mit18_06s10_pset9_s10_soln
TRANSCRIPT
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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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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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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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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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18.06 Linear Algebra
Spring 2010
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