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18.06ProblemSet7SolutionsTotal: 100points

Prob.16,Sec.5.2,Pg.265: Fn isthedeterminantofthe1,1,1tridiagonalmatrixofordern:11 111 1 0

11 11 11= 2 1 1

1 = 3 = 4.F2 F3 F4= = =1 1 11 1

0 1Expand incofactorstoshowthatFn =Fn1+Fn2. ThesedeterminantsareFibonaccinumbers1,2,3,5,8,13, . . .. The sequence usually starts 1,1,2,3 (with two 1s) so our Fn is the usualFn+1.Solution (seepg.535,4pts.): The1,1cofactorofthenbynmatrix isFn1. The1,2cofactor

hasa1incolumn1,withcofactorFn2. Multiplyby(1)1+2 andalso(1)fromthe1,2entrytofindFn =Fn1+Fn2 (sothesedeterminantsareFibonaccinumbers).Prob.32, Sec.5.2, Pg.268: Cofactors of the 1,3,1 matrices in Problem 21 give a recursionSn = 3Sn1Sn2. AmazinglythatrecursionproduceseverysecondFibonaccinumber. Here isthechallenge.Show that Sn is theFibonaccinumber F2n+2 byprovingF2n+2 = 3F2nF2n2. KeepusingFibonaccisruleFk =Fk1+Fk2 startingwithk= 2n+2.Solution (seepg.535,12pts.): ToshowthatF2n+2 = 3F2nF2n2,keepusingFibonaccisrule:

F2n+2 =F2n+1+F2n =F2n+Fn1+F2n = 2F2n+ (F2nF2n2) = 3F2nF2n2.Prob.33,Sec.5.2,Pg.268: ThesymmetricPascalmatriceshavedeterminant1. IfIsubtract1fromthen,nentry,whydoesthedeterminantbecomezero? (Userule3orcofactors.)

1 1 1 1 1 1 1 1det

1 2 3 41 3 6 10

=1(known) det

1 2 3 41 3 6 10

=0(toexplain).1 4 10 20 1 4 10 19

Solution (seepg.535, 12pts.): Thedifference from20to19multiplies itscofactor, which isthedeterminantofthe3by3Pascalmatrix,soequalto1. Thusthedetdropsby1.Prob.8,Sec.5.3,Pg.279: FindthecofactorsofAandmultiplyACT tofinddetA:

1 1 4 6 3 0 andACT = .A=1 2 2 and C= 1 2 5

Ifyouchangethat4to100,whyisdetAunchanged?

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]

Solution

pset7-s10-soln: page2

(seepg.536,4pts.): StraightforwardcomputationyieldsC anddetA=3: 6 3 0 3 0 0 This is(detA)I anddetA=3. andACT

13 16 2

0 3 0. The1,3cofactorofA is0.1C= =0 0 3 Multiplyingby4orby100: nochange.

Prob.28, Sec.5.3, Pg.281: Spherical coordinates ,, satisfy x = sincos and y =sinsinandz=cos. Findthe3by3matrixofpartialderivatives: x/,x/,x/ inrow1. SimplifyitsdeterminanttoJ =2sin. Thend V insphericalcoordinatesis2sindddthevolumeofan infinitesimalcoordinatebox.Solution (4pts.): Therowsareformedbythepartialsofx,y,z withrespectto,,:

sincos coscos sinsinsinsin cossin sincos.cos sin 0

Expanding itsdeterminantJ alongthebottomrow,wegetJ =cos(2cossin)(cos2+sin2) +2sin3(cos2+sin2)

=2sin(cos2+sin2) =2sin.Prob.40,Sec.5.3,Pg.282: SupposeAisa5by5matrix. Itsentriesinrow1multiplydeterminants(cofactors)inrows25togivethedeterminant. CanyouguessaJacobiformulafordetAusing 2 by 2 determinants from rows 12 times 3 by 3 determinants from rows 35? Test yourformula

on

the

1,2,1tridiagonalmatrixthathasdeterminant6.Solution (12pts.): Agoodguess fordetA isthesum,overallpairsi,j withi < j,of(1)i+j+1

timesthe2by2determinantformedfromrows12andcolumnsi,jtimesthe3by3determinant)formed from rows 35 and the complementary columns (this formula is more commonly named5(

2

afterLaplacethanJacobi). Thereare terms. Inthegivencase,onlythefirsttwoarenonzero:2

22

2 1 1 12 1 2

detA= 1 21 1 21 1 =(3)(4)(2)(3)=6.1 1 1

Prob.41, Sec.5.3, Pg.282: The 2 by 2 matrix AB = (2by3)(3by2) has a CauchyBinetformulafordetAB:

detAB= sumof(2by2determinants inA)(2by2determinants inB).(a)Guesswhich2by2determinantstousefromAandB.(b)TestyourformulawhentherowsofAare1,2,3and1,4,7withB=AT.Solution (12pts.): (a)A goodguess is thesum, overall pairs i,j with i < j, oftheproductof

the2by2determinantsformedfromcolumnsi,j ofAandrowsi,j ofB.1 1[ [ ]

1 2 3 14 30

2 4

=(b)First,AAT =11

. SodetAAT =924900=24.1 4 7 30 66

3+

71 31 724 1 12 4 1 13 7 + 2 34 7 2 43 7 =4 + 16 + 4=24.Ontheotherhand,

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pset7-s10-soln: page3

Prob.19, Sec.6.1, Pg.295: A 3 by 3 matrix B is known to have eigenvalues 0,1,2. This isinformationisenoughtofindthreeofthese(givetheanswerswherepossible):

(a)the

rank

of

B,

(b)thedeterminantofBTB,(c)theeigenvaluesofBTB,(d)theeigenvaluesof(B2+I)1.

Solution (4pts.): (a)Therankisatmost2sinceB issingularas0isaneigenvalue. Therankisnot0sinceB isnot0asB hasanonzeroeigenvalue. Therank isnot1sincearank-1matrixhasonlyonenonzeroeigenvalueaseveryeigenvectorlies inthecolumnspace. Thustherankis2.

(b)WehavedetBTB=detBTdetB=(detB)2 = 0 1 2 = 0. (c)ThereisnotenoughinformationtofindtheeigenvaluesofBTB. Forexample,

0 0 0 1 0ifB= 1 , thenBTB= 1 ; ifB= 1 , thenBTB= 2 .2 4 2 4

However,theeigenvaluesofatriangularmatrixare itsdiagonalentries.(d) IfAx=x,thenx=A1x; also,anypolynomialp(t)yieldsp(A)x=p()x. Hencethe

eigenvaluesof(B2+I)1 are1/(02+1)and1/(12+1)and1/(22+1),or1and1/2and1/5.Prob.29,Sec.6.1,Pg.296: (Review)FindtheeigenvaluesofA, B,andC:

1 2 3 0 0 1 2 2 2A=

0 4 5

and B=

0 2 0

and C=

2 2 2

.

0 0 6 3 0 0 2 2 2Solution (4pts.): Sincetheeigenvaluesofatriangularmatrixare itsdiagonalentries,theeigen

valuesofAare1,4,6. SincethecharacteristicpolynomialofB isdet(BI) = ()(2)()1(2)3=(2)(23),

theeigenvaluesofBare2,3. SinceC is6timestheprojectiononto(1,1,1),theeigenvaluesofC are6,0,0.Prob.6, Sec.6.2, Pg.308: Describe all matrices S that diagonalize this matrix A (find alleigenvectors):

[ ]4 0A= ,1 2

ThendescribeallmatricesthatdiagonalizeA1.Solution (seepg.537,4pts.): ThecolumnsofS arenonzeromultiplesof(2,1)and(0,1): either

order. SameforA1. Indeed,sincetheeigenvaluesofatriangularmatrixare itsdiagonalentries,theeigenvaluesofAare4,2. Further,(2,1)and(0,1)obviouslyspanthenullspacesof[ ] [ ]

0 0 2 0and .

1 2 1 0Prob.16,Sec.6.2,Pg.309: (Recommended)FindandS todiagonalizeA1 inProblem15:

[ ].6 .9A1 = ..4 .1

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pset7-s10-soln: page4

What is the limitofk ask ? What isthe limit of SkS1? Inthecolumnsofthematrixyouseethe .Solution (4pts.): Thecolumnssumto1;hence,A1I issingular,andso1isaneigenvalue. The

twoeigenvaluessumto0.6+0.1;sotheotherone is0.3. Further,thenullspacesof[ ] [ ]0.4 0.9 0.9 0.9and

0.4 0.9 0.4 0.4areobviouslyspannedby(9,4)and(1,1). Therefore,

=[

1 ]and S=

[9 1]

and k[

1 ]and0.3 4 1 0

[

4 1] [

0]

9 + 4[4 9]

13[4 4]

SkS1 9 1 1 1 1 1 = 1 9 9 .Inthecolumnsofthe lastmatrixyouseethesteadystatevector.Prob.37,Sec.6.2,Pg.311: ThetransposeofA=SS1 isAT = (S1)TST. Theeigenvectorsin ATy = y are the columns of that matrix (S1)T. They are often called left eigenvectors.HowdoyoumultiplymatricestofindthisformulaforA?

Sumofrank-1matrices A=SS1 =1x1y1T+ +nxnynT. Solution (seepg.539,12pts.): ColumnsofStimesrowsofS1 willgiverrank-1matrices(r=

rankof

A).

Challengeproblem: inMATLAB(andinGNUOctave),thecommandA=toepliz(v)producesasymmetricmatrixinwhicheachdescendingdiagonal(fromlefttoright)isconstantandthefirstrowisv. Forinstance,ifv= [010001],thentoepliz(v)isthematrixwith1sonbothsidesofthemaindiagonalandonthefarcorners,and0selsewhere. Moregenerally,letv(n)bethevectorinRn witha1 inthesecondandlastplacesand0selsewhere,and letA(n)=toepliz(v(n)).

(a)Experimentwithn= 5, . . . ,12inMATLABtoseetherepeatingpatternofdetA(n).(b)ExpanddetA(n)intermsofcofactorsofthefirstrowandintermsofcofactorsofthefirst

column. UsetheknowndeterminantCn ofproblem5.2.13torecoverthepatternfoundinpart(a).Solution (12pts.): (a)Theoutput2,

4,2,0,2,

4,2,0isreturnedbythislineofcode:

for n = 5:12; v=zeros(1,n); v(2)=1; v(n)=1; det(toeplitz(v)), endfor.(b)ExpanddetA(n)alongthefirstrowandthendownbothfirstcolumnstoget{

detA(n) =Cn2(1)n+ (1)n+1+ (1)n+1(1)nCn2 whereCn = 0, nodd;(1)n/2, neven.

ThusdetA(n)=2(Cn(1)n),whichrecoversthepatternfound inpart(a).

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18.06 Linear Algebra

Spring 2010

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