4.10 Properties of Matrix Transformations PDF

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270

Chapter 4 General Vector Spaces

(b) Show that if T : R 3 →R 3 is an orthogonal projection onto one of the coordinate axes, then for every vector x in R 3 , the vectors T(x) and x − T(x) are orthogonal.

(c) Make a sketch showing x and x − T(x) in the case where T is the orthogonal projection onto the x -axis.

4.10 Properties of MatrixTransformations In this section we will discuss properties of matrix transformations. We will show, for example, that if several matrix transformations are performed in succession, then the same result can be obtained by a single matrix transformation that is chosen appropriately. We will also explore the relationship between the invertibility of a matrix and properties of the corresponding transformation.

Compositions of Matrix Transformations

Suppose that TA is a matrix transformation from Rn to Rk and TB is a matrix transformation from Rk to Rm . If x is a vector in Rn , then TA maps this vector into a vector TA (x) in Rk , and TB , in turn, maps that vector into the vector TB (TA (x)) in Rm . This process creates a transformation from Rn to Rm that we call the composition of T B with TA and denote by the symbol

TB ◦ TA

which is read “TB circle TA .” As illustrated in Figure 4.10.1, the transformation TA in the formula is performed first; that is, (1)

(TB ◦ TA )(x) = TB (TA (x)) This composition is itself a matrix transformation since

(TB ◦ TA )(x) = TB (TA (x)) = B(TA (x)) = B(Ax) = (BA)x

which shows that it is multiplication by BA. This is expressed by the formula (2)

TB ◦ TA = TBA TA Rn

Figure 4.10.1

x

TB Rk

TA(x)

Rm

TB (TA(x))

TB ° TA

Compositions can be defined for any finite succession of matrix transformations whose domains and ranges have the appropriate dimensions. For example, to extend Formula (2) to three factors, consider the matrix transformations

TA : Rn → Rk , T B : Rk → Rl , TC : Rl → Rm

We define the composition (TC ◦ TB ◦ TA ): Rn →Rm by

(TC ◦ TB ◦ TA )(x) = TC (TB (TA (x)))

As above, it can be shown that this is a matrix transformation whose standard matrix is CBA and that TC ◦ TB ◦ TA = TCBA (3) Sometimes we will want to refer to the standard matrix for a matrix transformation

T : Rn →Rm without giving a name to the matrix itself. In such cases we will denote the standard matrix for T by the symbol [T ]. Thus, the equation T (x) = [T ]x

4.10 Properties of MatrixTransformations

271

states that T (x) is the product of the standard matrix [T ] and the column vector x. For example, if T1 : Rn →Rk and if T2 : Rk →Rm , then Formula (2) can be restated as (4)

[T2 ◦ T1 ] = [T2 ][T1 ] Similarly, Formula (3) can be restated as

(5)

[T3 ◦ T2 ◦ T1 ] = [T3 ][T2 ][T1 ] E X A M P L E 1 Composition Is Not Commutative

Let T1 : R2 →R2 be the reflection about the line y = x , and let T2 : R2 →R2 be the orthogonal projection onto the y -axis. Figure 4.10.2 illustrates graphically that T1 ◦ T2 and T2 ◦ T1 have different effects on a vector x. This same conclusion can be reached by showing that the standard matrices for T1 and T2 do not commute:

WARNING Just as it is not gen-

erally true for matrices that AB = BA, so it is not generally true that

TB ◦ TA = TA ◦ TB That is, order matters when matrix transformations are composed. In those special cases where the order does not matter we say that the linear transformations commute .

[T1 ◦ T2 ] = [T1 ][T2 ] =



0 1

1 0



0 0 1 = 1 0 0

[T2 ◦ T1 ] = [T2 ][T1 ] =



0 0

0 1



1 0 0 = 0 1 0

0 0

0 1













so [T2 ◦ T1 ] = [T1 ◦ T2 ]. y

y T1(x)

y=x

y=x

T2 (T1 (x))

x

T2 (x)

x x

x T1 (T2 (x))

T2 ° T1

Figure 4.10.2

T1 ° T2

y T 2 (T1 (x))

E X A M P L E 2 Composition of Rotations Is Commutative

T1 (x)

θ1 + θ 2

θ2

x

Let T1 : R2 →R2 and T2 : R2 →R2 be the matrix operators that rotate vectors about the origin through the angles θ1 and θ2 , respectively. Thus the operation

(T2 ◦ T1 )(x) = T2 (T1 (x))

θ1 x

Figure 4.10.3

first rotates x through the angle θ1 , then rotates T1 (x) through the angle θ2 . It follows that the net effect of T2 ◦ T1 is to rotate each vector in R2 through the angle θ1 + θ2 (Figure 4.10.3). The standard matrices for these matrix operators, which are

[T1 ] =

   cos θ2 − sin θ2 − sin θ1 , , [T2 ] = cos θ1 sin θ2 cos θ2   cos(θ1 + θ2 ) − sin(θ1 + θ2 ) [T2 ◦ T1 ] = sin(θ1 + θ2 ) cos(θ1 + θ2 ) 

cos θ1 sin θ1

should satisfy (4). With the help of some basic trigonometric identities, we can confirm that this is so as follows:

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Chapter 4 General Vector Spaces

Using the notation Rθ for a rotation of R 2 about the origin through an angle θ , the computation in Example 2 shows that

[T2 ][T1 ] =



cos θ2 sin θ2

− sin θ2  

cos θ1

cos θ2

− sin θ1

sin θ1

cos θ1



−(cos θ2 sin θ1 + sin θ2 cos θ1 ) = sin θ2 cos θ1 + cos θ2 sin θ1 − sin θ2 sin θ1 + cos θ2 cos θ1   cos(θ1 + θ2 ) − sin(θ1 + θ2 ) = sin(θ1 + θ2 ) cos(θ1 + θ2 )

Rθ1 Rθ2 = Rθ1 +θ2

This makes sense since rotating a vector through an angle θ1 and then rotating the resulting vector through an angle θ2 is the same as rotating the original vector through the angle θ1 + θ2 .



cos θ2 cos θ1 − sin θ2 sin θ1



= [T2 ◦ T1 ] E X A M P L E 3 Composition of Two Reflections

Let T1 : R2 →R2 be the reflection about the y -axis, and let T2 : R2 →R2 be the reflection about the x -axis. In this case T1 ◦ T2 and T2 ◦ T1 are the same; both map every vector x = (x, y) into its negative −x = (−x, −y) (Figure 4.10.4):

(T1 ◦ T2 )(x, y) = T1 (x, −y) = (−x, −y)

(T2 ◦ T1 )(x, y) = T2 (−x, y) = (−x, −y) The equality of T1 ◦ T2 and T2 ◦ T1 can also be deduced by showing that the standard matrices for T1 and T2 commute:      −1 0 −1 0 1 0 = [T1 ◦ T2 ] = [T1 ][T2 ] = 0 −1 0 1 0 −1      1 0 −1 0 −1 0 [T2 ◦ T1 ] = [T2 ][T1 ] = = 0 −1 0 −1 0 1

The operator T (x) = −x on R2 or R3 is called the reflection about the origin. As the foregoing computations show, the standard matrix for this operator on R2 is

[T ] =



−1 0



0

−1

y

y (x, y)

(x, y )

(–x, y )

x

T1(x) x

x

T2 (x)

T1(T2 (x)) (–x, –y)

Figure 4.10.4

x

(x, –y )

(–x, –y )

T2 (T1 (x))

T1 ° T2

T2 ° T1

E X A M P L E 4 Composition of Three Transformations

Find the standard matrix for the operator T : R3 →R3 that first rotates a vector counterclockwise about the z-axis through an angle θ , then reflects the resulting vector about the yz-plane, and then projects that vector orthogonally onto the xy-plane. Solution The operator

T can be expressed as the composition

T = T3 ◦ T2 ◦ T1 where T1 is the rotation about the z-axis, T2 is the reflection about the yz-plane, and T3 is the orthogonal projection onto the xy -plane. From Tables 6, 2, and 4 of Section 4.9, the standard matrices for these operators are

4.10 Properties of MatrixTransformations

⎡ cos θ ⎢ sin θ [T1 ] = ⎣

⎤

− sin θ cos θ

0

⎤

⎡−1 0 0 ⎡1 0 ⎢ ⎢ ⎥ ⎥ 0 ⎣ 0 1 0 ⎣0 ⎦, [T2 ] = ⎦, [T3 ] = 0 1 0 0 1

0

0 1 0

Thus, it follows from (5) that the standard matrix for T is

⎡

1 ⎢ [T ] = ⎣0 0

0 1 0

⎡ − cos θ ⎢ = ⎣ sin θ

⎤⎡

0

One-to-One Matrix Transformations

⎤⎡

−1 0 0 cos θ 0 ⎥⎢ ⎥⎢ 0 ⎦ ⎣ 0 1 0⎦ ⎣sin θ 0 0 0 1 0 sin θ cos θ 0

− sin θ cos θ 0

⎤

0 ⎥ 0⎦ 0

273

⎤

0 ⎥ 0⎦ 0

⎤

0 ⎥ 0⎦ 1

Our next objective is to establish a link between the invertibility of a matrix A and properties of the corresponding matrix transformation TA .

TA : Rn →Rm is said to be one-to-one if TA maps distinct vectors (points) in R into distinct vectors (points) in Rm .

DEFINITION 1 A matrix transformation n

(See Figure 4.10.5.) This idea can be expressed in various ways. For example, you should be able to see that the following are just restatements of Definition 1: 1. TA is one-to-one if for each vector b in the range of A there is exactly one vector x in Rn such that TA x = b. 2. TA is one-to-one if the equality TA (u) = TA (v) implies that u = v.

Rn

Rm

Figure 4.10.5

Rn

Rm

One-to-one

Not one-to-one

Rotation operators on R2 are one-to-one since distinct vectors that are rotated through the same angle have distinct images (Figure 4.10.6). In contrast, the orthogonal projection of R2 onto the x-axis is not one-to-one because it maps distinct points on the same vertical line into the same point (Figure 4.10.7). y

T(v) T(u)

y P

θ v θ u

Q x

M

Figure 4.10.6 Distinct vectors u and v are rotated into distinct vectors T (u) and T (v).

x

Figure 4.10.7 The distinct points P and Q are mapped into the same point M .

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Chapter 4 General Vector Spaces

Kernel and Range

In the discussion leading up to Theorem 4.2.5 we introduced the notion of the “kernel” of a matrix transformation. The following definition formalizes this idea and defines the companion notion of “range.” DEFINITION 2 If TA : Rn →Rm is a matrix transformation, then the set of all vectors

in Rn that TA maps into 0 is called the kernel of TA and is denoted by ker(TA ). The set of all vectors in Rm that are images under this transformation of at least one vector in Rn is called the range of TA and is denoted by R(TA ). In brief: ker (TA ) = null space of A

(6)

R(TA ) = column space of A

(7)

The key to solving a mathematical problem is often adopting the right point of view; and this is why, in linear algebra, we develop different ways of thinking about the same vector space. For example, if A is an m × n matrix, here are three ways of viewing the same subspace of Rn : • Matrix view: the null space of A • System view: the solution space of Ax = 0 • Transformation view: the kernel of TA

and here are three ways of viewing the same subspace of Rm : • Matrix view: the column space of A • System view: all b in Rm for which Ax = b is consistent • Transformation view: the range of TA

In the special case of a linear operator TA : Rn →Rn , the following theorem establishes fundamental relationships between the invertibility of A and properties of TA .

THEOREM 4.10.1 If

A is an n × n matrix and TA : Rn →Rn is the corresponding

matrix operator, then the following statements are equivalent. (a) A is invertible. (b) The kernel of TA is {0}. (c) The range of TA is Rn . (d ) TA is one-to-one.

Proof We can prove this theorem by establishing the chain of implications (a )

⇒ (b ) ⇒ (c) ⇒ (d ) ⇒ (a ). We will prove the first two implications and leave the rest as exercises.

(a) ⇒ (b) Assume that A is invertible. It follows from parts (a) and (b) of Theorem 4.8.8 that the system Ax = 0 has only the trivial solution and hence that the null space of A is {0}. Formula (6) now implies that the kernel of TA is {0}. (b) ⇒ (c) Assume that the kernel of TA is {0}. It follows from Formula (6) that the null space of A is {0} and hence that A has nullity 0. This in turn implies that the rank of A is n and hence that the column space of A is all of Rn . Formula (7) now implies that the range of TA is Rn .

4.10 Properties of MatrixTransformations

275

E X A M P L E 5 The Rotation Operator on R 2 Is One-to-One

As was illustrated in Figure 4.10.6, the operator T : R2 →R2 that rotates vectors through an angle θ is one-to-one. In accordance with parts (a) and (d) of Theorem 4.10.1, show that the standard matrix for T is invertible. Solution We will show that the standard matrix for

T is invertible by showing that its determinant is nonzero. From Table 5 of Section 4.9 the standard matrix for T is 

cos θ [T ] = sin θ

− sin θ cos θ



This matrix is invertible because

 cos θ det[T ] =  sin θ

 − sin θ  = cos2 θ + sin2 θ = 1 = 0 cos θ 

E X A M P L E 6 Projection Operators Are Not One-to-One

As illustrated in Figure 4.10.7, the operator T : R2 →R2 that projects onto the x -axis in the xy -plane is not one-to-one. In accordance with parts (a) and (d ) of Theorem 4.10.1, show that the standard matrix for T is not invertible. Solution We will show that the standard matrix for T is not invertible by showing that its determinant is zero. From Table 3 of Section 4.9 the standard matrix for T is

[T ] =



1

0

0

0



Since det[T ] = 0, the operator T is not one-to-one.

Inverse of a One-to-One Matrix Operator

If TA : Rn →Rn is a one-to-one matrix operator, then it follows from Theorem 4.10.1 that A is invertible. The matrix operator

TA−1 : Rn →Rn that corresponds to A−1 is called the inverse operator or (more simply) the inverse of TA . This terminology is appropriate because TA and TA−1 cancel the effect of each other in the sense that if x is any vector in Rn , then

TA (TA−1 (x)) = AA−1 x = I x = x

TA−1 (TA (x)) = A−1 Ax = I x = x

y

or, equivalently, TA

TA ◦ TA−1 = TAA−1 = TI

s x to w ma p w

TA −1 ◦ TA = TA −1 A = TI

x TA–1

to x ma ps w

x

Figure 4.10.8

From a more geometric viewpoint, if w is the image of x under TA , then TA−1 maps w backinto x, since TA−1 (w) = TA−1 (TA (x)) = x

This is illustrated in Figure 4.10.8 for R2 .

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Chapter 4 General Vector Spaces

Before considering examples, it will be helpful to touch on some notational matters. If TA : Rn →Rn is a one-to-one matrix operator, and if TA−1 : Rn →Rn is its inverse, then the standard matrices for these operators are related by the equation

TA−1 = TA−1

(8)

In cases where it is preferable not to assign a name to the matrix, we can express this equation as

[T −1 ] = [T ]−1

(9)

E X A M P L E 7 Standard Matrix for T −1

Let T : R2 →R2 be the operator that rotates each vector in R2 through the angle θ , so from Table 5 of Section 4.9,

[T ] =



cos θ

− sin θ cos θ

sin θ



(10)

It is evident geometrically that to undo the effect of T , one must rotate each vector in R2 through the angle −θ . But this is exactly what the operator T −1 does, since the standard matrix for T −1 is

[T −1 ] = [T ]−1 =



cos θ

sin θ

− sin θ

cos θ



=



cos(−θ ) − sin(−θ ) sin(−θ ) cos(−θ )



(verify), which is the standard matrix for a rotation through the angle −θ . E X A M P L E 8 Finding T −1

Show that the operator T : R2 →R2 defined by the equations

w 1 = 2 x1 + x2

w 2 = 3x1 + 4 x2

is one-to-one, and find T −1 (w1 , w2 ). Solution The matrix form of these equations is

    2 1 x1 w1 = w2 3 4 x2

 so the standard matrix for T is





2 1 [T ] = 3 4

This matrix is invertible (so T is one-to-one) and the standard matrix for T −1 is

⎡

Thus

4 5

[T −1 ] = [T ]−1 = ⎣ − 35

5

from which we conclude that

T −1 (w1 , w2 ) =

2 5

⎤

⎦

⎡ ⎤ ⎤   4 1 w − −51 w1 w 1 2 5 5 ⎦ ⎦ =⎣ 2 2 3 w2 w w + − 5 2 5 5 1

⎡  4 w 5 1 =⎣ [T −1 ] w2 −3 

− 51

4 5

w1 − 15 w2 , − 35 w1 + 52 w2



4.10 Properties of MatrixTransformations

More on the Equivalence Theorem

277

As our final result in this section, we will add parts (b), (c), and (d) of Theorem 4.10.1 to Theorem 4.8.8. THEOREM 4.10.2 Equivalent Statements

If A is an n × n matrix, then the following statements are equivalent. (a) A is invertible. (b) (c)

Ax = 0 has only the trivial solution.

The reduced row echelon form of A is In .

(d )

A is expressible as a product of elementary matrices.

(e)

Ax = b is consistent for every n × 1 matrix b.

( f ) Ax = b has exactly one solution for every n × 1 matrix b.

( g)

det(A) = 0.

(h)

The column vectors of A are linearly independent.

(i)

The row vectors of A are linearly independent.

( j)

The column vectors of A span Rn .

(k)

The row vectors of A span Rn .

(l ) The column vectors of A form a basis for Rn . (m) The row vectors of A form a basis for Rn . (n)

A has rank n.

(o) A has nullity 0. ( p) The orthogonal complement of the null space of A is Rn . (q)

The orthogonal complement of the row space of A is {0}.

(r) (s)

The kernel of TA is {0}. The range of TA is Rn .

(t)

TA is one-to-one.

Exercise Set 4.10 In Exercises 1–4, determine whether the operators T1 and T2 commute; that is, whether T1 ◦ T2 = T2 ◦ T1 .

1. (a) T1 : R 2 →R 2 is the reflection about the line y = x , and T2 : R 2 →R 2 is the orthogonal projection onto the x -axis. (b) T1 : R 2 →R 2 is the reflection about the x -axis, and T2 : R 2 →R 2 is the reflection about the line y = x . 2

2

2. (a) T1 : R →R is the orthogonal projection onto the x -axis, and T2 : R 2 →R 2 is the orthogonal projection onto the y -axis. 2

2

(b) T1 : R →R is the rotation about the origin through an angle of π/4, and T2 : R 2 →R 2 is the reflection about the y -axis. 3. T1 : R 3 →R 3 is a dilation with factor k, and T2 : R 3 →R 3 is a contraction with factor 1/k. 4. T1 : R 3 →R 3 is the rotation about the x -axis through an angle θ1 , and T2 : R 3 →R 3 is the rotation about the z-axis through an angle θ2 .

In Exercises 5–6, let TA and TB bet the operators whose standard matrices are given. Find the ...