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Mathematical Economics:Lecture 17

Yu Ren

WISE, Xiamen University

December 12, 2011

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Chapter 23: Eigenvalues and Dynamics

Outline

1 Chapter 23: Eigenvalues and Dynamics

Yu Ren Mathematical Economics: Lecture 17

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New Section

Chapter 23:Eigenvalues and

Dynamics


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Definitions

Eigenvalue: Let A be a square matrix. Aneigenvalue of A is a number r which makesdet(A− rI) = 0


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Example 23.1Example 23.1 Consider the matrix

A =

3 1 11 3 11 1 3

Subtracting 2 from each diagonal entrytransforms A into the singular matrix

A =

1 1 11 1 11 1 1

Therefore, 2 is an eigenvalue of A.


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Example 23.2

Example 23.2 Let’s look for the eigenvalues of

the diagonal matrix B =

(2 00 3

)Subtracting a

2 from each of the diagonal entries yields the

singular matrix(

0 00 1

)subtracting a 3 from

each of the diagonal entries yields the singular

matrix(−1 00 0

). Therefore, 2 and 3 are

eigenvalues od the(

2 00 3

).


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Theorem

Theorem 23.1 The diagonal entries of adiagonal matrix D are eigenvalues of D.Theorem 23.2 A square matrix A is singularif and only if 0 is an eigenvalue of A.


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Example 23.3

Example 23.3 Consider the matrix

B =

(1 −1−1 1

). Since the first row is the

negative of the second, B is a singular matrixand, therefore, 0 is an eigenvalue of B. We canuse the observation in Example 23.1 to find asecond eigenvalue, because subtracting 2 fromeach diagonal entry of B yields the singular

matrix(−1 −1−1 −1

). We conclude that 0 and 2

are eigenvalues of B.


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Example 23.4

Example 23.4 A matrix M whose entries arenonnegative and whose columns (or rows) eachadd to 1, such as

B =

(1/4 2/33/4 1/3

)ia called a Markov matrix. As we will see inSection 23.6, Markov matrices play a major rolein the dynamics of economic systems.


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Example 23.4If we subtract a 1 from each diagonal entry ofthe Markov matrix, then each column of thetransformed matrix:

M − 1I =

(−3/4 2/33/4 −2/3

)adds up to 0. But if the columns of a squarematrix add up to (0, . . . ,0), the rows are linearlydependent and the matrix must be singular. Itfollows that r = 1 is an eigenvalue of matrix B.The same argument shows that r = 1 is aneigenvalue of every Markov matrix.


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Theorem

Theorem 23.3 Let A be an n× n matrix and let rbe a scalar. Then, the following statements areequivalent:

(a) Subtracting r from each diagonal entryof A transforms A into a singular matrix.(b) A− rI is a singular matrix(c) det(A− rI)=0(d) (A− rI)V = 0 for some nonzero vectorV .(e) AV = rV for some nonzero vector V .


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Theorem

V is called eigenvector corresponding toeigenvalue r


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Example 23.5

Example 23.5 Find the eigenvalues andeigenvectors of the matrix

A =

(−1 3+2 0

)Its characteristic polynomial is

det(A−rI) = det(−1− r 3

2 0− r

)= (r+3)(r−2).


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Example 23.5

The eigenvalues of A are the roots of thecharacteristic polynomial: −3 and 2. To find thecorresponding eigenvectors, use statement d ofTheorem 23.3. First, subtract eigenvalue −3from the diagonal entries of A and solve

(A− (−3)I)v =

(2 32 3

)(v1

v2

)=

(00

)for v1 and v2.


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Example 23.5For a 2× 2 matrix, these equations are easilysolved just by looking carefully at the equation.For example, we can take v1 = 3 and v2 = −2and conclude that one eigenvector is

( 3−2

).

There are other eigenvalues, such as(1−2/3

),

(−32

),

(−3/2

1

)In general, one chooses the ′′simplest ′′ of thenonzero candidates. The set of all solutions oflinear equation−including v = 0−is called theeigenspace of A with respect to −3.


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Example 23.5

To find the eigenvector for eigenvalue r = 2,subtract 2 from the diagonal entries of A

(A− 2I)v =

(−3 32 −2

)(v1

v2

)=

(00

)The simplest solution is

(11

);but any multiple of(1

1

)is also an eigenvector for 2. The eigenspace

for eigenvalue 2 is the diagonal line in R2.


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Example 23.6Example 23.6 Example 23.6 Let’s compute theeigenvalues and eigenvectors of the 3X3 matrix

B =

1 0 20 5 03 0 2

Its characteristic equation is

det

1− r 0 20 5− r 03 0 2− r

= (5−r)(r−4)(r +1)


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Example 23.6Therefore, the eigenvalues of B are r = 5,4,−1.To compute an eigenvector corresponding tor = 5, we compute the nullspace of (B − 5I);that is, we solve the system

(B − 5I)v =

−4 0 20 0 03 0 −3

v1v2v3

=

−4v1 + 2v20

3v1 − 3v3

=

000

,

whose solution is v1 = v3 = 0, v2 = anything.Yu Ren Mathematical Economics: Lecture 17

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Example 23.6

So, we’ll take v1 =

010

as an eigenvector for

r = 5.


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Example 23.6

To find an eigenvector for r = 4, solve

(B − 4I)v =

−3 0 20 1 03 0 −2

v1v2v3

=

−4v1 + 2v20

3v1 − 3v3

=

000

.


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Example 23.6This system reduces to the two equations

−3v1 + 2v3 = 0v2 = 0

A simple eigenvector for r = 4 is

203

. This

same method yields the eigenvector

10−1

for eigenvalue r = −1.


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Solving Linear Difference Equations

yn+1 = ayn

y1 = ay0

yn = any0


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xn+1 = axn + byn

yn+1 = cxn + dyn


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Example 23.7

Example 23.7 Consider the coupled system ofdifference equations generated by a Lesliemodel with b1 = 1,b2 = 4 and d1 = 0.5

xn+1 = 1xn + 4yn

yn+1 = 0.5xn + 0yn


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Example 23.7The right change of coordinates for solving thissystem is

X =16

x +13

y

Y = −16

x +23

y

whose inverse transformation is

x = 4X − 2Yy = X + Y


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Example 23.7

In matrix form,(XY

)=

( 16

13

−16

23

)(xy

)and (

xy

)=

(4 −21 1

)(XY

)These two matrices are necessarily inverse toeach other.


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Example 23.7

Use these transformations

Xn+1 =16

xn+1 +13

yn+1 =16

(xn + 4yn) +13

(12

xn

)Yn+1 = −1

6xn+1 +

23

yn+1 = −16

(xn + 4yn) +23

(12

xn

)

using the change of coordinates and thendifference equation.


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Example 23.7

Simplifying,

Xn+1 =13

xn +23

yn =13

(4Xn − 2Yn) +23

(Xn + Yn)

= 2Xn

Yn+1 =16

xn −23

yn =16

(4Xn − 2Yn)− 23

(Xn + Yn)

= −Yn


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Example 23.7then

Xn+1 = 2Xn

Yn+1 = −Yn

is completely uncoupled and, so, can be easilysolved as two one-dimensional equations:

Xn = 2nc1

Yn = (−1)nc2


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Example 23.7Finally,

xn = 4Xn − 2Yn = 4 · 2nc1 − 2 · (−1)nc2

yn = Xn + Yn = 2nc1 + (−1)nc2

which can be written in vector notation as(xnyn

)=

(4 · 2nc1 − 2 · (−1)nc2

2nc1 + (−1)nc2

)= c12n

(41

)+ c2(−1)n

(−21

).


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Example 23.7The constants c1 and c2 are determined by theinitial amounts x0 and y0.

x0 = 4c1 − 2c2

y0 = c1 + c2

This system of equations can be solved in theusual way:(

c1c2

)=

(4 −21 1

)−1( x0y0

)=

( 16

13

−16

23

)(x0y0

).


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zt+1 = Azn

z = PZ


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Zn+1 = P−1zn+1

= P−1(Azn)

= (P−1A)zn

= (P−1A)PZn

= (P−1AP)Zn


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Solution: Theorem 23.4 Let A be a k × kmatrix. Let r1, r2, · · · , rk be eigenvalues of A, andV1,V2, · · · ,Vk the corresponding eigenvalues.Form the matrix P = [V1,V2, · · · ,Vk ] whosecolumns are these k eigenvectors. If P isinvertible, then P−1AP = diag(rk). Conversely, ifP−1AP is a diagonal matrix D, the columns of Pmust be eigenvectors of A and the diagonalentries of D must be eigenvalues of A.


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Revisit example 23.7 specify r , P, A


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Theorem 23.5 Let r1, · · · , rh be h distincteigenvalues of the k × k matrix A. let V1, · · · ,Vhbe corresponding eigenvectors. Then V1, · · · ,Vhare linearly independent, that is, no one of themcan be written as a linear combination of theothers.


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Theorem 23.6 Let A be a k × k matrix withk distinct real eigenvalues r1, · · · , rk andcorresponding eigenvectors V1, · · · ,Vk .The general solution of the system ofdifference equations zn+1 = Azn iszn = c1rn

1 v1 + c2rn2 v2 + · · ·+ ck rn

k vk

Page 594 and Theorem 23.7: AlternativeApproach


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Example 23.8

Example 23.8 We computed in Example 23.5

that the eigenvalues of C =

(−1 3+2 0

)are

−3,2 with corresponding eigenvectors(

3−2

)and

(11

)


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Example 23.8

By Theorem 23.4(3 1−2 1

)−1( −1 32 0

)(3 1−2 1

)=

(−3 00 2

)


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Example 23.8

By Theorem 23.7,(−1 32 0

)n

=(3 1−2 1

)((−3)n 0

0 2n

)(0.2 −0.20.4 0.6

).


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Stability

Stability of equilibriaTheorem 23.8 If the k × k matrix A has kdistinct real eigenvalues, then everysolution of the general system of lineardifference equations (23) tends to 0 if andonly if all the eigenvalues of A haveabsolute value less than 1.


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Properties of Eigenvalues

p(r) has k distinct, real rootsp(r) has some repeated rootsp(r) has some complex roots


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Example 23.10

Example 23.10

(a) For matrix(−4 2−1 −1

)the characteristic polynomial isp1(r) = (−4− r)(−1− r) = r2 + 5r + 6,whose roots are the distinct, real numberr = −3,−2.


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Example 23.10

(b) 4 0 −20 3 03 0 −1

isp2(r) = (4−r)(3−r)(−1−r) = (3−r)(r2−3r +2),whose roots are the distinct, real numberr = 1,2,3.


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Example 23.10

(c) (4 1−1 2

)The characteristic polynomial of the matrix isp3(r) = (4− r)(2− r) + 1 = (r − 3)2,whose roots are r = 3,3, 3 is a root of p3(r) ofmultiplicity two.


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Example 23.10

(d) (3 00 3

)The characteristic polynomial of the matrix isp4(r) = (3− r)2

Once again, 3 is an eigenvalue of multiplicitytwo.


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Example 23.10

(e) (0 2−1 2

)The characteristic polynomial of the matrix isp5(r) = −r(2− r) + 2 = r2 − 2r + 2,Its roots are the complex numbersr = 1 + i ,1− i .


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Repeated Eigenvalues

Theorem 23.9 Let A be a k × k matrix witheigenvalues r1, r2, · · · , rk . Then(a) r1 + r2 + · · ·+ rk = tr(A)(b) r1 · r2 · · · rk = det(A)


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Example 23.11

Example 23.11 To find the eigenvalues of the4X4 matrix

B =

4 1 1 11 4 1 11 1 4 11 1 1 4

,

subtract 3 from each of the diagonal entries ofB.


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Example 23.11

The result is the singular matrix

B − 3I =

1 1 1 11 1 1 11 1 1 11 1 1 1

,


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Example 23.11

Vector v = (v1, v2, v3, v4) is an eigenvector ofeigenvalue 3 if and only if (B − 3I)v = 0 if andonly if v1 = v2 = v3 = v4 = 0. The vectors

v1 =

100−1

, v2 =

010−1

, v3 =

001−1

are three linearly independent eigenvectors ofeigenvalue 3.


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Example 23.11

Therefore, 3 ia an eigenvalue of B of multiplicityat least 3. Use the fact that the sum of theeigenvalues of B is 16, the trace of B, toconclude that the fourth eigenvalue of B is16− (3 + 3 + 3) = 7.


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Nondiagonalizable matrix (page 601)Theorem 23.10: Let A be a 2× 2 matrixwith two equal eigenvalues. Then, A isdiagonizable if and only if A is alreadydiagonal.


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almost diagonal matrix:(r ∗ 10 r ∗

)Is this form simple enough that we canalways easily solve the transformedsystem?Is this “almost diagonal” form achieve asP−1AP for any defective matrix A?


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P−1AP =

(r ∗ 10 r ∗

)A[v1 v2] = [v1 v2]

(r ∗ 10 r ∗

)


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[Av1 Av2] = [r ∗v1 v1 + r ∗v2]

(A− r ∗I)v1 = 0(A− r ∗I)v2 = v1


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Example 23.13

Example 23.13 Consider the nondiagonalizablematrix

A =

(4 1−1 2

)of Example 23.10c, whose eigenvalues arer = 3,3. As we computed earlier, it has one

independent eigenvector v1 =

(1−1

).


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Example 23.13

Its generalized eigenvector will be a solution v2of

(A−3I)v2 = v1, or(

1 1−1 −1

)(v21v22

)=

(1−1

)Take v21 = 1, v22 = 0, for example. Then form

P = [v1 v2] =

(1 1−1 0

)


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Example 23.13

and check that

P−1AP =

(0 −11 1

)(4 1−1 2

)(1 1−1 0

)=

(3 10 3

)


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Theorem 23.11 let A be a 2× 2 matrix withequal eigenvalue r = r ∗, r ∗. Then, (a) either Ahas two independent eigenvectorscorresponding to r ∗, in which case A is diagonalmatrix r ∗I or (b) A has only one independenteigenvector, say V1. In this case, there is ageneralized eigenvector V2 such that(A− r ∗I)V2 = V1. If P ≡ [V1V2] then

P−1AP =

(r ∗ 10 r ∗

)


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Example 23.14

Example 23.14 The characteristic polynomial ofthe matrix

A =

4 2 −41 4 −31 1 0

is p(r) = (r − 3)2(2− r), its eigenvalues arer = 3,3,2.


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Example 23.14

For eigenvalues r = 2, the solution space of

(A− 2I)v =

2 2 −41 2 −31 1 −2

v1v2v3

=

000

is the one-dimensional space spanned by

v1 =

111

.


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Example 23.14

For eigenvalue r = 3, the solution space of

(A− 3I)v =

1 2 −41 1 −31 1 −3

v1v2v3

=

000

is the one-dimensional space spanned by

v2 =

211

.


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Example 23.14

There is only one independent eigenvectorcorresponding to the eigenvalue of multiplicitytwo. We need one more vector v3 independentof v1, v2 to form the change of coordinate matrixP = [v1, v2, v3]. Take v3 to be a generalizedeigenvector for the eigenvalue r = 3−a solutionto the system

(A−3I)v3 = v2, or

1 2 −41 1 −31 1 −3

v31v32v33

=

211


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Example 23.14

By inspection, we can take v3 =

010

.


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Example 23.14

Let

P = [v1v2v3] =

1 2 01 1 11 1 0

Then,

P−1AP =

2 0 00 3 10 0 3

.


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page 606Theorem 23.12 Suppose that A is a 2× 2matrix with multiple eigenvalue r and onlyone independent eigenvector V1. Let V2 bea generalized eigenvector corresponding toV1 and r . Then general solution of thesystem of difference equation zn+1 = Azn iszn = (c0rn + nc1rn−1)V1 + c1rnV2


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Example 23.15

Example 23.15 The linear system of differenceequations corresponding to thenondiagonalizable matrix in Example 23.13 is(

xn+1yn+1

)=

(4 1−1 2

)(xnyn

)


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Example 23.15

By Theorem 23.12, its general solution is(xnyn

)= (c03n + c1n3n−1)

(1−1

)+ c13n

(10

)=

(c03n + c1(n3n−1 + 3n)−c0 − c1n3n−1

).


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Complex eigenvalues

Example 23.16 The eigenvalues of(

a bc d

),

the general 2X2 matrix, are the roots of itscharacteristic equation

p(r) = r2 − (a + d)r + (ad − bc),


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Example 23.16

namely,

r = 12(a + d)± 1

2

√(a + d)2 − 4(ad − bc)

= 12(a + d)± 1

2

√(a− d)2 + 4bc

If (a− d)2 + 4bc < 0, then the roots are thecomplex numbers

r1 = 12(a + d)± i 1

2

√|(a + d)2 + 4bc|

r2 = r̄1 = 12(a + d)±−i 1

2

√|(a + d)2 + 4bc|


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Complex eigenvalues

1 (37) (38)2 Theorem 23.13 Let A be a k × k matrix with

real entries. If r = α+ iβ is an eigenvalue ofA, so is its conjugate r̄ = α− iβ. If U + iV isan eigenvector for α + iβ, then u − iv is aneigenvector for α− iβ. If k is odd, A musthave at least one real eigenvalue.


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Example 23.17

Example 23.17 For matrix A =

(1 1−9 1

), the

characteristic polynomial is

p(r) = r2 − 2r + 10,

whose roots are r = 1 + 3i ,1− 3i . Aneigenvector for r = 1 + 3i is a solution w of

(A−(1+3i)I)w =

(−3i 1−9 −3i

)(w1w2

)=

(00

)


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Example 23.17

This matrix may not look singular, but itsdeterminant is zero and its second row is −3itimes its first row. Using the first row of thismatrix, we conclude that an eigenvector is asolution w of the equatio

−3iw1 + w2 = 0;

for example, w =

(1−3i

), which we write as(

10

)+ i(

03

).


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Example 23.17

By Theorem 23.13, an eigenvector foreigenvalue 1− 3i is

w̄ =

(10

)− i(

03

)=

(1−3i

)


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Example 23.17

From the change of coordinate matrix P whosecolumns are these two eigenvectors:

P =

(1 13i −3i

)Then, applying Theorem 8.5.4

P−1 = − 16i

(−3i −1−3i 1

)=

( 12 −

16 i

12

16 i

),


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Example 23.17

and we calculate that

P−1AP =

(1 + 3i 0

0 1− 3i

)just as if we had been working with realeigenvalues and real eigenvectors.


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Complex eigenvalues

(40) (41) Figure 23.2Theorem 23.14Let A be a real 2× 2 matrixwith complex eigenvalues α∗ ± iβ∗ andcorresponding complex eigenvectorsu∗ ± iv∗. Write the eigenvalues α∗ ± iβ∗ inpolar coordinates as r ∗(cos θ∗ + i sin θ∗).Then, the general solution of the differenceequation zn+1 = Azn is

zn = r ∗n[(C1 cos nθ∗ − C2 sin nθ∗)u∗

− (C2 cos nθ∗ + C1 sin nθ∗)v∗]


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Example 23.18

Example 23.18 In example 23.17, we computed

that the eigenvalues of A =

(1 1−9 1

)are

1± 3i with corresponding eigenvectors(10

)± i(

03

).


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Example 23.18

In polar coordinates,

1+3i =√

10(

1√10

+ i3√10

)=√

10(cos θ∗+sin θ∗),

where θ∗ = arccos(1/√

10) ≈ 71.565o or 1.249radians.


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Example 23.18

The general solution of

xn+1 = xn + yn

yn+1 = −9xn + yn

is


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Example 23.18

(xnyn

)= (√

10)n(

c1 cos nθ∗ − c2 sin nθ∗

−3c2 cos nθ∗ − 3c1 sin nθ∗

)


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Example 23.19Example 23.19 Consider the Leslie model of anorganism that lives for three years, with deathrates d1 = 0.2,d2 = 0.6, and d3 = 1. Supposethat only third year individuals can reproduce,with birth rate b3 = 1.6. The correspondingLeslie matrix 0 0 1.6

0.8 0 00 0.4 0

its characteristic polynomial is p(r) = r3− 0.512,with roots r = 0.8 and −0.4± i

√0.48.


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Example 23.19

An eigenvector for r = 0.8 is

221

; an

eigenvector is

r = −0.4∓ i√

0.48 = −0.4(1± i√

3)

= −0.4(cosπ

3± i sin

π

3)

is 1∓ i√

31± i

√3

−1

=

11−1

± i

−√3+√

30


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Example 23.19The general solution is xn

ynzn

= c1(0.8)n

221

+ (−0.4)n(c1 cos n

π

3− c2 sin n

π

3)

11−1

− (−0.4)n(c2 cos n

π

3+ c1 sin n

π

3)

−√3+√

30


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Markov Process

Definition: a stochastic process is a rule whichgives the probability that the system (or anindividual in this system) will be in state i at timen + 1, given the probabilities of its being in thevarious states in previous periods. Thisprobability could depend on the whole previoushistory of the system. WhenP(Sn+1|Sn, · · · ,S1) = P(Sn+1|Sn), the process iscalled a Markov process.


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Markov Process

Transition matrix or Markov matrix: m11 · · · m1k... . . . ...

mk1 · · · mkk

, mij is called the transition

probabilities from state j at time n to state i attime n + 1.

∑i mij = 1


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Example 23.20

Example 23.20 Consider the employmentmodel of Section 6.2. In this model, eachperson in the population under study is eitheremployed or unemployed. The two states of thismodel are employed and unemployed. Let x1(n)denote the fraction of the study population thatis employed at the end of time period n andx2(n) denote the fraction unemployed.


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Example 23.20

Suppose that an employed person has a 90percent probability of being unemployed nextperiod (and, therefore, a 10 percent probabilityof being unemployed next period) and that anunemployed person has a 40 percent probabilityof being employed one period from now (and,therefore, a 60 percent probability of beingunemployed).


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Example 23.20

The corresponding dynamics are

x1(n + 1) = 0.9x1(n) + 0.4x2(n)

x2(n + 1) = 0.1x1(n) + 0.6x2(n)

or (x1(n + 1)x2(n + 1)

)=

(0.9 0.40.1 0.6

)(x1(n)x2(n)

)


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Example 23.20

By the argument of Example 23.4, oneeigenvalue of system is r = 1. By using thetrace result of Theorem 23.9, we conclude thatthe other eigenvalue is r = 1.5− 1.0 = 0.5.


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Example 23.20

To compute the corresponding eigenvectors, wesolve(−0.1 0.40.1 −0.4

)(αβ

)=

(00

)or(αβ

)=

(41

)and(

0.4 0.40.1 0.1

)(αβ

)=

(00

)or(αβ

)=

(1−1

)


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Example 23.20

We conclude by Theorem 23.6 that the generalsolution of system is(

x1n

x2n

)= c1

(41

)· 1n + c2

(1−1

)· 0.5n


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Example 23.20

Since 1n and 0.5n → 0 as n→∞, in thelong-run, the solution of the Markov system

tends to w1 = c1

(41

). Since w1 should be a

probability vector whose components sum to 1,take c1 to equal the reciprocal 1/5 of sum of thecomponents of w1. We conclude that the

solution of system tends to(

0.8−0.2

)as

n→∞, and that our assumptions lead to along-run unemployment rate of 20 percent inthis community.


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Regular Markov matrix

Regular Markov matrix if M r has only positiveentries for some integer r . If r = 1,M is called apositive matrix.


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Regular Markov matrix

Theorem 23.15 Let M be a regular Markovmatrix. Then, (a) 1 is an eigenvalue of M ofmultiplicity 1 (b) every other eigenvalue r of Msatisfies |r | < 1 (c) eigenvalue 1 has aneigenvector w1 with strictly positive componentsand (d) if we write V1 for w1 divided by the sumof its components, then V1 is a probability vectorand each of solution x(n) of x(n + 1) = Mx(n)tends to V1 as n→∞.


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Example 23.21

Example 23.21 Suppose that American familiesare classified as urban, suburban, or rural andthat each year: 20 percent of the urban familiesmove to the suburbs and 5 percent move torural areas; 2 percent of the suburban dwellersmove to urban areas and 8 percent move torural areas; 10 percent of the rural familiesmove to urban areas and 20 percent move tosuburban areas.


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Example 23.21

Let Un,Sn, and Rn denote the fractions of thepopulation classified as urban, suburban, andrural, respectively, n years from now. Then, thedata of this problem lead to the Markov system. Un+1

Sn+1Rn+1

=

0.75 0.02 0.10.8 0.9 0.2

0.05 0.08 0.7

UnSnRn


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Example 23.21

The eigenvalues are this Markov matrix are 1,0.7, and 0.65, and the correspondingeigenvectors are 2/15

10/153/15

,

8−5−3

,

10−1

,

where the eigenvector for r = 1 has beennormalized so that components sum to 1.


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Example 23.21

The general solution is UnSnRn

=

2/1510/153/15

+c2

8−5−3

0.7n+c3

10−1

0.65n,

which converges to (2/15, 10/15, 3/15) asn→∞. We conclude that in the long-run 2/15of the population will be living in cities, 2/3 in thesuburbs, and 1/5 in rural areas.


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Example 23.22Example 23.22 The eigenvalues of thesymmetric matrix

B =

3 1 −11 3 −1−1 −1 5

are r1 = 2, r2 = 3, r3 = 6. Correspondingeigenvectors are

v1 =

−110

, v2 =

111

, v3 =

11−2


math


Example 23.22

As Theorem 23.16 indicates, these vectors areperpendicular to each other. Divide eacheigenvector by its length to generate a set of′′normalized eigenvectors′′:

u1 =1√2

−110

,u2 =1√3

111

,u3 =1√6

11−2


math


Example 23.22and make these three orthonormal vectors −vectors which are orthogonal and have length 1− the columns of the orthogonal matrix −1/

√2 1/

√3 1/

√6

1/√

2 1/√

3 1/√

60 1/

√3 −2/

√2

Then, Q−1 = QT and

QT BQ =

2 0 00 3 00 0 6


math


Theorem 23.16 Let A be a k × k symmetricmatrix. Then, (a) all k roots of the characteristicequation det(A− rI) = 0 are real numbers. (b)eigenvectors corresponding to distincteigenvalues are orthogonal; and even if A hasmultiple eigenvalues, there is a nonsingularmatrix P whose columns w1, · · · ,wk areeigenvectors of A such that (i) w1, · · · ,wk aremutually orthogonal to each other, (ii) P−1 = PT

(iii) P−1AP = PT AP =

r1 0 · · · 00 r2 · · · 0... ... . . . ...0 0 · · · rk


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A matrix P which satisfies the conditionP−1 = PT , or PT P = I is called orthogonalmatrix.


math


Example 23.23

Example 23.23 Let’s diagonalize a symmetricmatrix with nondistinct eigenvalues. Considerthe 4X4 symmetric matrix

C =

3 1 1 11 3 1 11 1 3 11 1 1 3


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Example 23.23By the methods of Section 23.3, the eigenvaluesof C are, by inspection 2,2,2 and 6. The set ofeigenvectors for 2, the eigenspace ofeigenvectors 2, is the three-dimensionalnullspace of the matrix

C − 2I =

1 1 1 11 1 1 11 1 1 11 1 1 1

,

the space of{(u1,u2,u3,u4) : u1 + u2 + u3 + u4 = 0}.


math


Example 23.23

Three independent vectors in this eigenspaceare

v1 =

−1100

, v2 =

−1010

, v3 =

−1001


math


Example 23.23In order to construct an orthogonal matrix P sothat the product PT CP = P−1CP is diagonal, weneed to find three orthogonal vectors w1,w2,w3which span the same subspace as theindependent vectors v1, v2, v3. The followingprocedure, called the Gram-SchmidtOrthogonalization Process, will accomplish thistask. Let w1 = v1. Define

w2 = v2 −w1 · v2

w1 · w1w1,

w3 = v3 −w1 · v3

w1 · w1w1 −

w2 · v3

w2 · w2w2

and so on.Yu Ren Mathematical Economics: Lecture 17

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Example 23.23

The w ′i s so constructed are mutually orthogonal.By construction, w1,w2,w3 span the samespace as v1, v2, v3. The application of thisprocess to the eigenvectors v1, v2, v3 yields theorthogonal vectors

w1 =

−1100

,w2 =

−1/2−1/2

10

,w3 =

−1/3−1/3−1/3

1


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Example 23.23

Finally, normalize these three vectors and makethem the first three columns of an orthogonalmatrix whose fourth column is the normalizedeigenvector for r = 6:

Q =

−1/√

2 −1/√

6 −1/√

12 1/21/√

2 −1/√

6 −1/√

12 1/20 2/

√6 −1/

√12 1/2

0 0 3/√

12 1/2


math


Example 23.23

Check that QT = Q−1 and

QT CQ =

2 0 0 00 2 0 00 0 2 00 0 0 6


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Theorem 23.17

Theorem 23.17 Let A be a symmetric matrix,then(a) A is positive definite if and only if all theeigenvalues of A are >0(b) A is negative definite if and only if all theeigenvalues of A are <0(c) A is positive semidefinite if and only if all theeigenvalues of A are ≥0(d) A is negative semidefinite if and only if all theeigenvalues of A are ≤ 0


math


Theorem 23.18

Theorem 23.18 Let A be a symmetric matrix.Then, the following are equivalent (a) A ispositive definite (b) ∃ a nonsigular matrix B s.t.A = BT B (c) ∃ a nonsigular matrix Q s.t.QT AQ = I .


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