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18.06Pset9SolutionsProblem

6.5,

#25:

With

positive

pivots

in

D,

the

factorization

A

=

LDL

Tbecomes LDDLT. (Square roots of the pivots give D =DD.) Then C =

DLT yieldstheCholeskyfactorization A=CTC whichissymmetrizedLU.3 1 4 8

From C= findA. From A= findC=chol(A).0 2 8 25

Solution (4points)FromC,weobtain

A=CTC= 3 0 3 1 = 9 3 .

1 2 0 2 3 54 0 1 0

Conversely,thegivenAquicklydiagonalizesto viaL= : thus0 9 2 1

2 4C=chol(A) =DLT = .

0 3

Problem 6.5, #26: In the Cholesky factorization A = CTC, with CT = LD,thesquarerootsofthepivotsareonthediagonalofC. FindC (uppertriangular)for

9 0 0 1 1 1A= 0 1 2 and A= 1 2 2

0 2 8 1 2 7.Solution (4points)ForthefirstmatrixA,wehave

1 0 0 9 0 0 1 0 0 3 0 0A= 0 1 0 0 1 0 0 1 2 C= 0 1 0

0 2 1 0 0 4 0 0 1 0 2 2whileforthesecondmatrixwehave

1 0 0 1 0 0 1 1 1 1 1 1A= 1 1 0 0 1 0 0 1 1 C= 0 1 1

51 1 1 0 0 5 0 0 1 0 01

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2Problem6.5,#27: ThesymmetricfactorizationA=LDLT meansthatxTAx=xTLDLTx:

a b x

1 0 a 0 1 b/a x

x y = x yc d y b/a 1 0 (ac

b2)/a 0 1 y

The leftside isax2 + 2bxy+cy2. Therightside isa(x+aby)2 + y2. Thesecond

pivotcompletesthesquare! Testwitha= 2, b= 4, c=10.Solution (4 points) Evaluating out the right side gives ax2 + 2bxy+cy2, so the

b2entry in the space given is ca, i.e. the second pivot. For the given values, we

have2x2 + 8xy+10y2 =2(x+ 2y)2 + 2y2 asdesired.Problem6.5,#29: ForF1(x, y) =x4/4 +x2+x2y+y2 andF2(x, y) =x3+xyx,findthesecond-derivativematricesH1 andH2:

2F 2Fx2 xy

H= 2F 2F .yx y2H1 ispositive-definitesoF1 isconcaveup(=convex). FindtheminimumpointofF1 andthethesaddlepointofF2 (lookonlywherethefirstderivativesarezero).Solution (4points)ForF1(x, y),wefirstsolveforthestationarypoint

F1 F1=x3 + 2x+ 2xy= 0, =x2 + 2y= 0

x yFrom (2), we have y =x2/2. Plug this into (1), we have 2x = 0 and hence theonlycriticalpointisx=y=0. Atthispoint,

2

F 2

Fx2 xy 3x3 + 2 + 2y 2x 2 0H1 = 2F 2F = = .2x 2 0 2yx y2

Itispositivedefiniteandhence(0,0)isaminimalpointofF1(x, y).REMARK:TheproblemforF1 =x4/4 +x2y+y2 asoriginallystated,youget

acurveofminimax2 + 2y=0,andH1 isonlypositivesemidefinite.ForF2(x, y),wefirstsolveforthestationarypoint

F2 2 F2x = 3x +y1 = 0, y =x= 0

Thisimpliesthaty=1. Atthispoint(0,1),2F 2F x2 xy 6x 1 0 1H2 = 2F 2F = = .1 0 1 0

yx y2

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3TheeigenvaluesofH2 at(0,1)isthesolutiontodet(H2I) =21,whichare1 =1and2 =1. Theyarewithoppositesignsandhence(0,1)isasaddlepointofF2(x, y).Problem 6.5, #32: A group of nonsingular matrices include AB and A1 if itincludesAandB. Productsand inversesstay inthegroup. Whichofthesearegroups(asin2.7.37)? Inventasubgroupoftwoofthesegroups(notI byitself=thesmallestgroup).

(a) PositivedefinitesymmetricmatricesA.(b) OrthogonalmatricesQ.(c) AllexponentialsetA ofafixedmatrixA.(d) MatricesP withpositiveeigenvalues.(e) MatricesDwithdeterminant1.

Solution (12points)First,notethatallbutthefirstandfourtharegroups(assum1 2

ing we are only referring to square matrices in (b)): on the other hand, 2 3and 1

2 24 are both positive definite and symmetric, but their product is notsymmetric. Intersections of these groups give the simplest examples of subgroups(forinstance,orthogonalmatriceswithdeterminant1,calledthespecialorthogonalmatrices),thoughtherearemanyothers.Problem 6.5, #33: When A and B are symmetric positive definite, AB mightnotevenbesymmetric. Butitseigenvaluesarestillpositive. StartfromABx=xandtakedotproductswithBx. Thenprove >0.Solution (12 points) Taking dot products, we get (ABx)TBx= (Bx)TA(Bx) on

the leftand(x)TBx=xTBx. SinceB ispositivedefinite,xTBx>0,andsinceAispositivedefinite,(Bx)TA(Bx)istoo(Bxisjustanothervector). Thus,mustbepositiveaswell.Problem6.5,#34: Writedownthe5by5sinematrixS fromWorkedExample6.5D,containingtheeigenvectorsofKwhenn=5andh= 1/6. MultiplyK timesS toseethefivepositiveeigenvalues.

Theirsumshouldequalthetrace10. TheirproductshouldbedetK=6.

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4Solution (12points)S isthematrix

3/2 3/20

3/2

3/2

1 0 1 0 13/2 3/2 3/2 3/2 3/2

13/

/2

2 3/2 1 1/2S=

1/2 3/2 1 3/2 1/2Thefiveeigenvalues(correspondingtothecolumns)are23,1,2,3,and2 +3,whichaddupto10andmultiplyto6asdesired.Problem 6.5, #35: If A has full column rank, and C is positive-definite, showthatATCAispositivedefinite.Solution (12 points) Since C is positive-definite, yTCy >0 for any y = 0 inRn.

Now,

weneed

to

show

that

zT

AT

CAz

>0for

any

z

= 0 inRn. We can rewriteitaszTATCAz= (Az)TC(Az). SinceAhas fullcolumnrank,N(A) ={0}and inparticular, Az=0 inRn. Therefore, wehave(Az)TC(Az)>0. This impliesthatATCAispositivedefinite.Problem 8.1, #3: In the free-free case when ATCA in equation (9) is singular,addthethreeequationsATCAu=f toshowthatweneedf1+f2+f3 =0. FindasolutiontoATCAu=f whentheforcesf = (1,0,1)balancethemselves. Findallsolutions!Solution (4points)Dotproductingourformulawith(1,1,1)gives

c2 c2 0 f1u1 = 1 1 1 1 1 1 f2c2 c2 +c3 c30 u2 f3c3 c3 u3

0 =f1 +f2 +f3Substitutingf = (1,0,1)givesthetwoequationsc2(u1u2) =1, c3(u3u2) = 1(themiddleequationisredundant),withasolution(c

2

1,0, c3

1). Allothersolutionsaregivenbyaddingmultiplesof(1,1,1),whichspansthenullspace.Problem8.1,#5: Inthefixed-freeproblem,thematrixAissquareandinvertible.We can solve ATy = f separately from Au =e. Do the same for the differentialequation:

dySolve =f(x) with y(1)=0. Graph y(x) if f(x) = 1.

dx

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5xSolution (4 points) y(x)= f(x)dx and if f(x)=1 then y(x) = 1x. You

1cangraphthis.Problem8.1,#7: Forfivespringsandfourmasseswithbothendsfixed,whatarethematricesAandC andK? WithC=I solveKu=ones(4).Solution (4points)Thematricesare

1 0 0 0 c1A=

, C=

c2

c3,

1 1 0 00 1 1 00 0 1 10 0 0 c41 c5

c1 +c2 c2 0 0

c2 c2 +c3

c3 0 K= 0 c3 c3 +c4 c40 0 c4 c4 +c5

=c5 =1givesInvertingK forc1 =

4 3 2 11 3 6 4 22 4 6 3K1 = 51 2 3 4

Multiplyingby(1,1,1,1)gives(2,3,3,2).Problem 8.1, #10: (MATLAB) Find the displacements u(1), . . . , u(100) of 100massesconnectedbyspringsallwithc=1. Eachforceisf(i) = 0.01. Printgraphsofuwith fixed-fixed and fixed-free ends. Note thatdiag(ones(n,1), d) is a matrixwithnonesalongthediagonald. Thisprintcommandwillgraphavectoru:

plot(u,+); xlabel(massnumber); ylabel(movement); print

Solution (12points)>> E = diag(ones(99,1),1);>> K = 2*eye(100)-E-E;>> f = 0.01*ones(100, 1); u = K\f;>> plot(u,+); xlabel(mass number); ylabel(movement); print>> K(100,100) = 1; u = K\f;>> plot(u,+); xlabel(mass number); ylabel(movement); print

http:///reader/full/f(i)=0.01http:///reader/full/f(i)=0.01http:///reader/full/f(i)=0.01http:///reader/full/f(i)=0.01http:///reader/full/f(i)=0.01http:///reader/full/f(i)=0.01http:///reader/full/f(i)=0.01

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6

Figure1. Fixed-fixed

0 10 20 30 40 50 60 70 80 90 1000

2

4

6

8

10

12

14

mass number

mo

vement

Problem 8.1, #11: (MATLAB) Chemical engineering has a first derivative ofdu/dx from fluid velocity as well as d2u/dx2 from diffusion. Replace du/dx byaforward difference, then a centered difference, then a backward difference, withx=

8

1. Graphyournumericalsolutionsofd2u dudx2 +10dx =1withu(0)=u(1)=0.

Solution (12points)>> E = diag(ones(6,1),1);>> K = 64*(2*eye(7) - E - E);>> D = 80*(-eye(7)+E);

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7

Figure2. Fixed-free

0 10 20 30 40 50 60 70 80 90 1000

10

20

30

40

50

60

mass number

mo

vement

>> forward = (K+D)\\ones(7,1)forward =

0.01250.02500.03760.04960.06410.06880.1125

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8>> backward = (K-D)\\ones(7,1)backward =

0.04310.05540.05390.04620.03590.02440.0123

>> centered = (K+D/2-D/2)\\ones(7,1)centered =

0.01250.02500.03740.04970.06130.06970.0644

>> plot(n,forward(n),n,backward(n),n,centered(n))

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9

Figure3. Overlayednumericalsolutions

1 2 3 4 5 6 70

0.02

0.04

0.06

0.08

0.1

0.12

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MIT OpenCourseWarehttp://ocw.mit.edu

18.06 Linear Algebra

Spring 2010

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