from the basis B to the basis C . c) Show that the matrices in parts (a) and (b) are inverses of each other. d) Compute the coordinate matrix [u]C , where u = (−5, 8, −5). e) Use the transition matrix to compute the coordinate matrix [u]B . f) Compute the coordinate matrix [u]B directly and compare it to your answer in part (e). 8. Consider the bases B = {p1 , p2 } and C = {q1 , q2 } of P1 , where p1 = 2, p2 = 3 + 2x, q1 = 6 + 3x and q2 = 10 + 2x. a) Find the transition matrix from the basis C to the basis B . b) Find the transition matrix from the basis B to the basis C . c) Show that the matrices in parts (a) and (b) are inverses of each other. d) Compute the coordinate matrix [p]C , where p = −4 + x. e) Use the transition matrix to compute the coordinate matrix [p]B . f) Compute the coordinate matrix [p]B directly and compare it to your answer in part (e). 9. Let V be the real vector space spanned by the functions f1 = sin x and f2 = cos x. a) Show that g1 = 2 sin x + cos x and g2 = 3 cos x form a basis of V . b) Find the transition matrix from the basis C = {g1 , g2 } to the basis B = {f1 , f2 } of V . c) Compute the coordinate matrix [f ]C , where f = 2 sin x − 5 cos x. d) Use the transition matrix to compute the coordinate matrix [f ]B . e) Compute the coordinate matrix [f ]B directly and compare it to your answer in part (d). 10. Let P be the transition matrix from a basis C to another basis B of a real vector space V . Explain why P is invertible. 11. For each of the following linear transformations T , find ker(T ) and R(T ), and verify the Rank-nullity theorem: 1 −1 3 a) T : R3 → R3 , with standard matrix A = 5 6 −4 . 74 2 b) T : P3 → P2 , where T (p(x)) = p (x), the formal derivative. 1 c) T : P1 → R, where T (p(x)) = p(x) dx. 0 12. For each of the following, determine whether the linear operator T : Rn → Rn is one-to-one. If so, find also the inverse linear operator T −1 : Rn → Rn : a) T (x1 , x2 , x3 , . . . , xn ) = (x2 , x1 , x3 , . . . , xn ) b) T (x1 , x2 , x3 , . . . , xn ) = (x2 , x3 , . . . , xn , x1 ) c) T (x1 , x2 , x3 , . . . , xn ) = (x2 , x2 , x3 , . . . , xn ) 13. Consider the operator T : R2 → R2 , where T (x1 , x2 ) = (x1 + kx2 , −x2 ) for every (x1 , x2 ) ∈ R2 . Here k ∈ R is fixed. a) Show that T is a linear operator. b) Show that T is one-to-one. c) Find the inverse linear operator T −1 : R2 → R2 . 14. Consider the linear transformation T : P2 → P1 , where T (p0 + p1 x + p2 x2 ) = (p0 + p2 ) + (2p0 + p1 )x for every polynomial p0 + p1 x + p2 x2 in P2 . a) Find the matrix for T with respect to the bases {1, x, x2 } and {1, x}. b) Find T (2 + 3x + 4x2 ) by using the matrix A. c) Use the matrix A to recover the formula T (p0 + p1 x + p2 x2 ) = (p0 + p2 ) + (2p0 + p1 )x. Chapter 8 : Linear Transformations page 31 of 34 c Linear Algebra W W L Chen, 1997, 2005 15. Consider the linear operator T : R2 → R2 , where T (x1 , x2 ) = (x1 −x2 , x1 +x2 ) for every (x1 , x2 ) ∈ R2 . a) Find the matrix A for T with respect to the basis {(1, 1), (−1, 0)} of R2 . b) Use the matrix A to recover the formula T (x1 , x2 ) = (x1 − x2 , x1 + x2 ). c) Is T one-to-one? If so, use the matrix A to find the inverse linear operator T −1 : R2 → R2 . 16. Consider the real vector space of all real sequences x = (x1 , x2 , x3 , . . .) such that the series ∞ xn n=1 is convergent. a) Show that the transformation T : V → R, given by ∞ T (x) = xn n=1 for every x ∈ V , is a linear transformation. b) Is the linear transformation T one-to-one? If so, give a proof. If not, find two distinct vectors x, y ∈ V such that T (x) = T (y). 17. Suppose that T1 : R2 → R2 and T2 : R2 → R2 are linear operators such that T1 (x1 , x2 ) = (x1 + x2 , x1 − x2 ) and T2 (x1 , x2 ) = (2x1 + x2 , x1 − 2x2 ) for every (x1 , x2 ) ∈ R2 . a) Show that T1 and T2 are one-to-one. − − b) Find the formulas for T1 1 , T2 1 and (T2 ◦ T1 )−1 . −1 − −1 c) Verify that (T2 ◦ T1 ) = T1 ◦ T2 1 . 18. Consider the transformation T : P1 → R2 , where T (p(x)) = (p(0), p(1)) for every polynomial p(x) in P1 . a) Find T (1 − 2x). b) Show that T is a linear transformation. c) Show that T is one-to-one. d) Find T −1 (2, 3), and sketch its graph. 19. Suppose that V and W are finite dimensional real vector spaces with dim V > dim W . Suppose further that T : V → W is a linear transformation. Explain why T cannot be one-to-one. 20. Suppose that 1 A = 2 6 3 0 −2 −1 5 4 is the matrix for a linear operator T : P2 → P2 with respect to the basis B = {p1 (x), p2 (x), p3 (x)} of P2 , where p1 (x) = 3x + 3x2 , p2 (x) = −1 + 3x + 2x2 and p3 (x) = 3 + 7x + 2x2 . a) Find [T (p1 (x))]B , [T (p2 (x))]B and [T (p3 (x))]B . b) Find T (p1 (x)), T (p2 (x)) and T (p3 (x)). c) Find a formula for T (p0 + p1 x + p2 x2 ). d) Use the formula in part (c) to compute T (1 + x2 ). 21. Suppose that B = {v1 , v2 , v3 , v4 } is a basis for a real vector space V . Suppose that T : V → V is a linear operator, with T (v1 ) = v2 , T (v2 ) = v4 , T (v3 ) = v1 and T (v4 ) = v3 . a) Find the matrix for T with respect to the basis B . b) Is T one-to-one? If so, describe its inverse. Chapter 8 : Linear Transformations page 32 of 34 c Linear Algebra W W L Chen, 1997, 2005 22. Let Pk denote the vector space of all polynomials with real coefficients and degree at most k . Consider P2 with basis B = {1, x, x2 } and P3 with basis C = {1, x, x2 , x3 }. We define T1 : P2 → P3 and T2 : P3 → P2 as follows. For every polynomial p(x) = a0 + a1 x + a2 x2 in P2 , we have T1 (p(x)) = xp(x) = a0 x + a1 x2 + a2 x3 . For every polynomial q (x) in P3 , we have T2 (q (x)) = q (x), the formal derivative of q (x) with respect to the variable x. a) Show that T1 : P2 → P3 and T2 : P3 → P2 are linear transformations. b) Find T1 (1), T1 (x), T1 (x2 ), and compute the matrix A1 for T1 : P2 → P3 with respect to the bases B and C . c) Find T2 (1), T2 (x), T2 (x2 ), T2 (x3 ), and compute the matrix A2 for T2 : P3 → P2 with respect to the bases C and B . d) Let T = T2 ◦ T1 . Find T (1), T (x), T (x2 ), and compute the matrix A for T : P2 → P2 with respect to the basis B . Verify that A = A2 A1 . 23. Suppose that T : V → V is a linear operator on a real vector space V with basis B . Suppose that for every v ∈ V , we have x1 − x2 + x3 x1 [T (v)]B = x1 + x2 and [v]B = x2 . x1 − x2 x3 a) Find the matrix for T with respect to the basis B . b) Is T one-to-one? If so, describe its inverse. 24. For each of the following, let V be the subspace with basis B = {f1 (x), f2 (x), f3 (x)} of the space of all real valued functions defined on R. Let T : V → V be defined by T (f (x)) = f (x) for every function f (x) in V . Find the matrix for T with respect to the basis B : a) f1 (x) = 1, f2 (x) = sin x, f3 (x) = cos x b) f1 (x) = e2x , f2 (x) = xe2x , f3 (x) = x2 e2x 25. Let P2 denote the vector space of all polynomials with real coefficients and degree at most 2, with basis B = {1, x, x2 }. Consider the linear operator T : P2 → P2 , where for every polynomial p(x) = a0 + a1 x + a2 x2 in P2 , we have T (p(x)) = p(2x + 1) = a0 + a1 (2x + 1) + a2 (2x + 1)2 . a) Find T (1), T (x), T (x2 ), and compute the matrix A for T with respect to the basis B . b) Use the matrix A to compute T (3 + x + 2x2 ). c) Check your calculations in part (b) by computing T (3 + x + 2x2 ) directly. d) What is the matrix for T ◦ T : P2 → P2 with respect to the basis B ? e) Consider a new basis B = {1 + x, 1 + x2 , x + x2 } of P2 . Using a change of basis matrix, compute the matrix for T with respect to the basis B . f) Check your answer in part (e) by computing the matrix directly. 26. Consider the linear operator T : P1 → P1 , where for every polynomial p(x) = p0 + p1 x in P1 , we have T (p(x)) = p0 + p1 (x + 1). Consider also the bases B = {6 + 3x, 10 + 2x} and B = {2, 3 + 2x} of P1 . a) Find the matrix for T with respect to the basis B . b) Use Proposition 8V to compute the matrix for T with respect to the basis B . 27. Suppose that V and W are finite dimensional real vector spaces. Suppose further that B and B are bases for V , and that C and C are bases for W . Show that for any linear transformation T : V → W , we have [I ]C ,C [T ]C ,B [I ]B,B = [T ]C ,B . 28. Prove Proposition 8W. 29. Prove Proposition 8X. Chapter 8 : Linear Transformations page 33 of 34 c Linear Algebra W W L Chen, 1997, 2005 30. For each of the following linear transformations T : R3 → R3 , find a basis B of R3 such that the matrix for T with respect to the basis B is a diagonal matrix: a) T (x1 , x2 , x3 ) = (−x2 + x3 , −x1 + x3 , x1 + x2 ) b) T (x1 , x2 , x3 ) = (4x1 + x3 , 2x1 + 3x2 + 2x3 , x1 + 4x3 ) 31. Consider the linear operator T : P2 → P2 , where T (p0 + p1 x + p2 x2 ) = (p0 − 6p1 + 12p2 ) + (13p1 − 30p2 )x + (9p1 − 20p2 )x2 . a) Find the eigenvalues of T . b) Find a basis B of P2 such that the matrix for T with respect to B is a diagonal matrix. Chapter 8 : Linear Transformations page 34 of 34 ...
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This note was uploaded on 06/13/2009 for the course TAM 455 taught by Professor Petrina during the Fall '08 term at Cornell.

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