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CHAPTER 6

LINEAR TRANSFORMATIONS

6.1 WHAT IS A LINEAR TRANSFORMA-

TION?

You know what a function is - its a RULE which

turns NUMBERS INTO OTHER NUMBERS: f(x) =

x2 means please turn 3 into 9, 12 into 144 and so

on.

Similarly a TRANSFORMATION is a rule whichturns VECTORS into other VECTORS. For exam-

ple, please rotate all 3-dimensional vectors through

an angle of 90 clockwise around thez-axis. A LIN-

EAR TRANSFORMATION T is one that ALSOsatisfies these rules: ifc is any scalar, and

u and

v

are vectors, then

T(cu ) =cT(

u ) and T(

u +

v ) =T(

u ) + T(

v ).

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EXAMPLE: Let I be the rule Iu =

u for all

u .

You can check that I is linear! Called IDENTITY

Linear Transformation.

EXAMPLE : Let D be the rule Du = 2

u for all

u .

D(cu ) = 2(c

u ) =c(2

u ) =cD

u

D(u +

v ) = 2(

u +

v ) = 2

u + 2

v = D

u +D

v

LINEAR!

Note: Usually we write D(u ) as just D

u .

6.2. THE BASIC BOX, AND THE MATRIX

OF A LINEAR TRANSFORMATION

The usual vectors i andj define a square:

Lets call this the BASIC BOX in two dimensions.

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Similarly, i, j, and k define the BASIC BOX in 3

dimensions.

Now let T be any linear transformation. You know

that any 2-dimensional vector can be written asai +

bj, for some numbers aand b. So for any vector, we

have

T(ai + bj) =aTi + bTj.

This formula tells us something very important: IF

I KNOW WHAT T DOES TO i and j, THEN I

KNOW EVERYTHING ABOUT T - because now I

can tell you what Tdoes to ANY vector.

EXAMPLE: Suppose I know that T(i) = i+ 14j

and T(j) = 14 i +j. Then what is T(2i + 3j)?

Answer: T(2i+ 3j) = 2Ti+ 3Tj = 2

i + 14j

+

314 i +

j

= 2i + 12j+ 34 i + 3

j = 114 i + 72

j.

Since Ti and Tj tell me everything I need to know,

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this means that I can tell you everything about T by

telling you WHAT IT DOES TO THE BASIC BOX.

EXAMPLE: Let Tbe the same transformation as

above, T(i) = i + 14j and T(j) = 14 i +

j.

The basic box has been squashed a bit! Pictures

of WHAT T DOES TO THE BASIC BOX tell us

everything about T!

EXAMPLE: IfD is the transformation Du = 2

u ,

then the Basic Box just gets stretched:

So every LT can be pictured by

seeing what it does to the Basic Box.

There is another way!

Let Ti =

a

c

and Tj =

b

d

. Then we DEFINE

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THE MATRIX OF TRELATIVE TO i, jas

a b

c d

,

that is, the first COLUMN tells us what happened

to i, and the second column tells us what happened

toj.

EXAMPLE: Let Ibe the identity transformation.

Then Ii=

i=

10

, Ij =

j =

01

, so the matrix of

the identity transformation relative to ij is

1 00 1

.

EXAMPLE: If Du = 2

u , then Di =

20

and

Dj =

02

so the matrix of D relative to i, j is2 00 2

.

EXAMPLE: IfTi= i+ 14j and Tj = 14 i+j, then

the matrix is

1

1

414 1

.

EXAMPLE: If Ti =j and Tj = i, the matrix is0 11 0

. Basic box is REFLECTED

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EXAMPLE: Suppose in 3 dimensions Ti= i+4j +

7k, Tj = 2i+ 5j+ 8k, Tk = 3i+ 6j+ 9k, then the

matrix is

1 2 34 5 67 8 9

, relative to ijk.

EXAMPLE: Suppose Ti = i+j + 2k and Tj =

i

3k. This is an example of a LT that eats 2-

dimensional vectors but PRODUCES 3-dimensional

vectors. But it still has a matrix,

1 11 0

2 3

. Its

just that this matrix is not a SQUARE MATRIX,

that is, it is not 2 by 2 or 3 by 3. Instead it is 3 by

2.

We shall say that a linear transformation is a 2-

dimensional L.T. if it eats 2-dimensional vectors AND

produces 2-dimensional vectors. A 2-dimensional

L.T. has a square, 2 by 2 matrix relative to i, j. Sim-

ilarly a 3-dimensional linear transformation has a 3

by 3 matrix. In Engineering applications, most lin-

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ear transformations are 2-dimensional or 3-dimensional,

so we are mainly interested in these two cases.

EXAMPLE: Suppose T is a linear transformation

that eats 3-dimensional vectors and produces

2-dimensional vectors according to the rule Ti= 2i,

Tj = i + j, Tk= i

j. What is its matrix?

Answer:

2 1 10 1 1

, a 2 by 3 matrix.

EXAMPLE: Suppose, in solid mechanics, you take

a flat square of rubber and SHEAR it, as shown.

In other words, you dontchange its volume,

you just push it like a pack

of cards. The base stays fixed but the top moves

a distance tan(). (The height remains the same, 1unit.) Clearly the shearing transformationSsatisfies

Si = i, Sj = i tan +j, so the matrix ofS relative

to i, j is

1 tan 0 1

.

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EXAMPLE: Suppose T i = i+j and Tj = i+

j. Matrix is

1 11 1

and basic box is SQUASHED

FLAT!

EXAMPLE: Rotations in the plane. Suppose you

ROTATE the whole plane through an angle (anti-

clockwise). Then simple trigonometry shows you

that

Ri= cos i + sin j

Rj = sin i + cos jSo the rotation matrix is

R() =

cos sin sin cos

.

Application: Suppose an object is moving on a cir-

cle at constant angular speed . What is its accel-

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eration?

Answer: Let its position vector at t = 0 b e

r0.Because the object is moving on a circle, its position

at a later time t is given by rotatingr0 by an angle

(t). So

r (t) =

cos sin sin cos

r0

Differentiate

d

rdt

=

sin cos cos

sin

r0by the chain rule. Here

is actually , so

d

rdt

=

sin cos cos sin

r0. Differentiate again,

d2

rdt2

= cos sin sin cos

2r0

= 2

cos sin sin cos

r0

.

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Substitute the equation for

r (t),

d

2

rdt2 = 2

r ,

which is formula you know from physics.

6.3. COMPOSITE TRANSFORMATIONS

AND MATRIX MULTIPLICATION.

You know what it means to take the COMPOSITE

of two functions: if f(u) = sin(u), and u(x) = x2,

then f

umeans: please do uFIRST, THEN f, so

f u(x) = sin(x2). NOTE THE ORDER!!u f(x) = sin2(x), NOT the same!

Similarly ifAand Bare linear transformations, then

AB means do B FIRST, then A.

NOTE: BE CAREFUL! According to our definition,

AandB both eat vectors and both produce vectors.

But then you have to take care that Acan eat what

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B produces!

EXAMPLE: Suppose A eats and produces 2-dimensionalvectors, andB eats and produces 3-dimensional vec-

tors. Then AB would not make sense!

EXAMPLE: Suppose B eats 2-d vectors and pro-

duces 3-d vectors (so its matrix relative to ijk looks

like this:

b11 b12b21 b22

b31 b32

, a 3 by 2 matrix) and suppose

A eats 3-d vectors and produces 2-d vectors. Then

AB DOES make sense, because A can eat what B

produces. (In this case, BA also makes sense.).

IMPORTANT FACT: Suppose aij is the matrix

of a linear transformationArelative to ijk, and sup-

pose bij is the matrix of the Linear Transformation

B relative to ijk. Suppose that AB makes sense.

Then the matrix of AB relative to ij or ijk is just

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the matrix product ofaij and bij .

EXAMPLE: What happens to the vector 12 if

we shear 45 parallel to the x axis and then rotate

90 anticlockwise? What if we do the same in the

reverse order?

Answer: Shear

1 tan 0 1

so in this case it is

1 10 1

. A rotation through

has matrix

cos sin sin cos

, so here it is

0 11 0

Hence

SHEAR, THEN ROTATE

0

1

1 0 1 1

0 1

= 0

11 1

ROTATE, THEN SHEAR

1 10 1

0 11 0

=

1 11 0

.

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So shear, then rotate

1

2

0 11 1

1

2=

23 .

Rotate, then shear12

1 11 0

12

=

11

Very different!

EXAMPLE: Suppose B is a LT with matrix 1 00 1

1 1

andAis a LT with matrix

0 1 11 1 0

.

What is the matrix ofAB? OfBA?

Answer: 0 1 11 1 0

1 00 11 1

=

1 01 1

=AB

2 by 3 3 by 2 2 by 2 1 00 1

1 1

0 1 11 1 0

=

0 1 11 1 0

1 0 1

3 by 2 2 by 3 3 by 3

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EXAMPLE: Suppose you take a piece of rubber in

2 dimensions and shear it parallel to the x axis by

degrees, and then shear it again by degrees. What

happens?

1 tan 0 1

1 tan 0 1

= 1 tan + tan

0 1

which is also a shear, but NOT through + !

The shear angles dont add up, since tan +tan =tan(+ ).

EXAMPLE: Rotate 90 aroundz-axis, then rotate

90 around x-axis in 3 dimensions. [Always anti-

clockwise unless otherwise stated.] Is it the same

if we reverse the order? Rotate about z axisi becomes j, j becomesi, k stays the same, so 0 1 01 0 0

0 0 1

. Rotate aboutxaxis,istays the same,

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j becomes k, k becomesj, so

1 0 00 0 10 1 0

, and

1 0 00 0 1

0 1 0

0 1 01 0 0

0 0 1

= 0

1 0

1 0 00 0 1

1 0 00 0 10 1 0

,

so the answer is NO!

6.4 DETERMINANTS

You probably know that the AREA of the box de-

fined by two vectors is

|

u

v |, magnitudeof the vector product.

If you dont know it, you can easily check it, since

the area of any parallelogram is given by

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AREA=HEIGHT Base= |

v

|sin |

u

|= |u | |v | sin = |u v |.

Similarly, the VOLUME of a three-dimensional par-

allelogram [called a PARALLELOPIPED!] is given

by

VOLUME = (AREA OF BASE) HEIGHT.

If you take any 3 vectors in 3 dimensions, sayu ,

v ,

w,

then they define a 3-dimensional parallelogram. The

area of the base is|u v |,height is

|w

| |sin

2 |

where is the angle between

u v and w, so VOLUMEdefined by

u ,

v ,

w is just

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|u v | |w| | sin

2

|

=

|u

v

| |w

| |cos

|=|u v w|.

[Check: Volume of Basic Box defined by ijk is

|i j k| = |k k| = 1, correct!

Now let T be any linear transformation in two di-

mensions. [This means that it acts on vectors in the

xyplane and turns them into other vectors in the xy

plane.]

We let Tact on the Basic Box, as usual.

Now Ti and Tj still lie in the same plane as i and

j, s o (Ti) (Tj) must be perpendicular to thatplane. Hence it must be some multiple of k. We

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define the DETERMINANT ofT to be that multi-

ple, that is, by definition, det(T) is the number given

[STRICTLY IN 2 DIMENSIONS] by

(Ti) (Tj) = det(T)k.

EXAMPLE: IfI= identity, then

Ii I

j =

i

j =

k= 1

k

so det(I) = 1.

EXAMPLE: Du = 2

u

Di

Dj = 4i

j = 4k

det(D) = 4

EXAMPLE: Ti= i + 14j, Tj = 14 i +

j,

Ti Tj =

i +1

4j

1

4i + j

= i j+ 1

16j i

= 1516i j = 1516 k det T = 1516 .

EXAMPLE: Ti=j, Tj = i,

Ti Tj =j i= k det T = 1

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EXAMPLE: Shear, Si= i, Sj = i tan + j,

Si Sj =k det S= 1.

EXAMPLE: Ti= i + j =Tj,

Ti Tj = 0 det T = 0.

EXAMPLE: Rotation

Ri Rj = (cos i + sin j) ( sin i + cos j)= (cos2 sin2 )k=k det(R) = 1.

The area of the Basic Box is initially|i j| = 1.After we let Tact on it, the area becomes

|Ti T

j| = |det T| |

k| = | det T|.

So

Final Area of Basic Box

Initial Area of Basic Box =

| det T|1

= | det T|

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so | det T| TELLS YOU THE AMOUNT BY WHICHAREAS ARE CHANGED BY T. So det T =

1

means that the area is UNCHANGED (Shears, ro-

tations, reflections) while det T= 0 means that the

Basic Box is squashed FLAT, zero area.

Take a general 2 by 2 matrix M =

a bc d

. We

know that this means Mi = ai+cj, Mj = bi+dj.

Hence Mi Mj =

ai + cj

bi + dj

= (ad bc)k, so

det

a bc d

=ad bc.

Check: det

2 00 2

= 4, det

1 tan 0 1

= 1,

det

cos sin sin cos

= 1, det

1 11 1

= 0.

IN THREE dimensions there is a similar gadget.

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The Basic Box is defined by ijk, and we can let any

3-dimensional L.T. act on it, to get a new box de-

fined by Ti, Tj, Tk. We define

det T =

Ti

Tj

Tk

where the dot is the scalar product, as usual. Since

|TiTj Tk| is the volume of the 3-dimensionalparallelogram defined by Ti, Tj, Tk, we see that

| det T| = Final Volume of Basic BoxInitial Volume of Basic Box

,

that is,|

det T| tells you how much T changes vol-

umes. IfTsquashes the Basic Box flat, then

det T = 0.

Just as det a b

c d = ad

bc, there is a formula

for the determinant of a 3 by 3 matrix. The usual

notation is this. We DEFINE

a b

c d

= det

a b

c d

=ad bc.

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Similarly

a11 a12 a13a21 a22 a23a31 a32 a33

is the determinant of

a11 a12 a13a21 a22 a23

a31 a32 a33

and there is a formula for it, as

follows:

a11 a12 a13a21 a22 a23a31 a32 a33

=a11

a22 a23a32 a33 a12

a21 a23a31 a33+ a13

a21 a22a31 a32 .

In other words, we can compute a three-dimensional

determinant if we know how to work out 2-dimensional

determinants.

COMMENTS:

[a] We worked along the top row. Actually, a THE-

OREM says that you can use ANY ROW OR ANYCOLUMN!

[b] How did I know that a12had to multiply the par-

ticular 2-dimensional determinant

a21 a23a31 a33

? Easy:

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I just struck out EVERYTHING IN THE SAME

ROW AND COLUMN as a12 :

a21

a23

a31 a33

and

just kept the survivors!

This is the pattern, for example if you expand along

the second row you will get

a21

a12 a13 a32 a33

+ a22

a11 a13

a31 a33

a23

a11 a12 a31 a32

[c] What is the pattern of the + and signs? It is

an (Ang Moh ) CHESSBOARD, starting with a +

in the top left corner:+ + + + +

[d] You can do exactly the same thing in FOUR di-

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mensions, following this pattern, using

+ + + ++

+

+ +because now you know how to work out 3-dimensional

determinants. And so on!

Example:

1 1 01 1 12 0 0

= 1 1 10 0

1 1 12 0

+ 0 1 12 0

= 0 + 2 + 0 = 2

(expanding along the top row) or, if you use the sec-

ond row,

1 1 01 1 12 0 0

= 1

1 00 0+ 1

1 02 0 1

1 12 0

= 0 + 0 + 2 = 2

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or

1

1 0

1 1 12 0 0

= 1

1 10 0 1

1 00 0+ 2

1 01 1

= 0 + 0 + 2 = 2

(expanding down the first column).

Important Properties of Determinants

[a] Let S and Tbe two linear transformations such

that det Sand det T are defined. Then

det ST= det T S= (det S) (det T).

Therefore, det[ST U] = det[U ST] = det[T U S] and

so on: det doesnt care about the order. Remember

however that this DOES NOT mean that ST U =

U ST etc etc.

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[b] IfM is a square matrix, then

det MT = det M.

[c] Ifcis a number and M is an nbynmatrix, then

det(cM) =cn det M.

EXAMPLE: Remember from Section 2[g] of Chap-

ter 5 that an ORTHOGONAL matrix satisfies M MT =

I. So det(M MT) = det I = 1. But det(M MT) =

det(M)det(MT) = det(M) det(M) = (det M)2,thus

det M= 1

for any orthogonal matrix.

6.5. INVERSES.

If I give you a 3-dimensional vector

u and a 3-

dimensional linear transformation T, then T sends

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u to a particular vector, it never sends

u to two

DIFFERENT VECTORS! So this picture is impos-

sible:

Tu

u

Tu

But what about this picture:u

Tu =T

v

v

Can T send TWO DIFFERENT VECTORS TO

ONE? Yes!

1 0 00 0 00 0 0

010

= 000

= 1 0 00 0 00 0 0

001

So it can happen! Notice that this transformation

destroys j (and also k). In fact if

u = v and

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Tu = T

v , then T(

u v ) = 0, that is, Tw = 0

wherew IS NOT THE ZERO VECTOR. So if this

happens, T destroys everything in the w direction.

That is, T SQUASHES 3-dimensional space down

to two or even less dimensions. This means that

TLOSES INFORMATION it throws away all ofthe information stored in the w direction. ClearlyT

squashes the basic box down to zero volume, so

det T = 0

and we say T is SINGULAR.

SUMMARY: A SINGULAR LINEAR TRANSFOR-

MATION

[a] Maps two different vectors to one vector

[b] Destroys all of the vectors in at least one direction

[c] Loses all information associated with those direc-

tions

[d] Satisfies det T = 0.

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Conversely, a NON-SINGULAR transformation never

maps 2 vectors to one,

u Tuv Tv

Different

Therefore if I give you Tu , THERE IS EXACTLY

ONEu . The transformation that takes you fromT

u

back tou is called the INVERSE OF T. The idea is

that since a NON-SINGULAR linear transformation

does NOT destroy information, we can re-constructu if we are givenT

u . ClearlyTHAS AN INVERSE,

CALLED T1, if and only if det T= 0. All thisworks in other dimensions too.

EXAMPLE:

1 11 1

34

=

77

,

1 11 1

43

=

77

,

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two different vectors to one!

1 11 1

11 = 1 1

1 1 22

=

1 11 1

13.5913.59

=

00

It destroys everything

in that direction!Finally det

1 11 1

= 0.

So it is SINGULAR and

has NO inverse.

11

EXAMPLE: Take

0 11 0

and suppose it acts on

and

a

b

and sends them to the same vector,

so

0 11 0

= 0 11 0

a

b

.

Then

=

ba

=a

=b

=

a

b

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so

and

a

b

are the same this transforma-

tion never maps different vectors to the same vector.

No vector is destroyed, no information is lost, noth-

ing gets squashed! And det

0 11 0

= 1, NON-

SINGULAR.

How to FIND THE INVERSE.

By definition, T1 sends Tu to

u , i.e.

T1(T(u )) =

u =T(T1(

u )).

Butu =I

u (identity) so T1 satisfies

T1T =T T1 =I.

So to find the inverse of 0 11 0 we just have to

find a matrix

a b

c d

such that

a b

c d

0 11 0

=

1 00 1

, b= 1, a= 0, d= 0, c= 1 so answer is

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0 11 0

. In fact its easy to show that

a bc d

1

= 1ad bc

d bc a

.

For example, when we needed to find the matrix S

in Section 4 of Chapter 5, we needed to find a way

of solvingS

0.7 0.40.5 0.7

=I.

This just means that we need to inverse of

0.7 0.40.5 0.7

,

and the above formula does the job for us.

For bigger square matrices there are many tricks

for finding inverses. A general [BUT NOT VERY

PRACTICAL] method is as follows:

[a] Work out the matrix of COFACTORS. [A cofac-

tor is what you get when you work out the smaller

determinant obtained by striking out a row and a

column, for example the cofactor of 6 in

1 2 34 5 67 8 9

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is

1 27 8

=6. You can do this for each elementin a given matrix, to obtain a new matrix of the

same size. For example, the matrix of cofactors of 1 0 10 1 0

0 0 1

is

1 0 00 1 0

1 0 1

.

[b] Keep or reverse the signs of every element accord-

ing to+ + + + +

(you get

1 0 00 1 0

1 0 1

above.)

[c] Take the TRANSPOSE,

1 0 10 1 0

0 0 1

.

[d] Divide by the determinant of the original ma-

trix. THE RESULT IS THE DESIRED INVERSE.

1 0 10 1 00 0 1

in this example. Check:

1 0 10 1 0

0 0 1

1 0 10 1 0

0 0 1

=

1 0 00 1 0

0 0 1

.

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INVERSE OF A PRODUCT:

(AB)1 =B1A1

Note the order! Easily checked:

(AB)1

AB =B1

(A1

A)B=B1

IB =I.

APPLICATION: SOLVING LINEAR SYSTEMS.

Suppose you want to solve

x + 2y+ 3z = 14x + 5y+ 6z = 27x + 8y+ 9z= 4.

One way is to write it as

1 2 34 5 6

7 8 9

xy

z

=

12

4

.

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Then all you have to do is find

1 2 34 5 6

7 8 9

1

and

multiply it on both sides, so you getxy

z

=

1 2 34 5 6

7 8 9

1 12

4

= answer.

So this is a systematic way of solving such problems!

Now actually det

1 2 34 5 6

7 8 9

= 0, and you can see

why:

1 2 34 5 6

7 8 9

1

21

=

00

0

.So this transformation destroys everything in the di-

rection of

12

1

. In fact it squashes 3-dimensional

space down to a certain 2-dimensional space. [We

say that the matrix has RANK 2. If it had squashed

everything down to a 1-dimensional space, we would

say that it had RANK 1.] Now actually

12

4

. DOES

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NOT lie in that two-dimensional space. Since

1 2 34 5 67 8 9

squashes EVERYTHING into that two-dimensional

space, it is IMPOSSIBLE for

1 2 34 5 6

7 8 9

xy

z

to be

equal to

124

. Hence the system has NO solutions.

If we change

12

4

to

12

3

, this vector DOES lie

in the special 2-dimensional space, and the system

1 2 34 5 67 8 9

xyz

= 123

DOES have a solutionin fact it has infinitely many!

SUMMARY:

Any system of linear equations can be written as

Mr =

a

where M is a matrix,

r = the vector of variables,

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anda is a given vector. Suppose M is square.

[a] If det M= 0, there is exactly one solution,

r =M1

a .

[b] If det M= 0, there is probably no solution. But

if there is one, then there will be many.

PRACTICAL ENGINEERING PERSPECTIVE:

In the REAL world, NOTHING IS EVER EXACTLY

EQUAL TO ZERO! So if det M= 0, either [a] you

have made a mistake, OR [b] you are pretending that

your data are more accurate than they really are!

1 2 34 5 67 8 9

REALLY means

1.01 2.08 3.033.99 4.97 6.027.01 7.96 8.98

and of course the determinant of THIS is non-zero!

Actually, det = 0.597835!

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6.6 EIGENVECTORS AND EIGENVALUES.

Remember we said that a linear transformation USU-

ALLY changes the direction of a vector. But there

may be some special vectors which DONT have their

direction changed!

EXAMPLE:

1 22 2

clearly DOES change the

direction of i and j, since

12

is not parallel to i

and 22 is not parallel toj. BUT

1 22 2

21

=

42

= 2

21

which IS parallel to 21.

In general if a transformationTdoes not change the

direction of a vectoru , that is

Tu =

u

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for some (SCALAR), thenu is called an EIGEN-

VECTOR of T. The scalar is called the EIGEN-

VALUE ofu .

6.7 FINDING EIGENVALUES AND EIGEN-

VECTORS.

There is a systematic way of doing this. Take the

equation

Tu =

u

and write u =Iu , I = identity. Then

(T I)u = 0

Lets supposeu= 0[of course, 0is always an eigen-

vector, that is boring]. So the equation says that

T I SQUASHES everything in the u direction.Hence

det(T I) = 0.

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This is an equation which can be SOLVED to find

.

EXAMPLE: Find the eigenvalues of

1 22 2

:

det

1 22 2

1 00 1

= 0

det 1

2

2 2 = 0 (1 )(2 + ) 4 = 0 = 2 OR 3

So there are TWO answers for a 2 by 2 matrix. Sim-

ilarly, in general there are three answers for 3 by 3

matrices, etc.

What are the eigenvectors for = 2, = 3?

IMPORTANT POINT: Let

u be an eigenvectorofT. Then 2

u is also an eigenvector with the same

eigenvalue!

T(2u ) = 2T

u = 2

u = (2u ).

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Similarly 3u , 13.59

u etc are all eigenvectors! SO

YOU MUST NOT EXPECT A UNIQUE ANSWER!

OK, with that in mind, lets find an eigenvector for

= 2. Lets call an eigenvector

. Then

(T I)u = 0

1 22 2

= 0

1 2

2 4

= 0

+ 2= 0

2 4= 0But these equations are actually the SAME, so we

really only have ONE equation for 2 unknowns. We

arent surprised, because we did not expect a unique

answer anyway! We can just CHOOSE = 1 (or13.59 or whatever) and then solve for . Clearly

= 12 , so an eigenvector corresponding to = 2

is

112

. But if you said

21

or

10050

that is also

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correct!

What about =

3?

4 22 1

= 0

4 + 2= 0

2 + = 0

Again we can set = 1, then =2, so an eigen-vector corresponding to =3 is

12

or

24

or 10

20 etc.

EXAMPLE: Find the eigenvalues, and correspond-

ing eigenvectors, of

0 11 0

.

Answer: We have det

11

= 02 + 1 =

0 = i, i= 1.Eigenvector for i: we set

i 11 i

1

= 0

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i = 0=i so an eigenvector for i is 1

i

. For =

i we have

i 11 i

1

= 0

i = 0 = i so an eigenvector foriis 1

i. Note that a REAL matrix can have COM-

PLEX eigenvalues and eigenvectors! This is hap-

pening simply because

0 11 0

is a ROTATION

through 90, and of course such a transformation

leaves NO [real] vectors direction unchanged (apart

from the zero vector).

6.8. DIAGONAL FORM OF A LINEAR TRANS-

FORMATION.

Remember that we defined the matrix of a linear

transformation T WITH RESPECT TO i, j by let-

tingTact on iandj and then putting the results in

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the columns. So to say that T has matrix

a b

c d

with respect to i, j means that

Ti=ai + cj

Tj =bi + dj.

Whats so special about the two vectors i and j?

Nothing, except that EVERY vector in two dimen-

sions can be written as i + j for some , .

Now actually we only really use iandj for CONVE-

NIENCE. In fact, we can do this with ANY pair of

vectors u , v in two dimensions,

PROVIDED that they are not parallel.

That is, any vectorw can be

expressed as

w =u + v

for some scalars , . You can see this from the

diagramby stretching u to u and v to v , wecan make their sum equal to

w.

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We callu ,

v a BASIS for 2-dimensional vectors. Let

u =P11i + P21j = P11

P21

v =P12i + P22j =

P12P22

Then the transformation that takes

i, j

to (u ,

v )

has matrixP11 P12

P21 P22

=P. In order for

u ,

v to be

a basis, Pmust not squash the volume of the Basic

Box down to zero, since otherwiseu and

v will be

parallel. So we must have

det P= 0.

The same idea works in 3 dimensions: ANY set of

3 vectors forms a basis PROVIDED that the matrix

of components satisfies det P

= 0.

EXAMPLE: The pair of vectors

u =

10

,v =

11

forms a basis, because det

1 10 1

= 1 = 0.

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Now of course the COMPONENTS of a vector will

change if you choose a different basis. For example,12

= 1i + 2j BUT

12

= 1u + 2v .

Instead,

12

= u+ 2v , so the components of this

vector relative tou ,

v are

12

(

u,

v )

. Where did I

get these numbers?

As usual, set u = Pi, v = Pj where P =

1 10 1

.

We want to find , such that

12

=

u +

v .

Substitutingu = Pi,

v = Pj into this equation we

get

12

=Pi + Pj =P[i + j] =P

.

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We knowPis not singular, so we can take Pover to

the left side by multiplying both sides of this equa-

tion by the inverse ofP. So we get

=P1

12

and this is our answer: this is how we find

and

!So to get and we just have to work out

P1

12

=

1 10 1

12

=

12

,

that is, the components of this vector relative to u , v

are found as

12

(

u,

v )

=P1

12

(i,j)

THE COMPONENTS RELATIVE TO

u ,v ARE

OBTAINED BY MULTIPLYING P1 INTO THE

COMPONENTS RELATIVE TO i, j. Similarly for

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linear transformationsif a certain linear transfor-mation T has matrix

1 20

1

ij

relative to i, j it

will have a DIFFERENT matrix relative to

u ,v .

We have

1 20 1

(i,j)

12

(i,j)

=

52

(i,j)

That is, the matrix ofTrelative to i, j sends

12

(ij)

to

52

(ij)

. In the same way, the matrix ofT rela-

tive to (u ,

v ), which we dont know and want to find,

sends

12

(

u,

v )

to

72

(

u,

v )

, because these are

the components of these two vectors relative to u , v ,

as you can show by multiplyingP1 into

12

(ij)

and

52

(ij)

respectively.

So the unknown matrix we want satisfies

? ?? ?

(

u,

v )

12

(

u,

v )

=

72

(

u,

v )

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But we know

1

2(

u,

v ) =P1 1

2(i,j) and

72

(

u,

v )

=P1

52

(i,j)

so

? ?? ?

(

u,

v )

P1

12

(ij)

=P1

52

(ij)

.

Multiply both sides by P and get

P ? ?

? ?(

u,

v )P1 1

2(ij)

= 52

(ij)

Compare this with

1 20

1

(i,j)

12

(i,j)

=

5

2

(i,j)

1 20 1

(i,j)

=P

? ?? ?

(

u,

v )

P1

? ?? ?

(

u,

v )

=P1

1 20 1

(i,j)

P.

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[In the last step, we multiplied both sides on the

LEFT by P1, and on the RIGHT by P.]

We conclude that THE MATRIX OF T REL-

ATIVE TOu ,

v , IS OBTAINED BY MULTIPLY-

INGP1 ON THE LEFT ANDPON THE RIGHT

INTO THE MATRIX OF TRELATIVE TO i, j.

In this example,

? ?? ?

(

u,

v )

=

1 10 1

1 20 1

1 10 1

= 1 1

0 1 1 3

0 1 = 1 4

0 1 .

So now we know how to work out the matrix of any

linear transformation relative to ANY basis.

Now let Tbe a linear transformation in 2 dimensions,

with eigenvectors e1,

e2, eigenvalues 1, 2. Now

e1

ande2may or may not give a basis for 2-dimensional

space. But suppose they do.

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QUESTION: What is the matrix of T relative toe1,

e2?

ANSWER: As always, we see what T does to e1

ande2, and put the results into the columns!

By definition of eigenvectors and eigenvalues,

Te1=1

e1 =1

e1+ 0

e2

Te2=2

e2 = 0

e1+ 2

e2

So the matrix is

1 00 2

(

e1,

e2)

.

We say that a matrix of the form

a 00 d

or

0 00 00 0

is DIAGONAL. So we see that THE MATRIX OF

A TRANSFORMATION RELATIVE TO ITS OWN

EIGENVECTORS (assuming that these form a ba-sis) is DIAGONAL.

EXAMPLE: We know that the eigenvectors of1 22 2

are

112

and

12

. So here P =

1 112 2

,

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P1 = 25 2 1

12 1

,

P1

1 22 2

P = 2

5

2 1 12 1

1 22 2

1 112 2

= 25

2 1 12 1

2 31 6

= 2

5

5 0

0 152

=

2 00 3

as expected since the

eigenvalues are 2 and -3.

EXAMPLE: The shear matrix

1 tan 0 1

.

Eigenvalues: det

1 tan

0 1

= 0 (1)2 =

0

= 1. Only one eigenvector, namely

1

0, so

the eigenvectors DO NOT give us a basis in this case

NOT possible to diagonalize this matrix!

EXAMPLE: Suppose you want to do a reflection

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of the entire 2-dimensional plane around a straight

line that passes through the origin and makes an

angle of with the x-axis. Then the vector

cossin

lies along this line, so it is left unchanged by the

reflection; in other words, it is an eigenvector of

with eigenvalue 1. On the other hand, the vector sincos is perpendicular to the first vector [checkthat their scalar product is zero] so it is reflected into

its own negative by . That is, it is an eigenvector

with eigenvalue 1.

So has a matrix with these eigenvectors and

eigenvalues. The P matrix in this case is

P =

cos sin sin cos

,

and clearly P1 =

cos sin sin cos

, and since the

eigenvalues are1, we have

=

cos sin sin cos

1 00 1

cos sin sin cos

.

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Doing the matrix multiplication and using the trigono-

metric identities for cos2 and sin2, you will find

that

=

cos2 sin2sin2 cos2

.

Notice that the determinant is 1, as is typi-cal for a reflection. Check that this gives the right

answer for reflections around the 45 degree diagonal

and around the x-axis.

6.9 APPLICATION MARKOV CHAINS.We saw back in Section 3 of Chapter 5 that to predict

the weather 4 days from now, we needed the 4th

power of the matrix

0.6 0.30.4 0.7

.

But suppose I want the weather 30 days from now

I need M30! There is an easy way to work this

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out using eigenvalues.

Suppose I can diagonalize M, that is, I can write

P1M P =D=

1 00 2

for some matrixP. Then

M=P DP1

M2

= (P DP1

)(P DP1

)

=P DP1P DP1 =P D2P1

M3 =M M2 =P DP1P D2P1 =P D3P1 etc

M30 =P D30P1.

But D30 is very easy to work out it is just

301 00 302

.

Lets see how this works!

Eigenvectors and eigenvalues of

0.6 0.30.4 0.7

are

1

1

(eigenvalue 0.3) and

143

(eigenvalue 1) so

P =

1 11 43

, D=

0.3 00 1

, P1 =

47 3737

37

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D30 = (0.3)30 0

0 1

2

1016 0

0 1

so

M30 =

1 11 43

2 1016 0

0 1

47 3737

37

= 1

7

3 + 8 1016 3 6 10164 8 1016 4 + 6 1016

37

37

47

47

So if it is rainy today, the probability of rain tomor-

row is 60%, but the probability of rain 30 days from

now is only 3

7 43%. As we go forward in time, thefact that it rained today becomes less and less im-

portant! The probability of rain in 31 days is almost

the same as the probability of rain in 30 days!

6.10 APPLICATION: THE TRACE OF A

DIAGONALIZABLE MATRIX IS THE SUM

OF ITS EIGENVALUES.

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Let M be any square matrix. Then the TRACE

of M, denoted T rM, is defined as the sum of the

diagonal entries: T r

1 00 1

= 2, T r

1 2 34 5 67 8 9

=

15, T r

1 5 167 2 15

11 9 8

= 11, etc.

In general it is NOT true that T r(M N) =TrM T rN

BUT it is true that T rMN=T rNM.

Proof: T rM=i

Mii so

TrMN=i

j

MijNji =j

i

NjiMij =T rNM.

Hence T r(P1AP) = T r(AP P1) = T rA so THE

TRACE OF A MATRIX IS ALWAYS THE SAME

NO MATTER WHICH BASIS YOU USE! This iswhy the trace is interesting: it doesnt care which

basis you use. In particular, if A is diagonalizable,

T rA =T r

1 00 2

=1+ 2.

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So the trace is equal to the sum of the eigen-

values. This gives a quick check that you have not

made a mistake in working out the eigenvalues: they

have to add up to the same number as the trace of

the original matrix. Check this for the examples in

this chapter.