linalg ch6 - kaist

Chapter 6

Linear Transformations

6.1 Matrices as Transformations

Definition 6.1.1. A function or a transformation f : Rn → Rm, domain,

image, The set f(Rn) is the range of f . The space Rm is the codomain.

matrix transformation, identity transformation

Example 6.1.2. Consider a transformation T : R2 → R2 determined by one

of the following three conditions:

(1) T maps a vector x = (x1, x2) to T (x) = (x1 + x2, 2x1 − x2)

(2) T maps a vector x = (x1, x2) to T (x) = (2x1, 3x2)

(3) T maps a vector x = (x1, x2) to T (x) = (x21, x2)

Example 6.1.3. Let

A =

0 1

1 2

−2 3

and let TA be a linear transformation T : R3 → R2 determined by the following

TA(x) = Ax.

The component of T is

TA

([

x1

x2

])

=

x2

x1 + 2x2

−2x1 + 3x2

(6.1)

81

82 CHAPTER 6. LINEAR TRANSFORMATIONS

Definition 6.1.4. If for any vector u, v ∈ Rn and scalar α ∈ R, a function

T : Rn → Rm satisfies

(1) T (αu) = αT (u) (Homogeneity)

(2) T (u+ v) = T (u) + T (v) (Additivity)

then we say T is a linear transformation(mapping). In the special case

when m = n, T is called an operator.

Example 6.1.5. Express a given linear transformation T : Rn → Rm using

the standard basis vector.

Since any vector x in Rn can be written as x = x1e1 + x2e2 + · · ·+ xnen,

T is determined by the values of these vectors. Thus we see

T (x1e1 + x2e2 + · · ·+ xnen) = T (x1e1) + T (x2e2) + · · ·+ T (xnen)

= x1T (e1) + x2T (e2) + · · ·+ xnT (en)

= [T (e1)T (e2) · · · T (en)]

x1

x2...

xn

Since T (e1), T (e2), . . . , T (en) are vectors in Rm, we can write it as a linear

combination of e1, e2, . . . , em. Hence there exist numbers tij (1 ≤ i ≤ m, 1 ≤

j ≤ n) s.t.

T (e1) =

t11

t21...

tm1

, T (e2) =

t12

t22...

tm2

, · · · , T (en) =

t1n

t2n...

tmn

(6.2)

In other words,

T (ej) =m∑

i=1

tijei (1 ≤ j ≤ n) (6.3)

Hence

T (x1e1 + x2e2 + · · ·+ xnen) =

n∑

j=1

xjT (ej) =

m∑

i=1

n∑

j=1

tijxj

ei (6.4)

6.1. MATRICES AS TRANSFORMATIONS 83

The matrix having tij as ij-th component

A =

t11 t12 · · · t1n

t21 t22 · · · t2n...

.... . .

...

tm1 tm2 · · · tmn

is called the standard matrix for T . We have

T (x) = Ax =

t11 t12 · · · t1n

t21 t22 · · · t2n...

.... . .

...

tm1 tm2 · · · tmn

x1

x2...

xn

Conversely, assume any m×n matrix A = [tij] is given. Then it determines

a linear transformation TA : Rn → Rm by

TA(x) = Ax. (6.5)

Hence the set of all linear transformations T : Rn → Rm has one-one corre-

spondence with the m× n matrices as follows:

T 7→ A =[

ei · T (ej)]

1 ≤ i ≤ m

1 ≤ j ≤ n

Proposition 6.1.6. If B and A are two matrices that correspond to the trans-

formations T : Rn → Rm and S : Rp → R

n resp., and if C is the matrix of the

composite transformation T ◦ S, then it holds that

C = BA.

Example 6.1.7. For the given two linear transformations T : Rn → Rm,

S : Rp → Rn check that the proposition holds. 6.1.6 holds.

T (x, y, z) = (3y − z, x+ y)

S(s, t) = (2s− t, s+ 2t,−3s)


sol. The matrices for T and S are

B =

[

0 3 −1

1 1 0

]

, A =

2 −1

1 2

−3 0

Hence

BA =

[

0 3 −1

1 1 0

]

2 −1

1 2

−3 0

=

[

6 6

3 1

]

On the other hand,

(T ◦ S)(s, t) = T (2s− t, s+ 2t,−3s)

= (3(s + 2t)− (−3s), (2s − t) + (s+ 2t))

= (6s+ 6t, 3s + t)

So the matrix of the composite transformation T ◦ S is

C =

[

6 6

3 1

]

which equals BA.

Rotation about the origin

Consider the figure of rotating the standard vectors e1 and e2. Then we can

easily see that the matrix is given by

Rθ =[

T (e1) T (e2)]

=

[

cos θ − sin θ

sin θ cos θ

]

(6.6)

Example 6.1.8. Find the image of x = (√

32 , 12 ) under a rotation of θ = π/6

about the origin with.

Rπ/6x =

[√

32 −1

212

√

32

][√

3212

]

=

[

12√32

]

6.1. MATRICES AS TRANSFORMATIONS 85

(cos θ, sin θ)

θ

O x

y

Figure 6.1: Rotation of e1 and e2 by θ

θ

O x

y

x

T (x)

θ

2θ

(cos 2θ, sin 2θ)

O x

y

e1

π

2− θ

θ

2θ − π

2≡ ψ

(cosψ, sinψ)

O

y

e2

Figure 6.2: Reflection about a line/Reflection of e1 and e2

Reflection about a line through the origin

Now consider (See figure 6.2) the reflection of the standard vectors e1 and e2

w.r.t a line through the origin making angle θ with positive x-axis. Then we

can easily see that the matrix is given by

Hθ =[

T (e1) T (e2)]

=

[

cos 2θ cos(π2 − 2θ)

sin 2θ − sin(π2 − 2θ)

]

=

[

cos 2θ sin 2θ

sin 2θ − cos 2θ

]

(6.7)

Example 6.1.9. Find the image of x = (1, 1) under the reflection about the

line through origin that makes an angle π/6 with the x-axis is

sol.

Hπ/6x =

[

12

√

32√

32 −1

2

][

1

1

]

=

[

1+√

32√

3−12

]


θ

O x

y

x

Hθx

Pθx

Figure 6.3: Orthogonal Projection onto a line

Three typical Reflections

(1) Reflection w.r.t y-axis is given by the standard matrix

[

−1 0

0 1

]

(2) Reflection w.r.t x-axis is given by the standard matrix

[

1 0

0 −1

]

(3) Reflection w.r.t the line y = x is given by the standard matrix

[

0 1

1 0

]

Orthogonal Projection onto a line through the origin

The orthogonal projection(refer to figure 6.3) of x about a line through the

origin that makes an angle θ with the x-axis is given by

Pθx =1

2(Hθx+ x) =

1

2(Hθ + I)x.

Hence the the matrix representation(standard matrix) of the orthogonal pro-

jection is

Pθ =1

2(Hθ + I) (6.8)

By 6.7, we have

Pθ =

[

12 (1 + cos 2θ) 1

2 sin 2θ12 sin 2θ

12 (1− cos 2θ)

]

=

[

cos2 θ sin θ cos θ

sin θ cos θ sin2 θ

]

(6.9)

Example 6.1.10. Find the orthogonal projection of x = (1, 1) on the line

through origin with θ = π/12.

6.2. GEOMETRY OF LINEAR OPERATORS 87

sol.

Pπ/12x =

[

12(1 + cos π

6 )12 sin

π6

12 sin

π6

12(1− cos π

6 )

] [

1

1

]

=

[

12 (1 +

√

32 ) 1

414

12(1−

√

32 )

] [

1

1

]

.

Transformation of a unit square

Power Sequence

6.2 Geometry of Linear Operators

Norm Preserving Linear Operators

Consider the linear operators given by the following matrices

Rθ =

[

cos θ − sin θ

sin θ cos θ

]

, Hθ =

[

cos 2θ sin 2θ

sin 2θ − cos 2θ

]

, Pθ =

[

cos2 θ sin θ cos θ

sin θ cos θ sin2 θ

]

Among these, the rotation and reflection do not change the length of a vector

and the angle between two vectors.

We say these operators are length preserving and angle preserving.

Definition 6.2.1. A linear operator T : Rn → Rn with length preserving

property ‖T (x)‖ = ‖x‖ is called an orthogonal operator.

Theorem 6.2.2. If T : Rn → Rn is a linear operator, then the followings are

equivalent.

(1) ‖T (x)‖ = ‖x‖ for all x ∈ Rn.

(2) T (x) · T (y) = x · y for all x,y ∈ Rn.

Proof. (1) ⇒ (2). We observe the following identity (polarization identity):

x · y =1

4(‖x+ y‖2 − ‖x− y‖2) (6.10)

Hence


T (x) · T (y) =1

4(‖T (x) + T (y)‖2 − ‖T (x) − T (y)‖2)

=1

4(‖T (x+ y)‖2 − ‖T (x− y)‖2)

=1

4(‖x+ y‖2 − ‖x− y‖2)

= x · y.

(2) ⇒ (1). Put x = y in (2) to get (1).

Orthogonal Operators Preserve Angle and Orthogonality

Definition 6.2.3. A square matrix is said to be orthogonal if A−1 = AT .

If an orthogonal operator T is represented by the matrix A then

T (x) = Ax.

Since ‖T (x)‖ = ‖x‖ we see by above theorem

x · y = Ax ·Ay = xTATAy (6.11)

for all x and y. Thus it is easy to deduce ATA = I by putting x = ei,y =

ej, i, j = 1, 2, · · · , n. Thus we have

Theorem 6.2.4. A linear operator T : Rn → Rn is orthogonal if and only if

the standard matrix is orthogonal.

Example 6.2.5.

A =

1√

2− 1

√

31√

6

0 1√

32√

61√

21√

3− 1

√

6

is an orthogonal matrix.

If T : Rn → Rn is an orthogonal operator, then

θ = cos−1

(

T (x) · T (y)

‖T (x)‖‖T (y)‖

)

= cos−1

(

x · y

‖x‖‖y‖

)

. (6.12)

This implies orthogonal operator preserves angles.


Theorem 6.2.6. (1) The transpose of an orthogonal matrix is orthogonal.

(2) The inverse of an orthogonal matrix is orthogonal.

(3) The product of two orthogonal matrices is orthogonal.

(4) If A is orthogonal, then det(A) = ±1.

Proof. (3)

(AB)−1 = B−1A−1 = BTAT = (AB)T . (6.13)

Theorem 6.2.7. If A is an m×n is matrix, then the followings are equivalent.

(1) ATA = I.

(2) ‖Ax‖ = ‖x‖ for all x ∈ Rn.

(3) Ax ·Ay = x · y for all x,y ∈ Rn.

(4) The column vectors of A are orthonormal.

(5) The row vectors of A are orthonormal if m = n.

Proof. (2) ⇒ (3) requires polarization identity which holds for m× n matrix

also.

All Orthogonal Linear Operators on R2 are Rotations and Re-

flections

Theorem 6.2.8. If T : R2 → R2 is an orthogonal linear operator, then the

matrix representation(standard matrix) for T is of the form

Rθ =

[

cos θ − sin θ

sin θ cos θ

]

or Hθ/2 =

[

cos θ sin θ

sin θ − cos θ

]

(6.14)

Thus T is either a rotation(det(T ) = 1) or a reflection(det(T ) = −1).

Proof. ....

Contractions and Dilations on R2

The operator T (x, y) = (kx, ky) is called a scaling. It is called a contraction

if 0 ≤ k < 1 and called a dilation if k > 1.


O x

y

(1, 0) k > 0 k < 0

Figure 6.4: Shears along x

O x

y

(1, 0) k > 0 k < 0

Figure 6.5: Shears along y

Vertical Expansion and Compression on R2

The operator T (x, y) = (kx, y) is called an expansion in the x-direction if

k > 1, ( compression if k < 1). Similarly, the operator T (x, y) = (x, ky) is

called an expansion in the y-direction if k > 1, ( compression if k < 1).

Example 6.2.9. Refer to Tables 6.2.1 and 6.2.2 in p. 286. The contraction(or

a dilation), the expansion(or compression) in the x-direction, and expansion(or

compression) in the y-direction are given by the following matrices respectively.

[

k 0

0 k

]

,

[

k 0

0 1

]

, and

[

1 0

0 k

]

.

Shears

The operator T (x, y) = (x + ky, y) is a shear in the x-direction while the

operator T (x, y) = (x, y + kx) is a shear in the y-direction. They are

represented by[

1 k

0 1

]

, and

[

1 0

k 1

]

respectively.


Linear Operators on R3

Basically we study two kinds of operators.

(1) length preserving operators (orthogonal operators)

(2) length non-preserving operators(e.g., projections)

First, examples of length non-preserving operators are projections.

Projection onto xy-plane is given by T (x, y, z) = (x, y, 0). Thus the stan-

dard matrix is

A =

1 0 0

0 1 0

0 0 0

Projection onto other planes are given by similar form of matrices.

Now we study length preserving operators (orthogonal operators).

Theorem 6.2.10. If T : R3 → R3 is an orthogonal linear operator, then one

of the following holds.

(1) Rotations about line through the origin.

(2) Reflections about planes through the origin.

(3) A rotation about line through the origin followed by a reflection about

the plane through the origin that is perpendicular to the line.

In fact, a 3 × 3 orthogonal matrix A represents a rotation if det(A) = 1 and

represents a type 2 or type 3 operator if det(A) = −1.

Reflections about Coordinate Planes

Reflections about xy-plane, yz-plane and zx-plane are respectively given by

T (x, y, z) = (x, y,−z), T (x, y, z) = (−x, y, z) and T (x, y, z) = (x,−y, z).

Their standard matrices are

1 0 0

0 1 0

0 0 −1

,

−1 0 0

0 1 0

0 0 1

, and

1 0 0

0 −1 0

0 0 1


Rotations in R3

Consider the rotations in R3: The key point is to know the axis of rotation

and the orientation(direction of rotation). To find the axis of rotation, we

note that the vector along the axis is not changed under the rotation. Next

we consider how to choose direction of rotation. For example, we choose a

vector w in the plane and rotate it by T . Let

u = w × T (w)

Take the view point at the tip of the vector u(right handed rule). We call this

u an orientation of the rotation.

w × T (w)

θ

T (w)

w

W

Figure 6.6: Oriented axis of rotation

x

y

z

b

θ

Figure 6.7: The rotation about x-axis by θ

Two questions

There are two questions regarding the rotations in R3.


(1) Given a rotation, how to find the standard matrix representing the ro-

tation?

(2) Given a standard rotation matrix A, how to find the axis and angle of

rotation ?

To answer the first question, consider, for example, the rotation about x-

axis by θ. Under the given rotation, the vector e1 is fixed, while e2 and e3 are

mapped to (0, cos θ, sin θ) and (0,− sin θ, cos θ), resp. i.e.,

e1Rx,θ⇒

1

0

0

, e2

Rx,θ⇒

0

cos θ

sin θ

, e3

Rx,θ⇒

0

− sin θ

cos θ

Thus the rotation about x-axis by θ is given by the standard matrix

Rx,θ =

1 0 0

0 cos θ − sin θ

0 sin θ cos θ

To answer the second question, note that the axis remains fixed under the

matrix A, we have

(I −A)x = 0

Once we know the axis, choose a nonzero vector w perpendicular to the

axis and compute Aw. Finally, we obtain the angle of rotation by the formula:

cos θ =w ·Aw

‖w‖ ‖Aw‖. (6.15)

Example 6.2.11. (1) Show the matrix represents a rotation about a line

through the origin.

A =

1 0 0

0 0 −1

0 1 0

(2) Find the axis and angle of the rotation given by the matrix


(3) Repeat the question with

A =

0 1 0

1 0 0

0 0 −1

sol. (1) Since A is orthogonal and det(A) = 1, it is a rotation. To find the

axis of rotation, solve (I −A)x = 0, i.e.,

0 0 0

0 1 1

0 −1 1

x

y

z

=

0

0

0

⇒

x

y

z

=

1

0

0

Thus it is the rotation in yz-plane.

(2) To find the angle, choose any vector, say w = (0, 1, 0) in the yz-plane

and compute Aw = (0, 0, 1). Thus

cos θ =w ·Aw

‖w‖ ‖Aw‖= 0 (6.16)

and hence θ = π/2.

(3) (I − A)x = 0 gives x = (1, 1, 0). The plane of rotation is a plane

perpendicular to (1, 1, 0). i.e., vectors lying in the plane satisfies

x+ y = 0.

Hence we choose a vector as w = (1,−1, 0) and we see Aw = (−1, 1, 0). Hence

the angle is

θ = cos−1 w ·Aw

‖w‖ ‖Aw‖= cos−1

(

−2

2

)

= π. (6.17)

There is a formula: From Theorem 6.2.8 of the book, we have

tr(A) = (a2 + b2 + c2)(1 − cos θ) + 3 cos θ = 1 + 2 cos θ

from which we have

cos θ =tr(A)− 1

2. (6.18)

6.3. KERNEL AND RANGE OF A TRANSFORMATION 95

6.3 Kernel and Range of a Transformation

Kernel

Let T is a linear operator on Rn and v ∈ R

n. Let x = tv be a line through

the origin. Then

T (x) = T (tv) = tT (v).

We will investigate the geometry of the image. There are two possibilities:

(1) If T (v) = 0 then T (x) = 0 for all x ∈ Rn. So the image is the one point

0.

(2) If T (v) 6= 0 then the image T (x) is a line through the origin.

Similarly, if x = t1v1 + t2v2 is a plane through the origin, then the image

under T is the set

T (x) = t1T (v1) + t2T (v2)

Hence there are three possibilities:

(1) If T (v1) = 0 and T (v2) = 0, then T (x) = 0 for all x ∈ Rn. So the image

is the one point 0.

(2) If T (v1) 6= 0 and T (v2) 6= 0, then the image T (x) is a plane through the

origin.

(3) In the other cases, the image T (x) is the line through the origin.

Definition 6.3.1. If T is a linear operator Rn → R

m, then the set of all

vectors in Rn that are mapped to 0 by T is called kernel of T and denoted

by ker(T ).

Example 6.3.2. Find the kernel of the following linear operators in R3.

(1) The zero operator T0(x) = 0x = 0.

(2) The identity operator I(x) = Ix = x.

(3) The orthogonal projection on the xy-plane.

(4) A rotation about the line through the origin with angle θ.

Theorem 6.3.3. If T : Rn → R

m is a linear transformation, then the kernel

of T is a subspace of Rn.


Kernel of a Matrix Transformation

Theorem 6.3.4. The kernel of an m × n matrix transformation A is the

solution of Ax = 0.

Definition 6.3.5. If A is m× n matrix, then the solution space of the linear

system Ax = 0 (or, equivalently, the kernel of TA) is called the null space of

A and denoted by null(A).

Example 6.3.6. Find the null space of the following matrix.

A =

1 3 −2 0 2 0

2 6 −5 −2 4 −3

0 0 5 10 0 15

2 6 0 8 4 18

Reducing to row echelon form by Gaussian(Jordan) elimination, we obtain

1 3 0 4 2 0

0 0 1 2 0 0

0 0 0 0 0 1

0 0 0 0 0 0

Hence we get

x1 +3x2 +4x4 +2x5 = 0

x3 +2x4 = 0

x6 = 0

(6.19)

Solving

x1 = −3x2 − 4x4 − 2x5

x2 = −2x4

x6 = 0

(6.20)

Thus we have

x1 = −3r − 4s− 2t, x2 = r, x3 = −2s, x4 = s, x5 = t, x6 = 0

and we find the solutions of Ax = 0 are linear combinations of the following

vectors

[−3, 1, 0, 0, 0, 0], [−4, 0,−2, 1, 0, 0], [−2, 0, 0, 0, 1, 0].


m is a linear transformation, then T maps

6.3. KERNEL AND RANGE OF A TRANSFORMATION 97

subspaces of Rn into subspaces of Rm.

Range of a Linear Transformation

Definition 6.3.8. If T : Rn → Rm is a linear transformation, then the set of

all vectors in Rm that consist of all the images of T is called the range of T

and denoted by ran(T ).

Example 6.3.9. Find the range of the following linear operators in R3.

(1) The zero operator T0(x) = 0x = 0.

(2) The identity operator I(x) = Ix = x.

(3) The orthogonal projection on the xy-plane.

(4) A rotation about the line through the origin with angle θ.


m is a linear transformation, then ran(T )

is a subspace of Rm.

Range of a Matrix Transformation

Theorem 6.3.11. If A is an m×n matrix, then the range of the corresponding

linear transformation is the column space of A.

Example 6.3.12. Determine whether the vector b is in the range(column

space) of A when A and b are given as follows.

A =

1 −8 −7 −4

2 −3 −1 5

3 2 5 14

, and b =

8

−10

−28

sol. Reduce the augmented matrix to the echelon form:

1 0 1 4 −8

0 1 1 1 −2

0 0 0 0 0

From this we get

x1 = −8− s− 4t, x2 = −2− s− t, x3 = s, x4 = t.


Thus there is at least one solution which is

x1 = −8, x2 = −2, x3 = x4 = 0.

Thus b is in the range of A.

Existence and Uniqueness

Given a linear transformation T : Rn → Rm we have the following question:

(1) The existence question- Is every vector in Rm the image of at least one

vector in Rn ?

(2) Can two distinct vectors in Rn have the same image in R

m ?

Definition 6.3.13. A linear transformation T : Rn → Rm is said to be onto

if its range is entire codomain Rm; that is every vector in R

m is the image of

at least one vector in Rn.

Definition 6.3.14. A linear transformation T : Rn → Rm is said to be one-

to-one if T maps distinct vectors in Rn into distinct vectors in R

m.


m is a linear transformation, then the

following statements are equivalent:

(1) T is one-to-one.

(2) ker(T ) = 0.

One-to-one and Onto from the Viewpoint of Linear System

Theorem 6.3.16. If A is an m × n matrix, then the corresponding linear

transformation TA is one-to-one if and only if the linear system Ax = 0 has

trivial solution.

Theorem 6.3.17. If A is an m × n matrix, then the corresponding linear

transformation TA is onto if and only if the linear system Ax = b is consistent

for every b ∈ Rm.

Theorem 6.3.18. If A is an n × n matrix, then the corresponding linear

transformation TA is one-to-one if and only if it is onto.

6.4. COMPOSITIONAND INVERTIBILITY OF LINEAR TRANSFORMATIONS99

A Unifying Theorem- Theorem 6.3.15

Theorem 6.3.19. (1) Reduced echelon form is In.

(2) A is expressible as a product of elementary matrices.

(3) A is invertible.

(4) Ax = 0 has only trivial solution.

(5) Ax = b is consistent for any b ∈ Rn.

(6) Ax = b has exactly one solution for any b ∈ Rn.

(7) The column vectors are linearly independent.

(8) The row vectors are linearly independent.

(9) det(A) 6= 0.

(10) λ = 0 is not an eigenvalue of A.

(11) TA is one-to-one.

(12) TA is onto.

6.4 Composition and Invertibility of Linear Trans-

formations

Composition of Linear Transformations

If T1 : Rn → Rk and T2 : Rk → R

m are two linear transformations, then we

can consider their composition, called composition of T2 with T1.

(T2 ◦ T1)(x) = T2(T1(x)) (6.21)

Theorem 6.4.1. If T1 : Rn → R

k and T2 : Rk → R

m are two linear transfor-

mations, then (T2 ◦ T1) : Rn → R

m is also a linear transformation.

Theorem 6.4.2. If A : k×n and B : n×m are two linear matrices, then the

m× n matrix BA is the standard matrix of the corresponding composition of

linear transformations.


Example 6.4.3. Consider composition of two rotation transformations:

Rx,θ1 =

[

cos θ1 − sin θ1

sin θ1 cos θ1

]

and Rx,θ2 =

[

cos θ2 − sin θ2

sin θ2 cos θ2

]

(6.22)

Rx,θ2Rx,θ1 =

[

cos θ2 − sin θ2

sin θ2 cos θ2

] [

cos θ1 − sin θ1

sin θ1 cos θ1

]

=

[

cos θ3 − sin θ3

sin θ3 cos θ3

]

(6.23)

where θ3 = θ1 + θ2.

Thus Rx,θ2Rx,θ1 = Rx,θ1+θ2 .

Example 6.4.4. Consider composition of two reflections:

Hθ1 =

[

cos 2θ1 sin 2θ1

sin 2θ1 − cos 2θ1

]

and Hθ2 =

[

cos 2θ2 sin 2θ2


]

. (6.24)

Hθ2Hθ1 =

[

cos 2θ2 sin 2θ2


][

cos 2θ1 sin 2θ1


]

=

[

cos 2(θ2 − θ1) − sin 2(θ2 − θ1)

sin 2(θ2 − θ1) cos 2(θ2 − θ1)

]

Thus the composition of two reflections is a rotation:

Hθ2Hθ1 = R2(θ2−θ1)

This can be expected from the fact that the determinant of product of the

corresponding matrices is det(Hθ2Hθ1) = det(Hθ2)det(Hθ1) = (−1)(−1) = 1.

Example 6.4.5 (Composition is not a commutative operation). Consider the

composition of two matrices:

A1 =

[

1 2

0 1

]

and A2 =

[

0 1

1 0

]

. (6.25)

We see

A2A1 =

[

0 1

1 2

]

, while A1A2 =

[

2 1

1 0

]

6.4. COMPOSITIONAND INVERTIBILITY OF LINEAR TRANSFORMATIONS101

Composition of Three Transformations

If

T1 : Rn → R

k , T2 : Rk → R

ℓ and T3 : Rℓ → R

m

are three linear transformations, then we define their composition Rn → R

m

by

(T3 ◦ T2 ◦ T1)(x) = T3(T2(T1(x))) (6.26)

Factoring of a Linear Operators

Example 6.4.6. Consider

D =

[

λ1 0

0 λ2

]

=

[

1 0

0 λ2

][

λ1 0

0 1

]

= D1D2. (6.27)

Theorem 6.4.7. If A is an invertible 2 × 2 matrix, then the corresponding

linear operator is a composition of shears, compression, and expansions in the

direction of coordinate axes, and the reflection about the coordinate axes and

about the line y = x.

Example 6.4.8. Express the following as a composition of operator with

geometric meaning.

A =

[

1 2

3 4

]

Use elementary row operations:

[

1 2

0 −2

]

E21(−3)→

[

1 2

0 1

]

R2(−12)

→

[

1 2

0 1

]

(6.28)

So[

1 0

0 −12

] [

1 0

−3 1

]

A =

[

1 2

0 1

]

we have

A =

[

1 0

3 1

][

1 0

0 −2

] [

1 2

0 1

]

Inverse of a Linear Transformation

If T : Rn → Rm is a one-to-one linear transformation, then each vector w

in the range of T is the image of a unique vector x in Rn. We call it the


preimage of w. In this case we can define the inverse of T by

T−1(w) = x if and only if T (x) = w


n is a one-to-one linear operator, and if A is

the standard matrix of T , then A is invertible and A−1 is the standard matrix

of T−1.

Image of the Unit Square under a Linear Transformation

Example 6.4.10. Find the image of the unit square under the following linear

operators.

A[T (e1) T (e2)] =

[

x1 x2

y1 y2

]

.

Theorem 6.4.11. If T : R2 → R

2 is an invertible linear operator, then

the area of the parallelogram formed by T (e1) and T (e2) is det (A), where

A =[

T (e1) T (e2)]

.

linalg ch6 - kaist

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