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W W L CHEN

c

W W L Chen, 1997, 2008.

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 fromRn intoRm, we mean a function of the typeT :Rn →Rm, with domainRn and codomain Rm. For every vectorx∈Rn, the vectorT(x)∈Rm is called the image ofxunder the transformationT, and the set

R(T) ={T(x) :x∈Rn},

of all images underT, is called the range of the transformationT.

Remark.For our convenience later, we have chosen to use R(T) instead of the usualT(Rn) to denote

the range of the transformationT.

For everyx= (x1, . . . , xn)∈Rn, we can write

T(x) =T(x1, . . . , xn) = (y1, . . . , ym).

Here, for everyi= 1, . . . , m, we have

yi =Ti(x1, . . . , xn), (1)

whereTi:Rn→Ris a real valued function.

Definition. A transformation T : Rn Rm is called a linear transformation if there exists a real

matrix

A= 

a11 . . . a1n ..

. ...

am1 . . . amn 

(2)

such that for everyx= (x1, . . . , xn)∈Rn, we haveT(x1, . . . , xn) = (y1, . . . , ym), where

y1= a11x1+. . .+ a1nxn, ..

.

ym=am1x1+. . .+amnxn,

or, in matrix notation,

   y1 .. . ym   =  

a11 . . . a1n ..

. ...

am1 . . . amn     x1 .. . xn 

. (2)

The matrixA is called the standard matrix for the linear transformationT.

Remarks. (1) In other words, a transformation T : Rn Rm is linear if the equation (1) for every i= 1, . . . , mis linear.

(2) If we writex∈Rn andy∈Rm as column matrices, then (2) can be written in the formy=Ax, and so the linear transformation T can be interpreted as multiplication of x ∈ Rn by the standard matrixA.

Definition.A linear transformationT :RnRmis said to be a linear operator ifn=m. In this case,

we say thatT is a linear operator onRn.

Example 8.1.1.The linear transformationT :R5R3, defined by the equations

y1= 2x1+ 3x2+ 5x3+ 7x4−9x5, y2= 3x2+ 4x3 + 2x5, y3= x1 + 3x3−2x4 ,

can be expressed in matrix form as

  y1 y2 y3  =  

2 3 5 7 −9

0 3 4 0 2

1 0 3 −2 0

       x1 x2 x3 x4 x5      .

If (x1, x2, x3, x4, x5) = (1,0,1,0,1), then

  y1 y2 y3  =  

2 3 5 7 −9

0 3 4 0 2

1 0 3 −2 0

       1 0 1 0 1      =   −2 6 4  ,

so thatT(1,0,1,0,1) = (−2,6,4).

Example 8.1.2. Suppose thatA is the zerom×nmatrix. The linear transformation T :Rn Rm, whereT(x) =Axfor everyx∈Rn, is the zero transformation fromRn intoRm. ClearlyT(x) =0for everyx∈Rn.

Example 8.1.3.Suppose thatI is the identityn×nmatrix. The linear operatorT :Rn Rn, where

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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 forT is given by

A= (T(e1) . . . T(en) ),

whereT(ej)is a column matrix for everyj= 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 onR2. For everyx∈R2, we shall write x= (x1, x2).

Example 8.2.1.Consider reflection across thex2-axis, so thatT(x1, x2) = (−x1, x2). Clearly we have

T(e1) =

−1 0

and T(e2) =

0 1

,

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 thex1-axis and across the line

x1=x2are given respectively by

A=

1 0

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 acrossx2-axis ny1y2==−x1x2

−1 0

0 1

Reflection acrossx1-axis ny1y2==x1−x2

1 0

0 −1

Reflection acrossx1=x2 ny1y2==x2x1

0 1 1 0

Reflection across origin ny1y2==−x1−x2

−1 0

0 −1

Example 8.2.2.For orthogonal projection onto thex1-axis, we haveT(x1, x2) = (x1,0), with standard

matrix

A=

1 0 0 0

(4)

Similarly, the standard matrix for orthogonal projection onto thex2-axis is given by

A=

0 0 0 1

.

We give a summary in the table below:

Linear operator Equations Standard matrix

Orthogonal projection ontox1-axis ny1y2== 0x1

1 0 0 0

Orthogonal projection ontox2-axis

y1= 0

y2=x2

0 0 0 1

Example 8.2.3.For anticlockwise rotation by an angleθ, we haveT(x1, x2) = (y1, y2), where

y1+ iy2= (x1+ ix2)(cosθ+ i sinθ),

and so

y1 y2

=

cosθ −sinθ

sinθ cosθ

x1 x2

.

It follows that the standard matrix is given by

A=

cosθ −sinθ

sinθ cosθ

.

We give a summary in the table below:

Linear operator Equations Standard matrix

Anticlockwise rotation by angleθ

y1=x1cosθ−x2sinθ y2=x1sinθ+x2cosθ

cosθ −sinθ

sinθ cosθ

Example 8.2.4.For contraction or dilation by a non-negative scalark, we haveT(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 ifk >1, and can be extended to negative values ofkby noting that fork <0, we have

k 0 0 k

=

−1 0

0 −1

−k 0

0 −k

.

This describes contraction or dilation by non-negative scalar−kfollowed by reflection across the origin. We give a summary in the table below:

Linear operator Equations Standard matrix

Contraction or dilation by factork

y1=kx1 y2=kx2

k 0 0 k

(5)

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

A=

k 0 0 1

.

This can be extended to negative values ofkby noting that fork <0, we have

k 0 0 1

=

−1 0

0 1

−k 0

0 1

.

This describes expansion or compression in thex1-direction by positive factor−kfollowed by reflection across thex2-axis. Similarly, for expansion or compression in the x2-direction by a non-zero factor k, we have the standard matrix

A=

1 0 0 k

.

We give a summary in the table below:

Linear operator Equations Standard matrix

Expansion or compression inx1-direction

y1=kx1 y2=x2

k 0 0 1

Expansion or compression inx2-direction ny1y2==x1kx2

1 0 0 k

Example 8.2.6.For shears in thex1-direction with factork, we have T(x1, x2) = (x1+kx2, x2), with

standard matrix

A=

1 k

0 1

.

For the casek= 1, we have the following.

• • • •

• • • •

T

(k=1)

For the casek=−1, we have the following.

• • • •

• • • •

T

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Similarly, for shears in thex2-direction with factork, we have standard matrix A= 1 0 k 1 .

We give a summary in the table below:

Linear operator Equations Standard matrix

Shear inx1-direction

y1=x1+kx2 y2=x2

1 k

0 1

Shear inx2-direction ny1y2==x1kx1+x2

1 0

k 1

Example 8.2.7.Consider a linear operatorT :R2R2which consists of a reflection across thex2-axis,

followed by a shear in thex1-direction with factor 3 and then reflection across thex1-axis. To find the standard matrix, consider the effect ofT on a standard basis {e1,e2}ofR2. Note that

e1=

1 0 7→ −1 0 7→ −1 0 7→ −1 0

=T(e1),

e2=

0 1 7→ 0 1 7→ 3 1 7→ 3 −1

=T(e2),

so it follows from Proposition 8A that the standard matrix forT is

A=

−1 3

0 −1

.

Let us summarize the above and consider a few special cases. We have the following table of invertible linear operators withk6= 0. Clearly, ifAis the standard matrix for an invertible linear operatorT, then the inverse matrixA−1 is the standard matrix for the inverse linear operatorT−1.

Linear operatorT Standard matrixA Inverse matrixA−1 Linear operatorT−1

Reflection across linex1=x2

0 1 1 0 0 1 1 0 Reflection across linex1=x2

Expansion or compression inx1−direction

k 0 0 1

k−1 0

0 1

Expansion or compression inx1−direction

Expansion or compression inx2−direction

1 0 0 k

1 0

0 k−1

Expansion or compression inx2−direction

Shear inx1−direction

1 k 0 1 1 −k 0 1 Shear inx1−direction

Shear inx2−direction

1 0 k 1 1 0 −k 1 Shear inx2−direction

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matrixAby some elementary matrixE to give the productEA. We have the following table.

Elementary row operation Elementary matrixE

Interchanging the two rows

0 1 1 0

Multiplying row 1 by non-zero factork

k 0 0 1

Multiplying row 2 by non-zero factork

1 0 0 k

Addingktimes row 2 to row 1

1 k

0 1

Addingktimes row 1 to row 2

1 0

k 1

Now, we know that any invertible matrixAcan be reduced to the identity matrix by a finite number of elementary row operations. In other words, there exist a finite number of elementary matricesE1, . . . , Es of the types above with various non-zero values ofksuch that

Es. . . E1A=I,

so that

A=E1−1. . . E

−1

s . We have proved the following result.

PROPOSITION 8B.Suppose that the linear operatorT :R2→R2has standard matrixA, whereAis 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 operatorT :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 thatT(x1, x2) = (y1, y2). SinceAis invertible, we havex=A−1y, where

x=

x1 x2

and y=

y1 y2

.

The equation of a straight line is given byαx1+βx2=γ or, in matrix form, by

(α β)

x1 x2

= (γ).

Hence

(α β)A−1

y1 y2

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Let

(α′ β) = (α β)A−1.

Then

(α′ β)

y1 y2

= (γ).

In other words, the image underT of the straight lineαx1+βx2=γ isα′

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 ofαandβ.

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. ThenT =T2◦T1:Rn→Rk is also a linear transformation.

Proof.SinceT1andT2are linear transformations, they have standard matricesA1andA2respectively.

In other words, we haveT1(x) =A1x for everyx∈Rn andT2(y) =A2yfor everyy ∈Rm. It follows thatT(x) =T2(T1(x)) =A2A1xfor everyx∈Rn, so thatT has standard matrixA2A1.

Example 8.3.1. Suppose that T1 : R2 R2 is anticlockwise rotation by π/2 and T2 : R2 R2 is

orthogonal projection onto thex1-axis. Then the respective standard matrices are

A1=

0 −1

1 0

and A2=

1 0 0 0

.

It follows that the standard matrices forT2◦T1 andT1◦T2 are respectively

A2A1=

0 −1

0 0

and A1A2=

0 0 1 0

.

HenceT2◦T1andT1◦T2are not equal.

Example 8.3.2. Suppose that T1 : R2 R2 is anticlockwise rotation by θ and T2 : R2 R2 is

anticlockwise rotation byφ. Then the respective standard matrices are

A1=

cosθ −sinθ

sinθ cosθ

and A2=

cosφ −sinφ

sinφ cosφ

.

It follows that the standard matrix forT2◦T1is

A2A1=

cosφcosθ−sinφsinθ −cosφsinθ−sinφcosθ

sinφcosθ+ cosφsinθ cosφcosθ−sinφsinθ

=

cos(φ+θ) −sin(φ+θ) sin(φ+θ) cos(φ+θ)

.

HenceT2◦T1is anticlockwise rotation byφ+θ.

Example 8.3.3.The reader should check that inR2, reflection across thex1-axis followed by reflection

across thex2-axis gives reflection across the origin.

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Definition.A linear transformationT :RnRmis said to be one-to-one if for everyx

,x′′∈Rn, we havex′

=x′′

wheneverT(x′

) =T(x′′ ).

Example 8.3.4.If we consider linear operators T :R2R2, thenT is one-to-one precisely when the

standard matrix Ais invertible. To see this, suppose first of all thatA is invertible. IfT(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 existsx∈R2 such thatx6=0andAx=0. On the other hand, we clearly have

A0=0. It follows that T(x) =T(0), so thatT is not one-to-one.

PROPOSITION 8E.Suppose that the linear operator T :RnRn has standard matrixA. Then the

following statements are equivalent: (a) The matrixA is invertible.

(b) The linear operatorT is one-to-one.

(c) The range of T isRn; in other words, R(T) =Rn.

Proof.((a)(b)) Suppose thatT(x′

) =T(x′′

). ThenAx′ =Ax′′

. Multiplying on the left byA−1gives

x′ =x′′

.

((b)⇒(a)) Suppose thatT is one-to-one. Then the systemAx=0has unique solution x=0in Rn. It follows thatAcan be reduced by elementary row operations to the identity matrixI, and is therefore invertible.

((a)⇒(c)) For anyy∈Rn, clearlyx=A−1ysatisfiesAx=y, so thatT(x) =y.

((c)⇒(a)) Suppose that{e1, . . . ,en} is the standard basis for Rn. Letx1, . . . ,xn ∈Rn be chosen to satisfyT(xj) =ej, so thatAxj=ej, for everyj= 1, . . . , n. Write

C= (x1 . . . xn).

ThenAC=I, so thatAis invertible.

Definition.Suppose that the linear operatorT :RnRnhas standard matrixA, whereAis invertible. Then the linear operator T−1 : Rn Rn, defined by T−1(x) = A−1x for every x Rn, is called the inverse of the linear operatorT.

Remark.ClearlyT−1(T(x)) =xandT(T−1(x)) =xfor everyxRn.

Example 8.3.5. Consider the linear operator T : R2 R2, defined by T(x) = Ax for everyx ∈R2, where

A=

1 1 1 2

.

ClearlyAis invertible, and

A−1=

2 −1

−1 1

.

Hence the inverse linear operator isT−1:R2R2, defined byT−1(x) =A−1xfor every xR2.

Example 8.3.6. Suppose that T : R2 R2 is anticlockwise rotation by angle θ. The reader should

check thatT−1:R2R2 is anticlockwise rotation by angle 2πθ.

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PROPOSITION 8F. A transformation T : Rn → Rm is linear if and only if the following two conditions are satisfied:

(a) For everyu,v∈Rn, we haveT(u+v) =T(u) +T(v). (b) For everyu∈Rn andc∈R, we haveT(cu) =cT(u).

Proof.Suppose first of all thatT :RnRmis a linear transformation. LetAbe the standard matrix forT. Then for everyu,v∈Rn andc∈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 everyx∈Rn. Suppose that{e1, . . . ,en} is the standard basis forRn. As suggested by Proposition 8A, we write

A= (T(e1) . . . T(en) ),

whereT(ej) is a column matrix for everyj= 1, . . . , n. For any vector

x= 

x1

.. .

xn 

inRn, we have

Ax= (T(e1) . . . T(en) )

x1

.. .

xn 

=x1T(e1) +. . .+xnT(en).

Using (b) on each summand and then using (a) inductively, we obtain

Ax=T(x1e1) +. . .+T(xnen) =T(x1e1+. . .+xnen) =T(x)

as required.

To conclude our study of euclidean linear transformations, we briefly mention the problem of eigen-values 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 ofT if there exists a non-zero vectorx∈Rn such that T(x) =λx. This non-zero vector x∈Rn is called an eigenvector ofT corresponding to the eigenvalueλ.

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

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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 transformationT.

Definition. A transformation T : V W from a real vector spaceV into a real vector space W is

called a linear transformation if the following two conditions are satisfied: (LT1) For everyu,v∈V, we haveT(u+v) =T(u) +T(v).

(LT2) For everyu∈V andc∈R, we haveT(cu) =cT(u).

Definition.A linear transformationT :V V from a real vector spaceV into itself is called a linear

operator onV.

Example 8.4.1. Suppose thatV and W are two real vector spaces. The transformationT : V W,

whereT(u) =0for everyu∈V, is clearly linear, and is called the zero transformation fromV toW.

Example 8.4.2.Suppose thatV is a real vector space. The transformationI:V V, whereI(u) =u for everyu∈V, is clearly linear, and is called the identity operator onV.

Example 8.4.3.Suppose that V is a real vector space, and that k Ris fixed. The transformation T : V → V, whereT(u) = ku 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 thatV is a finite dimensional vector space, with basis{w1, . . . ,wn}. Define a transformationT :V →Rn as follows. For everyu∈V, there exists a unique vector (β1, . . . , βn)∈Rn such thatu=β1w1+. . .+βnwn. We let T(u) = (β1, . . . , βn). In other words, the transformationT gives the coordinates of any vector u∈V with respect to the given basis {w1, . . . ,wn}. Suppose now thatv=γ1w1+. . .+γnwn is another vector inV. Thenu+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, ifc∈R, thencu=cβ1w1+. . .+cβnwn, so that

T(cu) = (cβ1, . . . , cβn) =c(β1, . . . , βn) =cT(u).

HenceT is a linear transformation. We shall return to this in greater detail in the next section.

Example 8.4.5.Suppose thatPn denotes the vector space of all polynomials with real coefficients and

degree at mostn. Define a transformation T :Pn→Pn as follows. For every polynomial

p=p0+p1x+. . .+pnxn

inPn, we let

T(p) =pn+pn−1x+. . .+p0xn.

Suppose now thatq=q0+q1x+. . .+qnxn is another polynomial inPn. 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

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Also, for anyc∈R, we havecp=cp0+cp1x+. . .+cpnxn, so that

T(cp) =cpn+cpn−1x+. . .+cp0xn=c(pn+pn−1x+. . .+p0xn) =cT(p). HenceT is a linear transformation.

Example 8.4.6.LetV denote the vector space of all real valued functions differentiable everywhere inR,

and letW denote the vector space of all real valued functions defined onR. Consider the transformation

T :V →W, where T(f) =f′ for everyf V. It is easy to check from properties of derivatives thatT is a linear transformation.

Example 8.4.7.LetV denote the vector space of all real valued functions that are Riemann integrable

over the interval [0,1]. Consider the transformationT :V →R, where

T(f) = Z 1

0

f(x) dx

for everyf ∈V. It is easy to check from properties of the Riemann integral thatT is a linear transfor-mation.

Consider a linear transformationT :V →W from a finite dimensional real vector spaceV into a real vector spaceW. Suppose that {v1, . . . ,vn} is a basis ofV. Then everyu∈V can be written uniquely in the formu=β1v1+. . .+βnvn, whereβ1, . . . , βn∈R. It follows that

T(u) =T(β1v1+. . .+βnvn) =T(β1v1) +. . .+T(βnvn) =β1T(v1) +. . .+βnT(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 spaceV into a real vector spaceW. Suppose further that{v1, . . . ,vn} is a basis ofV. Then

T is completely determined byT(v1), . . . , T(vn).

Example 8.4.8.Consider a linear transformationT :P2R, whereT(1) = 1,T(x) = 2 andT(x2) = 3.

Since {1, x, x2} is a basis ofP2, 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 transformationT :R4R, where T(1,0,0,0) = 1,T(1,1,0,0) = 2, T(1,1,1,0) = 3 andT(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 ofR4, 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

andT2:W →U are linear transformations. ThenT =T2◦T1:V →U is also a linear transformation.

Proof.Suppose thatu,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, ifc∈R, then

T(cu) =T2(T1(cu)) =T2(cT1(u)) =cT2(T1(u)) =cT(u).

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8.5. Change of Basis

Suppose thatV is a real vector space, with basis B ={u1, . . . ,un}. Then every vectoru∈V can be written uniquely as a linear combination

u=β1u1+. . .+βnun, whereβ1, . . . , βn∈R. (3)

It follows that the vectorucan be identified with the vector (β1, . . . , βn)∈Rn.

Definition.Suppose thatu∈V and (3) holds. Then the matrix

[u]B= 

 

β1

.. .

βn 

 

is called the coordinate matrix ofurelative to the basisB={u1, . . . ,un}.

Example 8.5.1.The vectors

u1= (1,2,1,0), u2= (3,3,3,0), u3= (2,−10,0,0), 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=β1u1+β2u2+β3u3+β4u4.

In matrix notation, this becomes

 

x y z w

 =

 

1 3 2 −2

2 3 −10 1

1 3 0 −6

0 0 0 2

 

 

β1 β2 β3 β4

 ,

so that

[u]B= 

 

β1 β2 β3 β4

 =

 

1 3 2 −2

2 3 −10 1

1 3 0 −6

0 0 0 2

 

−1

 

x y z w

 .

Remark.Consider a functionφ:V Rn, whereφ(u) = [u]B for everyu∈V. It is not difficult to see that this function gives rise to a one-to-one correspondence between the elements ofV and the elements ofRn. 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) andφ(cu) =cφ(u) for everyu,v ∈V and c∈ R. Thusφ is a linear transformation, and preserves much of the structure ofV. We also say that V is isomorphic toRn. In practice, once we have made this identification between vectors and their coordinate matrices, then we can basically forget about the basisB and imagine that we are working inRn with the standard basis.

Clearly, if we change from one basisB={u1, . . . ,un}to another basisC={v1, . . . ,vn}ofV, then we also need to find a way of calculating [u]C in terms of [u]B for every vectoru∈V. To do this, note that each of the vectorsv1, . . . ,vncan be written uniquely as a linear combination of the vectorsu1, . . . ,un. Suppose that fori= 1, . . . , n, we have

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so that

[vi]B= 

a1i .. .

ani 

.

For everyu∈V, we can write

u=β1u1+. . .+βnun=γ1v1+. . .+γnvn, whereβ1, . . . , βn, γ1, . . . , γn∈R,

so that

[u]B= 

 

β1

.. .

βn 

 and [u]C = 

 

γ1

.. .

γn 

 .

Clearly

u=γ1v1+. . .+γnvn

=γ1(a11u1+. . .+an1un) +. . .+γn(a1nu1+. . .+annun) = (γ1a11+. . .+γna1n)u1+. . .+ (γ1an1+. . .+γnann)un =β1u1+. . .+βnun.

Hence

β1=γ1a11+. . .+γna1n, ..

.

βn=γ1an1+. . .+γnann.

Written in matrix notation, we have

 

β1

.. .

βn 

 =

a11 . . . a1n ..

. ...

an1 . . . ann 

 

 

γ1

.. .

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 spaceV. 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 basisB.

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 matrixP in Proposition 8J is sometimes called the transition matrix from the basisC

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Example 8.5.2.We know that with

u1= (1,2,1,0), u2= (3,3,3,0), u3= (2,−10,0,0), u4= (−2,1,−6,2),

and with

v1= (1,2,1,0), v2= (1,−1,1,0), v3= (1,0,−1,0), v4= (0,0,0,2),

bothB={u1,u2,u3,u4}andC={v1,v2,v3,v4}are bases ofR4. 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 [v4]B) =

 

1 −2 11 −27

0 1 −4 11

0 0 1 −2

0 0 0 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 [u4]C) =

 

1 2 −3 −1

0 1 4 −3

0 0 1 2

0 0 0 1

 .

Hence [u]C =Q[u]B for everyu∈R4. Note thatP Q=I. Now letu= (6,−1,2,2). We can check that u=v1+ 3v2+ 2v3+v4, so that

[u]C = 

  1 3 2 1

 .

Then

[u]B= 

 

1 −2 11 −27

0 1 −4 11

0 0 1 −2

0 0 0 1

 

  1 3 2 1

 =

 

−10 6 0 1

 .

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Example 8.5.3.Consider the vector spaceP2. It is not too difficult to check that

u1= 1 +x, u2= 1 +x2, u3=x+x2

form a basis ofP2. Letu= 1 + 4x−x2. Thenu=β1u1+β2u2+β3u3, 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

[u]B= 

 3

−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 ofP2. Alsou=γ1v1+γ2v2+γ3v3, where

1 + 4x−x2=γ1+γ2(1 +x) +γ3(1 +x+x2) = (γ1+γ2+γ3) + (γ2+γ3)x+γ3x2,

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

[u]C = 

−3 5

−1 

.

Next, note that

v1= 1 2u1+

1 2u2−

1 2u3, v2=u1,

v3= 1 2u1+

1 2u2+

1 2u3.

Hence

P = ( [v1]B [v2]B [v3]B) = 

1/2 1 1/2 1/2 0 1/2

−1/2 0 1/2 

.

To verify that [u]B=P[u]C, note that

 3

−2 1

= 

1/2 1 1/2 1/2 0 1/2

−1/2 0 1/2 

 

−3 5

−1 

.

8.6. Kernel and Range

Consider first of all a euclidean linear transformation T : Rn →Rm. Suppose that A is the standard matrix forT. Then the range of the transformationT is given by

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It follows thatR(T) is the set of all linear combinations of the columns of the matrixA, and is therefore the column space ofA. On the other hand, the set

{x∈Rn:Ax=0}

is the nullspace ofA.

Recall that the sum of the dimension of the nullspace ofAand dimension of the column space ofAis equal to the number of columns ofA. 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 thatT:V W is a linear transformation from a real vector spaceV into a real

vector spaceW. Then the set

ker(T) ={u∈V :T(u) =0}

is called the kernel ofT, and the set

R(T) ={T(u) :u∈V}

is called the range ofT.

Example 8.6.1.For a euclidean linear transformationT with standard matrixA, we have shown that

ker(T) is the nullspace ofA, whileR(T) is the column space ofA.

Example 8.6.2.Suppose thatT :V W is the zero transformation. Clearly we have ker(T) =V and R(T) ={0}.

Example 8.6.3.Suppose thatT :V V is the identity operator onV. Clearly we have ker(T) ={0}

andR(T) =V.

Example 8.6.4. Suppose thatT :R2 R2 is orthogonal projection onto the x1-axis. Then ker(T) is

thex2-axis, whileR(T) is thex1-axis.

Example 8.6.5.Suppose thatT :RnRn is one-to-one. Then ker(T) ={0} andR(T) =Rn, in view of Proposition 8E.

Example 8.6.6.Consider the linear transformation T :V W, whereV denotes the vector space of

all real valued functions differentiable everywhere in R, whereW denotes the space of all real valued functions defined inR, and whereT(f) =f′

for everyf ∈V. Then ker(T) is the set of all differentiable functions with derivative 0, and so is the set of all constant functions inR.

Example 8.6.7. Consider the linear transformationT : V R, where V denotes the vector space of

all real valued functions Riemann integrable over the interval [0,1], and where

T(f) = Z 1

0

f(x) dx

for every f ∈V. Then ker(T) is the set of all Riemann integrable functions in [0,1] with zero mean, whileR(T) =R.

PROPOSITION 8K.Suppose that T :V →W is a linear transformation from a real vector spaceV

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Proof.SinceT(0) =0, it follows that0ker(T)V and0R(T)W. For anyu,vker(T), we

have

T(u+v) =T(u) +T(v) =0+0=0,

so thatu+v∈ker(T). Suppose further thatc∈R. Then

T(cu) =cT(u) =c0=0,

so thatcu∈ker(T). Hence ker(T) is a subspace ofV. Suppose next thatw,z∈R(T). Then there exist u,v∈V such thatT(u) =wandT(v) =z. Hence

T(u+v) =T(u) +T(v) =w+z,

so thatw+z∈R(T). Suppose further thatc∈R. Then

T(cu) =cT(u) =cw,

so thatcw∈R(T). HenceR(T) is a subspace ofW.

To complete this section, we prove the following generalization of the Rank-nullity theorem.

PROPOSITION 8L.Suppose thatT :V →W is a linear transformation from ann-dimensional real vector spaceV into a real vector space W. Then

dim ker(T) + dimR(T) =n.

Proof.Suppose first of all that dim ker(T) =n. Then ker(T) =V, and soR(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 ofV, then it follows thatT(v1), . . . , T(vn) are linearly independent inW, for otherwise there exist c1, . . . , cn∈R, not all zero, such that

c1T(v1) +. . .+cnT(vn) =0,

so that T(c1v1+. . .+cnvn) =0, a contradiction since c1v1+. . .+cnvn 6= 0. On the other hand, elements ofR(T) are linear combinations ofT(v1), . . . , T(vn). Hence dimR(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}ofV. It suffices to show that

{T(vr+1), . . . , T(vn)} (4)

is a basis ofR(T). Suppose thatu∈V. Then there existβ1, . . . , βn∈Rsuch that

u=β1v1+. . .+βrvr+βr+1vr+1+. . .+βnvn,

so that

T(u) =β1T(v1) +. . .+βrT(vr) +βr+1T(vr+1) +. . .+βnT(vn) =βr+1T(vr+1) +. . .+βnT(vn).

It follows that (4) spansR(T). It remains to prove that its elements are linearly independent. Suppose thatcr+1, . . . , cn ∈Rand

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We need to show that

cr+1=. . .=cn= 0. (6)

By linearity, it follows from (5) thatT(cr+1vr+1+. . .+cnvn) =0, so that

cr+1vr+1+. . .+cnvn ∈ker(T).

Hence there existc1, . . . , cr∈Rsuch that

cr+1vr+1+. . .+cnvn=c1v1+. . .+crvr,

so that

c1v1+. . .+crvr−cr+1vr+1−. . .−cnvn =0.

Since{v1, . . . ,vn}is a basis ofV, it follows thatc1=. . .=cr=cr+1=. . .=cn= 0, so that (6) holds. This completes the proof.

Remark.We sometimes say that dimR(T) and dim ker(T) are respectively the rank and the nullity of

the linear transformationT.

8.7. Inverse Linear Transformations

In this section, we generalize some of the ideas first discussed in Section 8.3.

Definition.A linear transformationT :V W from a real vector spaceV into a real vector spaceW

is said to be one-to-one if for everyu′

,u′′

∈V, we haveu′ =u′′

wheneverT(u′

) =T(u′′ ).

The result below follows immediately from our definition.

PROPOSITION 8M.Suppose thatT :V →W is a linear transformation from a real vector space V

into a real vector spaceW. ThenT is one-to-one if and only ifker(T) ={0}.

Proof.() Clearly0ker(T). Suppose that ker(T)6={0}. Then there exists a non-zerovker(T).

It follows thatT(v) =T(0), and soT 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 ifu′−u′′=0; in other words, if and only ifu′ =u′′.

We have the following generalization of Proposition 8E.

PROPOSITION 8N.Suppose thatT :V →V is a linear operator on a finite-dimensional real vector spaceV. Then the following statements are equivalent:

(a) The linear operatorT is one-to-one. (b) We haveker(T) ={0}.

(c) The range of T isV; in other words, R(T) =V.

Proof.The equivalence of (a) and (b) is established by Proposition 8M. The equivalence of (b) and (c)

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Suppose thatT :V →W is a one-to-one linear transformation from a real vector spaceV into a real vector spaceW. Then for everyw∈R(T), there exists exactly oneu∈V such thatT(u) =w. We can therefore define a transformationT−1 :R(T)V by writingT−1(w) =u, whereuV is the unique

vector satisfyingT(u) =w.

PROPOSITION 8P.Suppose thatT :V →W is a one-to-one linear transformation from a real vector spaceV into a real vector space W. ThenT−1:R(T)V is a linear transformation.

Proof.Suppose thatw,z∈R(T). Then there existu,v∈V such that T−1(w) =uandT−1(z) =v.

It follows thatT(u) =wandT(v) =z, so thatT(u+v) =T(u) +T(v) =w+z, whence

T−1(w+z) =u+v=T−1(w) +T−1(z).

Suppose further thatc∈R. ThenT(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. Suppose further that T1:V →W

andT2: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=T−1

1 ◦T

−1 2 .

8.8. Matrices of General Linear Transformations

Suppose thatT :V →W is a linear transformation from a real vector spaceV to a real vector spaceW. Suppose further that the vector spacesV andW are finite dimensional, with dimV =nand dimW =m. We shall show that if we make use of a basisB ofV and a basis Cof W, then it is possible to describe

T indirectly in terms of some matrixA. The main idea is to make use of coordinate matrices relative to the basesBandC.

Let us recall some discussion in Section 8.5. Suppose that B ={v1, . . . ,vn} is a basis of V. Then every vectorv∈V can be written uniquely as a linear combination

v=β1v1+. . .+βnvn, whereβ1, . . . , βn∈R. (7)

The matrix

[v]B= 

 

β1

.. .

βn 

 (8)

is the coordinate matrix ofvrelative to the basisB.

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

one-to-one linear transformation, with range R(φ) = Rn. Furthermore, the inverse linear transformation

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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.

V W

Rn Rm

T

φ ψ

φ−1 ψ−1

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 matrixAin terms ofT and the basesBandC.

We know from Proposition 8A that

A= (S(e1) . . . S(en) ),

where{e1, . . . ,en} is the standard basis forRn. For everyj= 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 . . . [T(vn)]C). (9)

Definition.The matrixAgiven by (9) is called the matrix for the linear transformationT with respect

to the basesBandC.

We now have the following diagram of linear transformations.

V W

Rn Rm

T

φ ψ

S

φ−1 ψ−1

Hence we can writeT as the composition

T =ψ−1◦S◦φ:V →W.

For everyv∈V, we have the following:

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More precisely, ifv=β1v1+. . .+βnvn, then [v]B=    β1 .. . βn  

 and A[v]B=A    β1 .. . βn   =    γ1 .. . γm   ,

say, and soT(v) =ψ−1(A[v]B) =γ1w

1+. . .+γmwm. 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 andW are finite dimensional, with bases B andC

respectively, and thatAis the matrix for the linear transformation T with respect to the basesB andC. Then for everyv∈V, we haveT(v) =w, wherew∈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

ofT. In the case when T is the identity linear operator, we often chooseB 6=C since this represents a change of basis. In the case whenT is not the identity operator, we often choose B=Cfor the sake of convenience; we then say thatAis the matrix for the linear operatorT with respect to the basisB.

Example 8.8.1.Consider an operatorT :P3P3 on the real vector spaceP3 of all polynomials with

real coefficients and degree at most 3, where for every polynomialp(x) inP3, we haveT(p(x)) =xp′ (x), the product of x with the formal derivative p′

(x) of p(x). The reader is invited to check thatT is a linear operator. Now consider the basisB={1, x, x2, x3} ofP3. The matrix forT with respect toB is

given by

A= ( [T(1)]B [T(x)]B [T(x2)]B [T(x3)]B) = ( [0]B [x]B [2x2]B [3x3]B) =

 

0 0 0 0 0 1 0 0 0 0 2 0 0 0 0 3

 .

Suppose thatp(x) = 1 + 2x+ 4x2+ 3x3. Then

[p(x)]B=    1 2 4 3  

 and A[p(x)]B= 

 

0 0 0 0 0 1 0 0 0 0 2 0 0 0 0 3

      1 2 4 3   =    0 2 8 9   ,

so thatT(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, ifp(x) =p0+p1x+p2x2+p3x3, then

[p(x)]B=    p0 p1 p2 p3  

 and A[p(x)]B= 

 

0 0 0 0 0 1 0 0 0 0 2 0 0 0 0 3

      p0 p1 p2 p3   =    0 p1 2p2 3p3   ,

so thatT(p(x)) =p1x+ 2p2x2+ 3p3x3. Observe that

T(p(x)) =xp′(x) =x(p1+ 2p2x+ 3p3x2) =p1x+ 2p2x2+ 3p3x3,

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Example 8.8.2. Consider the linear operatorT :R2R2, given byT(x1, x2) = (2x1+x2, x1+ 3x2)

for every (x1, x2)∈R2. Consider also the basisB={(1,0),(1,1)} of R2. Then the matrix for T with respect toB is given by

A= ( [T(1,0)]B [T(1,1)]B) = ( [(2,1)]B [(3,4)]B) =

1 −1

1 4

.

Suppose that (x1, x2) = (3,2). Then

[(3,2)]B=

1 2

and A[(3,2)]B=

1 −1

1 4

1 2

=

−1 9

,

so thatT(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 thatT(x1, x2) = (x1−2x2)(1,0) + (x1+ 3x2)(1,1) = (2x1+x2, x1+ 3x2).

Example 8.8.3.Suppose that T :RnRmis a linear transformation. Suppose further thatB andC are the standard bases forRn andRm respectively. Then the matrix forT with respect to Band C is given by

A= ( [T(e1)]C . . . [T(en)]C) = (T(e1) . . . T(en) ),

so it follows from Proposition 8A thatAis simply the standard matrix forT.

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 basesB ={v1, . . . ,vn}, C ={w1, . . . ,wm} and

D={u1, . . . ,uk}. We then have the following diagram of linear transformations.

V W U

Rn Rm Rk

T1

φ

T2

ψ η

S1

φ−1

S2

ψ−1 η−1

Hereη:U →Rk, whereη(u) = [u]D for everyu∈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 forS1 andS2, so that they are respectively the matrix forT1 with respect toB and C and the matrix forT2with respect to C andD. Clearly

S2◦S1=η◦T2◦T1◦φ−1:

Rn →Rk.

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PROPOSITION 8T. Suppose thatT1 :V → W andT2 :W →U are linear transformations, where the real vector spacesV, W, U are finite dimensional, with basesB,C,Drespectively. Suppose further that

A1 is the matrix for the linear transformationT1 with respect to the basesB and C, and that A2 is the matrix for the linear transformationT2 with respect to the basesC andD. ThenA2A1 is the matrix for the linear transformationT2◦T1 with respect to the basesB andD.

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 forT1 with respect to the basis

B={1, x, x2, x3} ofP3 is given by

A1= 

 

0 0 0 0 0 1 0 0 0 0 2 0 0 0 0 3

 .

Consider next the linear operatorT2:P3→P3, where for every polynomialq(x) =q0+q1x+q2x2+q3x3

inP3, we have

T2(q(x)) =q(1 +x) =q0+q1(1 +x) +q2(1 +x)2+q3(1 +x)3.

We haveT2(1) = 1,T2(x) = 1 +x, T2(x2) = 1 + 2x+x2 andT2(x3) = 1 + 3x+ 3x2+x3, so that the

matrix forT2 with respect toB is given by

A2= ( [T2(1)]B [T2(x)]B [T2(x2)]B [T2(x3)]B) =

 

1 1 1 1 0 1 2 3 0 0 1 3 0 0 0 1

 .

Consider now the compositionT =T2◦T1:P3→P3. LetAdenote the matrix forT with respect toB. By Proposition 8T, we have

A=A2A1= 

 

1 1 1 1 0 1 2 3 0 0 1 3 0 0 0 1

     

0 0 0 0 0 1 0 0 0 0 2 0 0 0 0 3

  =   

0 1 2 3 0 1 4 9 0 0 2 9 0 0 0 3

 .

Suppose thatp(x) =p0+p1x+p2x2+p3x3. Then

[p(x)]B=    p0 p1 p2 p3  

 and A[p(x)]B= 

 

0 1 2 3 0 1 4 9 0 0 2 9 0 0 0 3

      p0 p1 p2 p3   =   

p1+ 2p2+ 3p3 p1+ 4p2+ 9p3

2p2+ 9p3

3p3

 ,

so thatT(p(x)) = (p1+ 2p2+ 3p3) + (p1+ 4p2+ 9p3)x+ (2p2+ 9p3)x2+ 3p3x3. We can check this directly

by noting that

T(p(x)) =T2(T1(p(x))) =T2(p1x+ 2p2x2+ 3p3x3) =p1(1 +x) + 2p2(1 +x)2+ 3p3(1 +x)3 = (p1+ 2p2+ 3p3) + (p1+ 4p2+ 9p3)x+ (2p2+ 9p3)x2+ 3p3x3.

Example 8.8.5. Consider the linear operatorT :R2R2, given byT(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)}ofR2 is given by

A=

1 −1

1 4

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Consider the linear operatorT2:R2R2. By Proposition 8T, the matrix forT2 with respect toB is

given by

A2=

1 −1

1 4

1 −1

1 4

=

0 −5 5 15

.

Suppose that (x1, x2)∈R2. Then

[(x1, x2)]B=

x1−x2 x2

and A2[(x1, x2)]B=

0 −5 5 15

x1−x2 x2

=

−5x2

5x1+ 10x2

,

so thatT(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 thatT :V →V is a linear operator on a finite dimensional real vector spaceV with basis B. Suppose further thatA is the matrix for the linear operatorT with respect to the basisB. Then T is one-to-one if and only ifAis invertible. Furthermore, ifT is one-to-one, then A−1

is the matrix for the inverse linear operatorT−1:V V with respect to the basis B.

Proof.Simply note thatT is one-to-one if and only if the systemAx=0has only the trivial solution x = 0. The last assertion follows easily from Proposition 8T, since if A′ denotes the matrix for the inverse linear operatorT−1 with respect toB, then we must haveAA=I, the matrix for the identity operatorT−1T with respect to B.

Example 8.8.6.Consider the linear operatorT :P3P3, where for everyq(x) =q0+q1x+q2x2+q3x3

inP3, 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 forT with respect to the basisB={1, x, x2, x3}is given by

A= 

 

1 1 1 1 0 1 2 3 0 0 1 3 0 0 0 1

 .

This matrix is invertible, so it follows thatT is one-to-one. Furthermore, it can be checked that

A−1= 

 

1 −1 1 −1

0 1 −2 3

0 0 1 −3

0 0 0 1

 .

Suppose thatp(x) =p0+p1x+p2x2+p3x3. Then

[p(x)]B=    p0 p1 p2 p3  

 and A

−1[p(x)]B=

 

1 −1 1 −1

0 1 −2 3

0 0 1 −3

0 0 0 1

      p0 p1 p2 p3   =   

p0−p1+p2−p3 p1−2p2+ 3p3

p2−3p3 p3

 ,

so that

T−1(p(x)) = (p0p1+p2p3) + (p12p2+ 3p3)x+ (p23p3)x2+p3x3

=p0+p1(x−1) +p2(x22x+ 1) +p3(x33x2+ 3x1)

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8.9. Change of Basis

Suppose that V is a finite dimensional real vector space, with one basis B={v1, . . . ,vn} and another basisB′ ={u1, . . . ,u

n}. Suppose that T :V →V is a linear operator onV. LetA 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 andT(v) =w, then

[w]B=A[v]B (10)

and

[w]B′ =A′[v]B′. (11)

We wish to find the relationship betweenA′ andA.

Recall Proposition 8J, that if

P= ( [u1]B . . . [un]B)

denotes the transition matrix from the basisB′ to the basisB, then

[v]B=P[v]B′ and [w]B=P[w]B′. (12)

Note that the matrixP can also be interpreted as the matrix for the identity operatorI:V →V with respect to the basesB′

andB. It is easy to see that the matrixP is invertible, and

P−1= ( [v1]B′ . . . [vn]B′)

denotes the transition matrix from the basisBto the basisB′

, and can also be interpreted as the matrix for the identity operatorI:V →V with respect to the basesBandB′

.

Combining (10) and (12), we conclude that

[w]B′ =P

−1[w]B=P−1A[v]B=P−1AP[v]B

′.

Comparing this with (11), we conclude that

P−1AP =A

. (13)

This implies that

A=P A′P−1. (14)

Remark.We can use the notation

A= [T]B and A′

= [T]B′

to denote thatAandA′ are the matrices forT with respect to the basisBand with respect to the basis

B′

respectively. We can also write

P = [I]B,B′

to denote thatP is the transition matrix from the basis B′

to the basisB, so that

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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 forT with respect to the basis B and with respect to the basisB′

respectively. Then

P−1AP =A

and A′

=P AP−1,

where

P = ( [u1]B . . . [un]B)

denotes the transition matrix from the basisB′ to the basisB.

Remarks.(1) We have the following picture.

v w

v w

[v]B! [w]B!

[v]B [w]B

T

I I

T

A!

P

A

P−1

(2) The idea can be extended to the case of linear transformationsT :V →W from a finite dimensional real vector space into another, with a change of basis inV and a change of basis inW.

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 polynomialp(x) =p0+p1x+p2x2+p3x3, we have T(p(x)) = (p0+p1) + (p1+p2)x+ (p2+p3)x2+ (p0+p3)x3. LetAdenote the matrix forT with respect to the basisB. ThenT(1) = 1 +x3,T(x) = 1 +x,T(x2) =x+x2 andT(x3) =x2+x3, and so

A= ( [T(1)]B [T(x)]B [T(x2)]B [T(x3)]B) =

 

1 1 0 0 0 1 1 0 0 0 1 1 1 0 0 1

 .

Next, note that the transition matrix from the basisB′ to the basisBis given by

P = ( [1]B [1 +x]B [1 +x+x2]B [1 +x+x2+x3]B) =

 

1 1 1 1 0 1 1 1 0 0 1 1 0 0 0 1

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It can be checked that

P−1= 

 

1 −1 0 0

0 1 −1 0

0 0 1 −1

0 0 0 1

 ,

and so

A′

=P−1AP =

 

1 −1 0 0

0 1 −1 0

0 0 1 −1

0 0 0 1

 

 

1 1 0 0 0 1 1 0 0 0 1 1 1 0 0 1

 

 

1 1 1 1 0 1 1 1 0 0 1 1 0 0 0 1

 =

 

1 1 0 0

0 1 1 0

−1 −1 0 0

1 1 1 2

 

is the matrix forT with respect to the basisB′

. 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 spaceV.

Then any real numberλ∈Ris called an eigenvalue ofT if there exists a non-zero vectorv∈V such that

T(v) =λv. This non-zero vectorv∈V is called an eigenvector ofT corresponding to the eigenvalueλ.

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 BofV. The starting point of our argument is the following theorem, the proof of which is left as an exercise.

PROPOSITION 8W.Suppose thatT :V →V is a linear operator on a finite dimensional real vector spaceV, with bases BandB′. Suppose further thatAandAare the matrices forT with respect to the

basisB and with respect to the basisB′ respectively. Then

(a) detA= detA′

; (b) AandA′

have the same rank;

(c) AandA′ have the same characteristic polynomial;

(d) AandA′ have the same eigenvalues; and

(e) the dimension of the eigenspace of Acorresponding to an eigenvalue λis equal to the dimension of the eigenspace ofA′

corresponding toλ.

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 spaceV. Suppose further thatAis the matrix for T with respect to a basisB of V. Then

(a) the eigenvalues of T are precisely the eigenvalues ofA; and

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Suppose now that A is the matrix for a linear operator T : V → V on a finite dimensional real vector spaceV with respect to a basisB={v1, . . . ,vn}. IfAcan be diagonalized, then there exists an invertible matrixP such that

P−1AP =D

is a diagonal matrix. Furthermore, the columns of P are eigenvectors of A, and so are the coordinate matrices of eigenvectors ofT with respect to the basisB. In other words,

P = ( [u1]B . . . [un]B),

whereB′ ={u1, . . . ,u

n}is a basis ofV consiting of eigenvectors ofT. Furthermore,P is the transition matrix from the basisB′ to the basisB. It follows that the matrix forT with respect to the basisBis given by

D= 

λ1

. ..

λn 

,

whereλ1, . . . , λn are the eigenvalues ofT.

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

polynomialp(x) =p0+p1x+p2x2, we haveT(p(x)) = (5p02p1) + (6p1+ 2p22p0)x+ (2p1+ 7p2)x2.

ThenT(1) = 5−2x,T(x) =−2 + 6x+ 2x2 andT(x2) = 2x+ 7x2, so that the matrix forT with respect

to the basisBis given by

A= ( [T(1)]B [T(x)]B [T(x2)]B) =

5 −2 0

−2 6 2

0 2 7

.

It is a simple exercise to show that the matrixAhas eigenvalues 3,6,9, with corresponding eigenvectors

x1=

 2 2

−1 

, x2=

 2

−1 2

, x3=

−1 2 2

,

so that writing

P = 

2 2 −1

2 −1 2

−1 2 2

,

we have

P−1AP = 

3 0 0 0 6 0 0 0 9

.

Now letB′

={p1(x), p2(x), p3(x)}, where

[p1(x)]B= 

 2 2

−1 

, [p2(x)]B= 

 2

−1 2

, [p3(x)]B= 

−1 2 2

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ThenP is the transition matrix from the basisB′

to the basisB, andD is the matrix forT with respect to the basisB′

. Clearlyp1(x) = 2 + 2x−x2,p2(x) = 2x+ 2x2andp3(x) =1 + 2x+ 2x2. Note now

that

Referências

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