Essential mathematics for economic analysis - Student’s Manual

115 Pages·2012·0.73 MB·English

Student’s Manual
Further Mathematics for
Economic Analysis
nd
2 edition
Arne Strøm
Knut Sydsæter
Atle Seierstad
Peter Hammond
For further supporting resources please visit:
www.pearsoned.co.uk/sydsaeter
Preface
This student’s solutions manual accompanies Further Mathematics for EconomicAnalysis (2nd edition, FT
⊂⊃
PrenticeHall,2008). Itsmainpurposeistoprovidemoredetailedsolutionstotheproblemsmarkedwith SM
inthetext. TheManualshouldbeusedinconjunctionwiththeanswersinthebook. Insomefewcasesonly
apartoftheproblemisdoneindetail,becausetherestfollowsthesamepattern.
Attheendofthismanualthereisalistofmisprintsandothererrorsinthebook,andevenoneortwoin
the errata list in the preliminary and incomplete version of this manual released in September this year. We
would greatly appreciate suggestions for improvements from our readers as well as help in weeding out the
inaccuraciesanderrorsthatprobablyremain.
OsloandCoventry,October2008
ArneStrøm ([email protected])
KnutSydsæter ([email protected])
AtleSeierstad ([email protected])
PeterHammond ([email protected])
Contents
1 TopicsinLinearAlgebra ............................................................ 1
2 MultivariableCalculus ............................................................. 12
3 StaticOptimization ............................................................... 23
4 TopicsinIntegration .............................................................. 38
5 DifferentialEquationsI:First-OrderEquationsinOneVariable............................. 43
6 DifferentialEquationsII:Second-OrderEquationsandSystemsinthePlane .................. 49
7 DifferentialEquationsIII:Higher-OrderEquations ...................................... 56
8 CalculusofVariations ............................................................. 60
9 ControlTheory: BasicTechniques ................................................... 65
10 ControlTheorywithManyVariables.................................................. 81
11 DifferenceEquations .............................................................. 92
12 DiscreteTimeOptimization ......................................................... 96
13 TopologyandSeparation .......................................................... 102
14 CorrespondencesandFixedPoints .................................................. 105
A Sets,Completeness,andConvergence ................................................ 107
B TrigonometricFunctions .......................................................... 109
Correctionstothebook ........................................................... 112
Version1.0 28102008 879
©ArneStrøm,KnutSydsæter,AtleSeierstad,andPeterHammond2008
CHAPTER 1 TOPICS IN LINEAR ALGEBRA 1
Chapter 1 Topics in Linear Algebra
1.2
⎛ ⎞
1 2 0
1.2.3 Let A = ⎝0 1 1⎠ be a matrix with the three given vectors as columns. Cofactor expansion of
1 0 1
|A|alongthefirstrowyields
(cid:6) (cid:6)
(cid:6) (cid:6) (cid:6) (cid:6) (cid:6) (cid:6)
(cid:6)1 2 0(cid:6) (cid:6) (cid:6) (cid:6) (cid:6)
(cid:6)(cid:6)0 1 1(cid:6)(cid:6) = 1(cid:6)(cid:6)1 1(cid:6)(cid:6)−2(cid:6)(cid:6)0 1(cid:6)(cid:6)+0 = 1−2(−1) = 3 (cid:4)= 0
(cid:6) (cid:6) 0 1 1 1
1 0 1
ByTheorem1.2.1thisshowsthatthegivenvectorsarelinearlyindependent.
1.2.6 Part (a) is just a special case of part (b), so we will only prove (b). To show that a , a , ..., a are
1 2 n
linearlyindependentitsufficestoshowthatifc ,c ,...,c arerealnumberssuchthat
1 2 n
c a +c a +···+c a = 0
1 1 2 2 n n
thenallthec havetobezero. Sosupposethatwehavesuchasetofrealnumbers. Thenforeachi = 1,
i
2,...,n,wehave
a ·(c a +c a +···+c a ) = a ·0 = 0 (1)
i 1 1 2 2 n n i
Sincea ·a = 0wheni (cid:4)= j,theleft-handsideof(1)reducestoa ·(c a ) = c (cid:5)a (cid:5)2. Hence,c (cid:5)a (cid:5)2 = 0.
i j i i i i i i i
Becausea (cid:4)= 0wehave(cid:5)a (cid:5) (cid:4)= 0,anditfollowsthatc = 0.
i i i
1.3
1.3.1 (a) Therankis1. Seetheanswerinthebook.
(cid:6) (cid:6)
(cid:6) (cid:6)
(b) Theminorformedfromthefirsttwocolumnsinthematrixis(cid:6)(cid:6)1 3(cid:6)(cid:6) = −6 (cid:4)= 0. Sincethisminor
2 0
isoforder2,therankofthematrixmustbeatleast2,andsincethematrixhasonlytworows,therank
cannotbegreaterthan2,sotherankequals2.
(cid:6) (cid:6)
(c) Thefirsttworowsandlasttwocolumnsofthematrixyieldtheminor(cid:6)(cid:6)(cid:6)−1 3(cid:6)(cid:6)(cid:6) = 5 (cid:4)= 0,sotherank
−4 7
of the matrix is at least 2. On the other hand, all the four minors of order 3 are zero, so the rank is less
than3. Hencetherankis2. (Itcanbeshownthatr = 3r +r ,wherer ,r ,andr aretherowsofthe
2 1 3 1 2 3
matrix.)
Analternativeargumentrunsasfollows: Therankofamatrixdoesnotchangeifweaddamultiple
ofonerowtoanotherrow,so
⎛ ⎞ ⎛ ⎞
1 2 −1 3 −2 1 1 2 −1 3
⎝ 2 4 −4 7⎠ ← ∼ ⎝0 0 −2 1⎠.
−1 −2 −1 −2 ← 0 0 −2 1
©ArneStrøm,KnutSydsæter,AtleSeierstad,andPeterHammond2008
2 CHAPTER 1 TOPICS IN LINEAR ALGEBRA
Here∼meansthatthelastmatrixisobtainedfromthefirstonebyelementaryrowoperations. Thelast
matrixobviouslyhasrank2,andthereforetheoriginalmatrixalsohasrank2.
(cid:6) (cid:6)
(cid:6) (cid:6)
(cid:6)1 3 0(cid:6)
(d) Thefirstthreecolumnsofthematrixyieldtheminor(cid:6)(cid:6)2 4 0(cid:6)(cid:6) = −4 (cid:4)= 0,sotherankis3.
(cid:6) (cid:6)
1 −1 2
(cid:6) (cid:6)
(cid:6) (cid:6)
(e) (cid:6)(cid:6) 2 1(cid:6)(cid:6) = 9 (cid:4)= 0,sotherankisatleast2.Allthefourminorsoforder3arezero,sotherankmust
−1 4
belessthan3. Hencetherankis2. (Thethreerows,r ,r ,andr ,ofthematrixarelinearlydependent,
1 2 3
becauser = −14r +9r .)
2 1 3
(f) Thedeterminantofthewholematrixiszero,sotherankmustbelessthan4. Ontheotherhand,the
firstthreerowsandthefirstthreecolumnsyieldtheminor
(cid:6) (cid:6)
(cid:6)(cid:6) 1 −2 −1(cid:6)(cid:6)
(cid:6)(cid:6) 2 1 1(cid:6)(cid:6) = −7 (cid:4)= 0
(cid:6) (cid:6)
−1 1 −1
sotherankisatleast3. Itfollowsthatthematrixhasrank3.
1.3.2 (a) Thedeterminantis(x +1)(x −2). Therankis3ifx (cid:4)= −1andx (cid:4)= 2. Therankis2ifx = −1
orx = 2.
⎛ ⎞
x 0 x2 −2
(a) By cofactor expansion along the first row, the determinant of the matrix A = ⎝ 0 1 1 ⎠
−1 x x −1
is
|A| = x ·(−1)−0·1+(x2 −2)·1 = x2 −x −2 = (x +1)(x −2)
If x (cid:4)= −1 and x (cid:4)= 2, then |A| (cid:4)= 0, so the rank of A equals 3. If x = −1 or x = 2, then |A| = 0 and
r(A) ≤ 2. Ontheotherhand,theminorwegetifwestrikeoutthefirstrowandthethirdcolumninAis
(cid:6) (cid:6)
(cid:6) (cid:6)
(cid:6)(cid:6) 0 1(cid:6)(cid:6) = 1 (cid:4)= 0forall x,sor(A)canneverbelessthan2.
−1 x
(cid:7)
2 ifx = −1orx = 2
Conclusion: r(A) =
3 otherwise
(b) Alittlecalculationshowsthatthedeterminantofthematrixist3+4t2−4t−16,andifwenotethat
thisexpressionhast +4asafactor,itfollowsthatthedeterminantis
t3+4t2 −4t −16 = t2(t +4)−4(t +4) = (t2 −4)(t +4) = (t +2)(t −2)(t +4)
Thus,ift doesnotequalanyofthenumbers−2,2,and−4,therankofthematrixis3.
(cid:6) (cid:6)
(cid:6) (cid:6)
Ifwestrikeoutthesecondrowandthefirstcolumnofthematrix,wegettheminor(cid:6)(cid:6)5 6 (cid:6)(cid:6) = 5t+14,
1 t +4
which is different from 0 for all the three special values of t that we found above, and thus the rank of
thematrixis (cid:7)
2 ift = −4,−2,or2
3 otherwise
(c) Thefirstandthirdrowsareidentical,asarethesecondandfourth. Butthefirsttworowsarealways
linearlyindependent. Sotherankis2forallvaluesofx,y,z,andw.
©ArneStrøm,KnutSydsæter,AtleSeierstad,andPeterHammond2008
CHAPTER 1 TOPICS IN LINEAR ALGEBRA 3
1.4
1.4.2 (a) Itisclearthatx = x = x = 0isasolution. Thedeterminantofthecoefficientmatrixis
1 2 3
(cid:6) (cid:6)
(cid:6)(cid:6)1 −1 1(cid:6)(cid:6)
D = (cid:6)(cid:6)1 2 −1(cid:6)(cid:6) = 9 (cid:4)= 0
(cid:6) (cid:6)
2 1 3
Hence, by Cramer’s rule the solution is unique and the system has 0 degrees of freedom. (This agrees
withTheorem 1.4.2: since the rank of the coefficient matrix is 3 and there are 3 unknowns, the system
has3−3 = 0degreesoffreedom.)
(b) By Gaussian elimination (or other means) we find that the solution is x = a, x = −a, x = −a,
1 2 3
andx = a,witha arbitrary. Thus,thereisonedegreeoffreedom. (Wecouldgetthenumberofdegrees
4
of freedom fromTheorem 1.4.2 in this case, too. The minor formed from the first three columns of the
coefficient matrix has determinant −3 (cid:4)= 0, so the rank of the coefficient matrix is 3. Since there are 4
unknowns,thesystemhas4−3 = 1degreeoffreedom.)
1.4.3 Thedeterminantofthecoefficientmatrixisa2−7a = a(a−7). Thus,ifa (cid:4)= 0anda (cid:4)= 7,thesystem
hasauniquesolution. Ifa = 0 ora = 7, therankofthecoefficientmatrixis2(why?), andthesystem
eitherhassolutionswith1degreeoffreedomorhasnosolutionsatall,dependingonthevalueofb. One
wayoffindingoutisbyGaussianelimination,i.e.byusingelementaryrowoperationsontheaugmented
coefficientmatrix:
⎛ ⎞ ⎛ ⎞
1 2 3 1 1 −3 1 2 3 1
⎝−1 a −21 2⎠ ← ∼ ⎝0 a+2 −18 3 ⎠ ←
3 7 a b ← 0 1 a−9 b−3 −(a+2)
⎛ ⎞
1 2 3 1
∼ ⎝0 0 −a2 +7a −ab−2b+3a+9⎠ ←
0 1 a−9 b−3 ←
⎛ ⎞
1 2 3 1
∼ ⎝0 1 a−9 b−3 ⎠
0 0 a(7−a) −ab−2b+3a+9
Thisconfirmsthataslongasa (cid:4)= 0anda (cid:4)= 7,thesystemhasauniquesolutionforanyvalueofb. Butif
a = 0ora = 7,thesystemhassolutionsifandonlyif−ab−2b+3a+9 = 0,andthenithassolutions
with1degreeoffreedom. (Therankofthecoefficientmatrixis2andthereare3unknowns.)
Ifa = 0,thenthesystemhassolutionsifandonlyifb = 9/2.
Ifa = 7,thenthesystemhassolutionsifandonlyif−9b+30 = 0,i.e.b = 10/3.
1.4.6 (a) |A | = (t −2)(t +3),sor(A ) = 3ift (cid:4)= 2andt (cid:4)= −3. Becausetheupperleft2×2minorof
t t
At is−1 (cid:4)= 0⎛,the⎞rankofAt canneverbelessthan⎛2,so⎞r(A2) = 2,r(A−3) = 2.
x 11
1
(b) Letx = ⎝x2⎠. ThevectorequationA−3x = ⎝ 3⎠isequivalenttotheequationsystem
x 6
3
x + 3x +2x = 11 (1)
1 2 3
2x + 5x −3x = 3 (2)
1 2 3
4x +10x −6x = 6 (3)
1 2 3
©ArneStrøm,KnutSydsæter,AtleSeierstad,andPeterHammond2008
4 CHAPTER 1 TOPICS IN LINEAR ALGEBRA
Equation(3)isobviouslyequivalentto(2),sowecanremoveit,andthenweareleftwiththetwoequations
x +3x +2x = 11 (1)
1 2 3
2x +5x −3x = 3 (2)
1 2 3
Wecanconsidertheseequationsasanequationsystemwithx andx astheunknowns:
1 2
x +3x = 11−2x (1(cid:9))
1 2 3
2x +5x = 3+3x (2(cid:9))
1 2 3
For each value of x this system has the unique solution x = 19x −46, x = −7x +19. Thus the
3 1 3 2 3
vectorequationA x = (11,3,6)(cid:9) hasthesolution
3
x = (19s −46, −7s +19, s)(cid:9)
wheres runsthroughallrealnumbers.
1.5
1.5.1 Forconvenience,letA,B,...,Fdenotethematricesgivenin(a),(b),...,(f),respectively.
(cid:8) (cid:9)
2 −7
(a) ThecharacteristicpolynomialofthematrixA = is
3 −8
(cid:6) (cid:6)
|A−λI| = (cid:6)(cid:6)(cid:6)2−λ −7 (cid:6)(cid:6)(cid:6) = λ2 +6λ+5 = (λ+1)(λ+5)
3 −8−λ
sotheeigenvalueso(cid:8)fA(cid:9)areλ1 = −1andλ2 = −5. Theeigenvectorscorrespondingtoaneigenvalueλ
x
arethevectorsx = (cid:4)= 0thatsatisfyAx = λx,i.e.
y
(cid:10)
2x −7y = −x
⇐⇒ 3x = 7y forλ = −1
3x −8y = −y
and (cid:10)
2x −7y = −5x
⇐⇒ 7x = 7y forλ = −5
3x −8y = −5y
(cid:8) (cid:9) (cid:8) (cid:9)
7 1
Thisgivesustheeigenvectors v = s and v = t , wheres andt arearbitraryrealnumbers
1 2
3 1
(differentfrom0).
(cid:6) (cid:6)
(b) The characteristic equation of B is |B−λI| = (cid:6)(cid:6)(cid:6)2−λ 4 (cid:6)(cid:6)(cid:6) = λ2 −8λ+20 = 0. B has two
−2 6−λ
complexeigenvalues,4±2i,andnorealeigenvalues.
(c) ThecharacteristicpolynomialofCis|C−λI| = λ2−25,andweseeimmediatelythattheeigenvalues
areλ = 5andλ = −5. Theeigenvectorsaredeterminedbytheequationsystems
1 2
(cid:11) (cid:11)
x +4y = 5x x +4y = −5x 3
⇐⇒ x = y and ⇐⇒ y = − x
6x − y = 5y 6x − y = −5y 2
©ArneStrøm,KnutSydsæter,AtleSeierstad,andPeterHammond2008
CHAPTER 1 TOPICS IN LINEAR ALGEBRA 5
respectively,sotheeigenvectorsare
(cid:8) (cid:9) (cid:8) (cid:9)
1 −2
v = s and v = t
1 2
1 3
wheres andt arearbitrarynonzeronumbers.
(d) ThecharacteristicpolynomialofDis
(cid:6) (cid:6)
(cid:6)(cid:6)2−λ 0 0 (cid:6)(cid:6)
|D−λI| = (cid:6)(cid:6) 0 3−λ 0 (cid:6)(cid:6) = (2−λ)(3−λ)(4−λ)
(cid:6) (cid:6)
0 0 4−λ
Theeigenvaluesareobviouslyλ = 2,λ = 3,λ = 4,andthecorrespondingeigenvectorsare
1 2 3
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 0 0
v = s⎝0⎠, v = t⎝1⎠, v = u⎝0⎠
1 2 3
0 0 1
where s, t, u are arbitrary nonzero numbers. (Note: The eigenvalues of a diagonal matrix are always
preciselythediagonalelements,and(multiplesof)thestandardunitvectorswillbeeigenvectors. Butif
twoormoreofthediagonalelementsareequal,therewillbeothereigenvectorsaswell.Anextremecase
istheidentitymatrixI : all(nonzero)n-vectorsareeigenvectorsforI .)
n n
(e) ThecharacteristicpolynomialofEis
(cid:6) (cid:6)
(cid:6)(cid:6)2−λ 1 −1 (cid:6)(cid:6)
(cid:6)(cid:6) 0 1−λ 1 (cid:6)(cid:6) = −λ3+λ2 +2λ = −λ(λ2 −λ−2)
(cid:6) (cid:6)
2 0 −2−λ
Theeigenvaluesaretherootsoftheequation−λ(λ2−λ−2) = 0,namelyλ = −1,λ = 0andλ = 2.
1 2 3
Theeigenvectorscorrespondingtoλ = −1aresolutionsof
1
⎧
⎪⎨2x1+x2 − x3 = −x1 (cid:16)x = 1x (cid:16)x = −x
Ex = −x ⇐⇒ x + x = −x ⇐⇒ 1 2 3 ⇐⇒ 2 1
⎪⎩ 2 3 2 x = −1x x = 2x
2x −2x = −x 2 2 3 3 1
1 3 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 1 2
sotheyareoftheformv = s⎝−1⎠. Similarly,v = t⎝−1⎠andv = u⎝1⎠aretheeigenvectors
1 2 3
2 1 1
correspondingtoλ = 0andλ = 2.
2 3
(f) ThecharacteristicpolynomialofFis
(cid:6) (cid:6)
(cid:6)(cid:6)1−λ −1 0 (cid:6)(cid:6)
(cid:6)(cid:6) −1 2−λ −1 (cid:6)(cid:6) = −λ3+4λ2 −3λ = −λ(λ2 −4λ+3) = −λ(λ−1)(λ−3)
(cid:6) (cid:6)
0 −1 1−λ
The eigenvalues are therefore λ = 0, λ = 1, and λ = 3. By the same method as above we find that
1 2 ⎛ ⎞ 3 ⎛ ⎞ ⎛ ⎞
1 −1 1
thecorrespondingeigenvectorsarev = s⎝1⎠,v = t⎝ 0⎠,andv = u⎝−2⎠.
1 2 3
1 1 1
©ArneStrøm,KnutSydsæter,AtleSeierstad,andPeterHammond2008
6 CHAPTER 1 TOPICS IN LINEAR ALGEBRA
⎛ ⎞
ax +ay
1.5.2 (a) X(cid:9)AX = X(cid:9)(AX) = (x, y, z)⎝ax +ay⎠ = (ax2 +ay2 +2axy +bz2) (a1×1matrix),
bz
⎛ ⎞ ⎛ ⎞
2a2 2a2 0 4a3 4a3 0
A2 = ⎝2a2 2a2 0 ⎠, A3 = ⎝4a3 4a3 0 ⎠
0 0 b2 0 0 b3
(cid:6) (cid:6)
(cid:6)(cid:6)a−λ a 0 (cid:6)(cid:6)
(b) ThecharacteristicpolynomialofAisp(λ) = (cid:6)(cid:6) a a−λ 0 (cid:6)(cid:6) = (λ2−2aλ)(b−λ),sothe
(cid:6) (cid:6)
0 0 b−λ
eigenvaluesofAareλ = 0,λ = 2a,λ = b.
1 2 3
(c) From (b) we get p(λ) = −λ3 +(2a +b)λ2 −2abλ. Using the expressions for A2 and A3 that we
foundinpart(a),itiseasytoshowthatp(A) = −A3+(2a+b)A2 −2abA = 0.
1.5.4 (a) TheformulainProblem1.9.7(b)yields
(cid:6) (cid:6)
(cid:6)4−λ 1 1 1 (cid:6)
(cid:6) (cid:6) (cid:8) (cid:9)
|A−λI| = (cid:6)(cid:6)(cid:6) 1 4−λ 1 1 (cid:6)(cid:6)(cid:6) = (3−λ)4 1+ 4 = (3−λ)3(7−λ)
(cid:6) 1 1 4−λ 1 (cid:6) 3−λ
(cid:6) (cid:6)
1 1 1 4−λ
Hence,theeigenvaluesofAareλ = λ = λ = 3,λ = 7.
1 2 3 4
(b) An eigenvector x = (x ,x ,x ,x )(cid:9) of A corresponding to the eigenvalue λ = 3 must satisfy the
1 2 3 4
equationsystem(A−3I)x = 0. The4equationsinthissystemareallthesame,namely
x +x +x +x = 0
1 2 3 4
Thesystemhassolutionswith4−1 = 3degreesoffreedom. Onesimplesetofsolutionsis
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 1 1
x1 = ⎜⎜⎝−01⎟⎟⎠, x2 = ⎜⎜⎝−01⎟⎟⎠. x3 = ⎜⎜⎝ 00⎟⎟⎠
0 0 −1
Thesethreevectorsareobviouslylinearlyindependentbecauseif
⎛ ⎞
c +c +c
1 2 3
c1x1+c2x2 +c3x3 = ⎜⎜⎝ −−cc1 ⎟⎟⎠
2
−c
3
isthezerovector,thenc = c = c = 0.
1 2 3
1.5.5 (c) If λ is an eigenvalue for C with an associated eigenvector x, then Cnx = λnx for every natural
numbern. IfC3 = C2+C,thenλ3x = λ2x+λx,so(λ3−λ2−λ)x = 0. Thenλ3−λ2−λ = 0,because
x (cid:4)= 0. IfC+I didnothaveaninverse,|C+I | = 0. Thenλ = −1wouldbeaneigenvalueforC,and
n n
sowewouldhaveλ3−λ2−λ = 0,whichisnottrueforλ = −1. Hence−1isnotaneigenvalueforC,
andconsequentlyC+I hasaninverse.
n
©ArneStrøm,KnutSydsæter,AtleSeierstad,andPeterHammond2008
CHAPTER 1 TOPICS IN LINEAR ALGEBRA 7
1.6
(cid:8) (cid:9)
2 1
1.6.1 (a) LetA = . ThecharacteristicpolynomialofAis
1 2
(cid:6) (cid:6)
p(λ) = (cid:6)(cid:6)(cid:6)2−λ 1 (cid:6)(cid:6)(cid:6) = (2−λ)2 −1 = λ2 −4λ+3 = (λ−1)(λ−3)
1 2−λ
Thus, the eigenvalues are λ = 1 and λ = 3. The associated eigenvectors with length 1 are uniquely
1 2
determineduptosignas (cid:8) √ (cid:9) (cid:8) √ (cid:9)
1 2 1 2
x = 2√ and x = 2√
1 −1 2 2 1 2
2 2
Thisyieldstheorthogonalmatrix (cid:8) √ √ (cid:9)
1 2 1 2
P = 2√ 2√
−1 2 1 2
2 2
(ItiseasytoverifythatP(cid:9)P = I,i.e.P−1 = P(cid:9).) Wethereforehavethediagonalization
(cid:8) √ √ (cid:9)(cid:8) (cid:9)(cid:8) √ √ (cid:9) (cid:8) (cid:9)
1 2 −1 2 2 1 1 2 1 2 1 0
P−1AP = 2√ 2√ 2√ 2√ =
1 2 1 2 1 2 −1 2 1 2 0 3
2 2 2 2
⎛ ⎞
1 1 0
(b) LetB = ⎝1 1 0⎠. ThecharacteristicpolynomialofBis
0 0 2
(cid:6) (cid:6)
(cid:6)(cid:6)1−λ 1 0 (cid:6)(cid:6) (cid:19) (cid:20)
(cid:6)(cid:6) 1 1−λ 0 (cid:6)(cid:6) = (2−λ) (1−λ)2 −1 = (2−λ)(λ2 −2λ) = −λ(λ−2)2
(cid:6) (cid:6)
0 0 2−λ
(use cofactor expansion of the first determinant along the last row or the last column). The eigenvalues
are λ = 0 and λ = λ = 2. It is easily seen that one eigenvector associated with the eigenvalue
1 2 3
λ = 0isx = (1,−1,0)(cid:9). Eigenvectorsx = (x ,x ,x )(cid:9) associatedwiththeeigenvalue2aregivenby
1 1 1 2 3
(B−2I)x = 0,i.e.
−x +x = 0
1 2
x −x = 0
1 2
0 = 0
Thisgivesx = x ,x arbitrary. Onesetoflinearlyindependenteigenvectorswithlength1isthen
1 2 3
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 1 0
1 1
√ ⎝−1⎠, √ ⎝1⎠, ⎝0⎠
2 2
0 0 1
Fortunately,thesethreevectorsaremutuallyorthogonal(thisisnotautomaticallytruefortwoeigenvectors
associatedwiththesameeigenvalue),andsowehaveasuitableorthogonalmatrix
⎛ √ √ ⎞
1 2 1 2 0
2√ 2√
⎜ ⎟
P = ⎝−1 2 1 2 0⎠
2 2
0 0 1
©ArneStrøm,KnutSydsæter,AtleSeierstad,andPeterHammond2008
8 CHAPTER 1 TOPICS IN LINEAR ALGEBRA
ItisnoweasytoverifythatP−1 = P(cid:9) and
⎛ ⎞
0 0 0
P−1BP = ⎝0 2 0⎠
0 0 2
⎛ ⎞
1 3 4
(c) ThecharacteristicpolynomialofC = ⎝3 1 0⎠is
4 0 1
(cid:6) (cid:6)
(cid:6)(cid:6)(cid:6)(cid:6)1−3 λ 1−3 λ 04 (cid:6)(cid:6)(cid:6)(cid:6) = 4(cid:6)(cid:6)(cid:6)(cid:6)3 1−λ(cid:6)(cid:6)(cid:6)(cid:6)+(1−λ)(cid:6)(cid:6)(cid:6)(cid:6)1−λ 3 (cid:6)(cid:6)(cid:6)(cid:6) = (1−λ)(cid:19)(1−λ)2 −25(cid:20)
(cid:6) (cid:6) 4 0 3 1−λ
4 0 1−λ
(cofactorexpansionalongthelastcolumn),andsotheeigenvaluesareλ = 1,λ = 6,andλ = −4.An
1 2 3
eigenvectorx = (x,y,z)(cid:9) correspondingtotheeigenvalueλmustsatisfy
x +3y +4z = λx
Cx = λx ⇐⇒ 3x + y = λy
4x + z = λz
Onesetofunnormalizedeigenvectorsis
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
0 5 −5
u = ⎝−4⎠, v = ⎝3⎠, w = ⎝ 3⎠
3 4 4
√
withlengths(cid:5)u(cid:5) = 5,(cid:5)v(cid:5) = (cid:5)w(cid:5) = 5 2,andacorrespondingorthogonalmatrixis
⎛ √ √ ⎞
0 5 2 − 5 2
10√ 10√
⎜ ⎟
P = ⎝−4 3 2 3 2⎠
5 10√ 10√
3 4 2 4 2
5 10 10
AstraightforwardcalculationconfirmsthatP(cid:9)CP = diag(1,6,−4) = diag(λ ,λ ,λ ).
1 2 3
1.6.5 ForthegivenA,wehaveA2 = 5A−5I.(cid:8)ThereforeA(cid:9)3 = A2A = (5A−5I)A = 5A2−5A = 20A−25I
50 75
andA4 = 20A2 −25A = 75A−100I = .
75 125
1.7
1.7.5 (a) ItisclearthatQ(x ,x ) ≥ 0forallx andx andthatQ(x ,x ) = 0onlyifx = x = 0,soQ
1 2 1 2 1 2 1 2
ispositivedefinite.
⎛ ⎞
5 0 1
⎝ ⎠
(b) ThesymmetriccoefficientmatrixofQis 0 2 1 . Theleadingprincipalminorsare
1 1 4
(cid:6) (cid:6)
(cid:6) (cid:6) (cid:6) (cid:6)
(cid:6) (cid:6) (cid:6)5 0 1(cid:6)
D1 = 5, D2 = (cid:6)(cid:6)5 0(cid:6)(cid:6) = 10, D3 = (cid:6)(cid:6)0 2 1(cid:6)(cid:6) = 33
0 2 (cid:6) (cid:6)
1 1 4
Since all the leading principal minors are positive, it follows from Theorem 1.7.1 that Q is positive
definite. (AnalternativewaytoseethisistowriteQasasumofsquares: Q(x ,x ,x ) = (x +x )2+
1 2 3 1 3
(x +x )2+4x2+x2+2x2 isobviouslynonnegativeanditiszeroonlyifallthesquaretermsarezero.
2 3 1 2 3
Butitisnotalwayseasytoseehowtorewriteaquadraticforminthisfashion.)
©ArneStrøm,KnutSydsæter,AtleSeierstad,andPeterHammond2008