CHAPTER 5

[Pages:84]5 CHAPTER

Linear Transformations

5.1 Linear Transformations

Note: Many different arguments may be used to prove nonlinearity; our solutions to Problems 1?23 provide a sampling.

!

Checking Linearity

1.

T(x, y) = xy

If

a f a f u = u1, u2 , v = v1, v2 , a f a fa f T(u + v) = T u1 + v1, u2 + v2 = u1 + v1 u2 + v2

T(u) + T(v) = u1u2 + v1v2 .

We see that

T(u + v) T(u) + T(v),

so T is not a linear transformation.

2.

T(x, y) = (x + y, 2y)

We can write this transformation in matrix form as

T(x,

y) = LNM01

21OQP

LNMxyOQP

=

LNMx

+ 2

yyOQP

.

Hence, T is a linear transformation.

3.

T(x, y) = (xy, 2y)

a f If we let u = u1, u2 , we have a f a f a f cT(u) = cT u1, u2 = c u1u2, 2u2 = cu1u2, 2cu2

351

352 CHAPTER 5 Linear Transformations

and

a f b g T(cu) = T cu1, cu2 = c2u1u2, 2cu2 .

Hence cT(u) T(cu), so T is not a linear transformation.

4.

T(x, y) = (x, 2, x + y)

Note that T(0, 0) = (0, 2, 0) . Linear transformations always map the zero vector into the zero vector (in their respective spaces), so T is not a linear transformation.

5.

T(x, y) = (x, 0, 0)

We let

a f a f u = u1, u2 , v = v1, v2 so b g b g b g b g b g b g b g T u + v = T u1 + v1, u2 + v2 = u1 + v1, 0, 0 = u1, 0, 0 + v1, 0, 0 = T u + T v

and

a f a f cT(u) = c u1, 0, 0 = cu1, 0, 0 = T(cu).

Hence, T is a linear transformation from R2 to R3.

6.

T(x, y) = (x, 1, y, 1)

Because T does not map the zero vector 0, 0 R2 into the zero vector 0, 0, 0, 0 R4, T is not a linear transformation.

7.

T( f ) = f (0)

If f and g are continuous functions on 0, 1 , then

T( f + g) = ( f + g)(0) = f (0) + g(0) = T( f ) + T(g)

and T(cf ) = (cf )(0) = cf (0) = cT( f ).

Hence, T is a linear transformation.

8.

T( f ) = - f

If f and g are continuous functions on 0, 1 , then

T( f + g) = -( f + g) = - f - g = T( f ) + T(g)

and T(cf ) = -cf = c(- f ) = cT( f ).

Hence, T is a linear transformation.

SECTION 5.1 Linear Transformations 353

9.

T( f ) = tf (t)

If f and g are continuous functions on 0, 1 , then

T( f + g) = t f (t) + g(t) = tf (t) + tg(t) = T( f ) + T(g)

and T(cf ) = t(cf (t)) = ctf (t) = cT( f ).

Hence, T is a linear transformation. 10. T( f ) = f + 2 f + 3 f

If we are given that f and g are continuous functions that have two continuous derivatives, then T( f + g) = ( f + g) + 2( f + g) + 3( f + g) = ( f + 2 f + 3 f ) + (g + 2g + 3g) = T( f ) + T(g)

and T(cf ) = (cf ) + 2(cf ) + 3(cf ) = c( f + 2 f + 3 f ) = cT( f ).

Hence, T is a linear transformation.

b g 11. T at2 + bt + c = 2at + b

If we introduce the two vectors

p = a1t2 + b1t + c1 q = a2t2 + b2t + c2

then

a f a f a f a f a f T(p + q) = T a1 + a2 t2 + b1 + b2 t + c1 + c2 = 2 a1 + a2 t + b1 + b2 a f a f = 2a1t + b1 + 2a2t + b2 = T(p) + T(q)

and

b g a f T(cp) = T ca1t2 + cb1t + cc1 = 2ca1t + cb1 = c 2a1t + b1 = cT(p).

Hence, the derivative transformation defined on P2 is a linear transformation.

b g 12. T at3 + bt2 + ct + d = a + b

If we introduce the two vectors

p = a1t3 + b1t2 + c1t + d1 q = a2t3 + b2t2 + c2t + d2

354 CHAPTER 5 Linear Transformations

then

a f a f a f a f a f a f T(p + q) = T a1 + a2 t3 + b1 + b2 t2 + c1 + c2 t + d1 + d2 = a1 + a2 + b1 + b2 a f a f = a1 + b1 + a2 + b2 = T(p) + T(q)

b g b g T(cp) = T c a1t3 + b1t2 + c1t + d1 = T ca1t3 + cb1t2 + cc1t + cd1 = ca1 + cb1 a f = c a1 + b1 = cT(p).

Hence, the derivative transformation defined on P3 is a linear transformation. 13. T(A) = AT . If we introduce two 2 ? 2 matrices B and C, we have

T(B + C) = (B + C)T = BT + CT = T(B) + T(C) T(kB) = (kB)T = kBT = kT(B).

Hence, the transformation defined on M22 is a linear transformation.

14.

T LNMac

dbOQP

=

a c

b d

Letting

A = LNMac dbOQP

be an arbitrary vector, we show the homogeneous property T(kA) = kT(A) fails because

LNM OQP T(kA) = T

ka kc

kb kd

ka = kc

kb kd

=

k 2ad

-

k 2cb

=

k2

det(A)

=

k 2T(A)

kT(A)

when k 1. Hence, T is not a linear transformation.

15. TLNMac dbOQP = TrLNMac dbOQP

Let

LNM OQP LNM OQP A =

a11 a21

a12 a22

, B=

b11 b21

b12 b22

so

LNM OQP A + B =

a11 + b11 a21 + b21

a12 + b12 a22 + b22

.

Then

a f a f a f a f T(A + B) = a11 + b11 + a22 + b22 = a11 + a22 + b11 + b22 = T(A) + T(B)

SECTION 5.1 Linear Transformations 355

and

LNM OQP T(kA) = T

ka kc

kb kd

= ka + kd = k(a + d) = kT(A).

Hence, T is a linear transformation on M22 . 16. T(x) = Ax

T(x + y) = A(x + y) = Ax + Ay = T(x) + T(y)

and T(kx) = A(kx) = kAx = kT(x) .

Hence, T is a linear transformation.

!

Integration

17.

z z T(kf ) =

bkf

a

(t)dt

=

k

b a

f

(t)dt

=

kT(

f

)

z z z T(f

+ g) =

b a

f (t) + g(t) dt =

b a

f

(t

)dt

+

b a

g(t

)dt

=

T(

f

)

+

T( g) .

Hence, T is a linear transformation.

!

Laying Linearity on the Line

18. T(x) = x

b g b g b g T x + y = x + y T x + T y = x + y

so T(x + y) T(x) + T(y) . Hence, T is not a linear transformation.

19. T(x) = ax + b

b g b g b g b g T kx = a kx + b = akx + b kT x = k ax + b = akx + kb

so T(kx) kT(x) . Hence, T is not a linear transformation.

20.

T(x)

=

1 ax +

b

Not linear because when b 0, the zero vector does not map into the zero vector. Even when b = 0 the transformation is not linear because the zero vector (the real number zero) does not map into the zero vector (the real number zero).

356 CHAPTER 5 Linear Transformations

21. T(x) = x2

Because

T(2 + 3) = T(5) = 25 T(2) + T(3) = 4 + 9 = 13

we have that T is not linear. (You can also find examples where the property T(cx) = cT(x) fails.)

22. T(x) = sin x

Because

T(kx) = sin(kx) kT(x) = k sin x

we have that

T(kx) kT(x)

so T is not a linear transformation. We could also simply note that

FH IK T

2

+

2

= T( ) = sin(0) = 0

but

TFH

2

IK

+

TFH

2

IK

=

sinFH

2

IK

+

sinFH

2

IK

=

1

+

1

=

2

.

23.

T(

x)

=

-

3x 2+

Finally, we have a linear transformation. Any mapping of the form T(x) = ax , where a is a nonzero constant, is a linear transformation because

T(x + y) = a(x + y) = ax + ay = T(x) + T(y) T(kx) = a(kx) = k(ax) = kT(x).

In

this

problem

we

have

the

nonzero

constant

a

=

-

2

3 +

.

!

Geometry of a Linear Transformation

24. Direct computation: the vectors x, 0 for x real constitute the x-axis and because x, 0 maps

into itself the x-axis maps into itself.

25.

Direct computation: the vector (0, y) lies on the y-axis and (2y,

y) lies on the line

y

=

x 2

,

so

the

transformation maps vectors on the y-axis onto vectors on the line

y=

x 2

.

SECTION 5.1 Linear Transformations 357

b g 26. Direct computation: the transformation T maps points (x, y) into x + 2 y, y . For example, the

b g b g unit square with corner (0, 0), (1, 0) , (0, 1) , and (1, 1) map into the parallelogram with corners

(0, 0), (1, 0) , 2, 1 and 3, 1 . This transformation is called a shear mapping in the direction y.

!

Geometric Interpretations in R2

27. T(x, y) = (x, - y)

This map reflects points about the x-axis. A matrix representation is

LNM01 -01OQP.

y

a f x2, - y2

a f x1, y1

x

a f x2, y2

a f x1, - y1

28. T(x, y) = (x, 0)

This map reflects points about the x-axis. A matrix representation is

LNM01 00OQP.

y

a f x1, y1

a f x2, 0 a f x2, y2

a f x1, 0

x

29. T(x, y) = (x, x)

This map reflects points vertically about the 45degree line y = x . A matrix representation is

LNM11 00OQP.

a f x2, y2

y

a f x1, y1 a f x1, y1

x

a f x2, y2

!

Composition of Linear Transformations

30. (ST)(u + v) = S(T(u + v)) = S T(u) + T(v) = S(T(u)) + S(T(v)) = ST(u) + ST(v)

ST(cu) = S(T(cu)) = S(cT(u)) = cS(T(u)) = cST(u)

358 CHAPTER 5 Linear Transformations

!

Find the Standard Matrix

31. T(x, y) = x + 2y

T maps the point (x, y) R2 into the real number x + 2y R . In matrix form,

LNM OQP T(x, y) = 1 2

x y

= x + 2y.

32. T(x, y) = (y, - x)

T maps the point (x, y) R2 into the point (y, - x) R2 . In matrix form,

T(x, y) = LNM-01 01OQP LNMxyOQP = LNM-yxOQP.

33. T(x, y) = (x + 2y, x - 2y)

T maps the point (x, y) R2 into the point (x + 2y, x - 2y) R2 . In matrix form,

T(x,

y) = LNM11

-22OQP

LNM

xyOQP

=

LNMxx

+ -

2 2

yyOQP.

34. T(x, y) = (x + 2y, x - 2y, y)

a f a f a f T maps the point x, y R2 in two dimensions into the new point T x, y = x + 2y, x - 2y, y R3.

In matrix form, the linear transformation T can be written

T(x, y) = LNMMM011

-221OQPPP

LNM

xyOQP

=

LNMMMxx

+ -

2 2

yyyOQPPP

.

35. T(x, y, z) = (x + 2y, x - 2y, x + y - 2z)

T maps (x, y, z) R3 into (x + 2y, x - 2y, x + y - 2z) R3 . In matrix form,

T(x,

y) = LNMMM111

2 -2

1

-200OQPPP

LNMMMxyzOQPPP

=

LNMMMx

+

x x y

+ - -

222yyzOQPPP

.

36.

a f T 1, 2, 3 = 1 + 3

a f b g T maps the point 1, 2, 3 R3 into the real number T 1, 2, 3 = 1 + 3 R . In matrix

form,

L O1 a f M P T 1, 2, 3 = 1 0 1 2 = 1 + 3 .

NMM QPP3

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