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18.06ProblemSet2SolutionTotal: 100points

Section2.5. Problem24: UseGauss-Jordaneliminationon [U I]tofindtheuppertriangular1

U1: 1 0 0a b

U U1 x1 x2 x3= 00 0 1 0 0 1

(4points): Rowreduce[U I]toget[I U1]asfollows(hereRi standsfortheithrow):

0 1 1 0=I c .

Solution

1 a b 1 0 0 (R1 =R1aR2)

=R2cR2)1

0 0 10 0 1 0 0

0 1 0bac a 0 1 c 0 1 0 (R2 1 0 c

0 0 1 0 0 1 1a acb

1

R1 1 0 0 1 =R1(bac)R3 0 1 0 0 1

0 0 1 0 0 c .

Section 2.5. Problem40: (Recommended)A isa4by4matrixwith 1sonthe diagonal anda,b,conthediagonalabove. FindA1 forthisbidiagonalmatrix.Solution (12points): Rowreduce[A I]toget[I A1]asfollows(hereRi standsfortheithrow):

1 a 0 0 1 0 0 00 1 b 0 0 1 0 00 0 1 c 0 0 1 00 0 0 1 0 0 0 1

(R1 =R1+aR2) 1 0 ab 0 1 a 0 0 (R2 =R2+bR2) 00 0 10 1 bc 0 1 b 0(R3 =R3+cR4) 00 0 0 1 0 0 0 10 0 1 c

(R1 =R1+abR3) 1 0 0 0 1 a ab abc(R2 =R2+bcR4)

0 1 0 0 0 1 b bc0 0 1 0 0 0 1 c0 0 0 1 0 0 0 1

.

Alternatively, write A = IN. Then N has a,b,c above the main diagonal, and all otherentriesequalto0. HenceA1 = (IN)1 =I+N+N2+N3 asN4 =0.Section2.6. Problem13: (Recommended)ComputeLandU forthesymmetricmatrix

a a a aa b b b

A= .a b c ca b c d

Findfourconditionsona,b,c,dtogetA=LU withfourpivots.

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pset2-s10-soln: page2

Solution (4 points): Eliminationsubtractsrow1 fromrows24, then row2 fromrows34, andfinally row 3 from row 4; the result is U. All the multipliers ij are equal to 1; so L is the lowertriangular

matrix

with

1s

on

the

diagonal

and

below

it.

a a a a a a a a 0 0 ba ba baba ba baA 0 0 00 0 bc da db

a a a a 1 0 0 0baba cbcbca ca0 ca

0 ba ba ba 1 1 0 0=U, L= .0 cb cb0 0 0 dc 1 1 1 1

ThepivotsarethenonzeroentriesonthediagonalofU. Sotherearefourpivotswhenthese fourconditionsaresatisfied: a=0,b=a,c=b,andd=c.Section2.6. Problem18: IfA=LDU andalso A=L1D1U1 withall factors invertible, thenL=L1 andD=D1 andU =U1. Thethreefactorsareunique.

Derive the equation L1

1LD = D1U1U1. Are the two sides triangular or diagonal? DeduceL=L1 andU =U1 (theyallhavediagonal1s). ThenD=D1.Solution (4points): NoticethatLDU =L1D1U1. Multiplyonthe leftbyL1 1 andontheright

byU1,gettingL

1

1LDUU1 =L1

1L1D1U1U1.

ButU U1 =I andL1

1L1 =I. ThusL1 1LD=D1U1U1,asdesired.TheleftsideL1 1LD islowertriangular,andtherightsideD1U1U1 isuppertriangular. But

theyreequal. Sotheyrebothdiagonal. HenceL1

1LandU1U1 arediagonaltoo. Buttheyhavediagonal 1s. So theyre both equal to I. Thus L =L1 and U = U1. Also L1 1LD =D1U1U1.ThusD=D1.Section2.6. Problem25: Forthe6by6seconddifferenceconstant-diagonalmatrixK,putthepivots and multipliers intoK =LU. (L and U will have onlytwo nonzero diagonals, because Khas three.) Find a formula for the i, j entry of L1, by software like MATLAB using inv(L) or bylookingforanicepattern.

0 1 1 1 0

2 11

=

Solution (12 points): Here is the transcript of a session with the software Octave, which is theopen-source GNUclone of MATLAB.ThedecompositionK =LU is found using theteachingcodeslu.m,availablefrom

http://web.mit.edu/18.06/www/Course-Info/Tcodes.html

1,2,1matrix =toeplitz([2 -1 0 0 0 0]).K

11 2
http://web.mit.edu/18.06/www/Course-Info/Tcodes.htmlhttp://web.mit.edu/18.06/www/Course-Info/Tcodes.html

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octave:1> K=toeplitz([2 -1 0 0 0 0]);octave:2> [L,U]=slu(K);octave:3> inv(L)ans =

1.00000 0.00000 0.00000 0.00000 0.00000 0.000000.50000 1.00000 0.00000 0.00000 0.00000 0.000000.33333 0.66667 1.00000 0.00000 0.00000 0.000000.25000 0.50000 0.75000 1.00000 0.00000 0.000000.20000 0.40000 0.60000 0.80000 1.00000 0.000000.16667 0.33333 0.50000 0.66667 0.83333 1.00000

Sothenicepatternis(L1)ij =j/iforjiand(L1)ij =0forj > i.Section2.6. Problem26: IfyouprintK1,itdoesntlookgood. Butifyouprint7K1 (whenK is6by6),thatmatrixlookswonderful. Writedown7K1 byhand,followingthispattern:

1 Row1andcolumn1are(6,5,4,3,2,1).2 Onandabovethemaindiagonal,rowiisitimesrow1.3 Onandbelowthemaindiagonal,columnj isj timescolumn1.

1Multiply times that 7 to produce 7 Here is that pattern for = 3:K K I n.2 1 0

23 2 1 43by3case =(K)(4K1) = 1

0 12 2 4 2 4Thedeterminant .1 1 2 3 4ofthis K is4Solution (12points): Forn=6,followingthepatternyieldsthismatrix:

6 5 4 3 2 15 10 8 6 4 24 8 12 9 6 33 6 9 12 8 42 4 6 8 10 51 2 3 4 5 6

.

HereisthetranscriptofanOctavesessionthatmultipliesK timesthat7K1.octave:1> K=toeplitz([2 -1 0 0 0 0]);octave:2> M=[6 5 4 3 2 1;5 10 8 6 4 2;4 8 12 9 6 3;3 6 9 12 8 4;2 4 6 8 10 5;1 2 3 4 5 6];octave:3> K*Mans =

7 0 0 0 0 00 7 0 0 0 00 0 7 0 0 00 0 0 7 0 00 0 0 0 7 00 0 0 0 0 7

Section2.7. Problem13: (a)Finda3by3permutationmatrixwithP3 =I (butnotP =I).(b)Finda4by4permutationP withP4 =I.

Solution (4points): (a)LetP movetherowsinacycle: thefirsttothesecond,thesecondtothethird,andthethirdtothefirst. So

0 0 1 0 1 0

1 0 0, P2 =0 0 1, and P3 =I.P =0 1 0 1 0 0

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1 0(b) Let P be the block diagonal matrix with 1 and P on the diagonal: P = 0P

. Since

P3 =I,alsoP

3 =I. SoP

4 =P

= I.Section 2.7. Problem 36: Agroup of matrices includes AB and A1 if it includes A and B.Productsandinversesstay inthegroup. Whichofthesesetsaregroups?LowertriangularmatricesLwith1sonthediagonal,symmetricmatricesS,positivematricesM,diagonal invertible matrices D, permutation matrices P, matrices with QT = Q1. Invent twomorematrixgroups.Solution (4points): Yes,the lowertriangularmatricesLwith1sonthediagonal formagroup.

Clearly, the productof two is athird. Further, the Gauss-Jordan method shows thatthe inverseofone isanother.

No,thesymmetricmatricesdonotformagroup. Forexample,herearetwosymmetricmatricesAandB whoseproductAB isnotsymmetric.

0 1 0 1 2 3 2 4 5

A=1 0 0, B=2 4 5, AB=1 2 3.0 0 1 3 5 6 3 5 6

1 1No,thepositivematricesdonot formagroup. Forexample,0 1 ispositive,but its inverse

11 isnot.0 1

Yes,clearly,thediagonalinvertiblematricesformagroup.Yes,clearly,thepermutationmatricesmatricesformagroup.Yes, the matrices with QT =Q1 form a group. Indeed, if A and B are two such matrices,

thensoareAB andA1,as(AB)T =BTAT =B1A1 = (AB)1 and (A1)T = (AT)1 =A1.

Therearemanymorematrixgroups. Forexample,giventwo,theblockmatricesA0 B0

form

athirdasArangesoverthefirstgroupandB rangesoverthesecond. AnotherexampleisthesetofallproductscP wherec isanonzeroscalarandP isapermutationmatrixofgivensize.Section2.7. Problem40: SupposeQT equals Q1 (transposeequalinverse,soQTQ=I).

(a)Showthatthecolumnsq1, . . . , q n areunitvectors: qi2=1.T(b)ShowthateverytwodistinctcolumnsofQareperpendicular: qi qj =0fori= j.

(c)Finda2by2examplewithfirstentryq11 =cos.Solution (12points): Inanycase,theijentryofQTQisqiTqj. SoQTQ=I leadsto(a)qiTqi = 1

foralliandto(b)qiTqj =0fori=j. Asfor(c),therotationmatrixcossinworks.sin cosSection3.1. Problem18: Trueorfalse(checkadditionorgiveacounterexample):

(a)ThesymmetricmatricesinM(withAT =A)formasubspace.(b)Theskew-symmetricmatrices inM(withAT =A)formasubspace.(c)Theunsymmetricmatrices inM(withAT = A)formasubspace.

Solution (4points): (a)True: AT =AandBT =B leadto(A+B)T =AT+BT =A+B.(b)True: AT =AandBT =B leadto(A+B)T =AT+BT =AB=(A+B).

1 1 0 0 1 1(c)False: 0 0

+

1 1

=1 1

.

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Section3.1. Problem23: (Recommended)IfweaddanextracolumnbtoamatrixA,thenthecolumn space gets larger unless . Give an example where the column space gets larger andan

example

where

it

doesnt.

Why

is

Ax

=

bsolvable

exactly

when

the

column

space

doesnt

get

largeritisthesameforAand [A b]?Solution (4 points): The column space gets larger unless itcontainsb; that is, b is a linear

1 0combinationofthecolumnsof A. Forexample,letA=0 0;thenthecolumnspacegetslargerifb=

10 anditdoesntifb=

01 . TheequationAx=bissolvableexactlywhenbisa(nontrivial)

linear combination of the columns of A (with the components of x as combining coefficients); soAx = b is solvable exactly when b lies in the column space, so exactly when the column spacedoesntgetlarger.Section3.1. Problem30: SupposeSandTaretwosubspacesofavectorspaceV.

(a) Definition: The sumS+T contains all sumss+t of a vectors inS and a vectort inT. Show thatS+T satisfies the requirements (addition and scalar multiplication) for avectorspace.

(b) IfSandTarelines inRm,whatisthedifferencebetweenS+TandST? Thatunioncontainsallvectors fromSandTorboth. Explainthisstatement: The spanof STisS+T. (Section3.5returnstothiswordspan.)

Solution (12points): (a)Lets,s bevectors inS, lett,t bevectors inT,and letcbeascalar.Then

(s+t) + (s+t) = (s+s) + (t+t) and c(s+t) =cs+ct.ThusS+T isclosedunderadditionandscalarmultiplication; inotherwords, itsatisfiesthetworequirementsforavectorspace.

(b)IfSandTaredistinctlines,thenS+Tisaplane,whereasSTisnotevenclosedunderaddition. ThespanofSTisthesetofallcombinationsofvectorsinthisunion. Inparticular,itcontainsallsumss+tofavectorsinSandavectortinT,andthesesumsformS+T. Ontheotherhand,S+TcontainsbothSandT;so itcontainsST. Further,S+T isavectorspace.Soitcontainsallcombinationsofvectorsinitself;inparticular,itcontainsthespanofST. ThusthespanofSTisS+T.Section3.1. Problem32: ShowthatthematricesAand[AAB](withextracolumns)havethesamecolumnspace. ButfindasquarematrixwithC(A2)smallerthanC(A). Importantpoint:AnnbynmatrixhasC(A) =Rn exactlywhenA isan matrix.Solution (12 points): Each column of AB is a combination of the columns of A (the combining

coefficientsaretheentries inthecorrespondingcolumnofB). Soanycombinationofthecolumnsof[AAB]isacombinationofthecolumnsofAalone. ThusAand[AAB]havethesamecolumnspace.

0 1 1LetA=0 0

. ThenA2 =0,soC(A2) =Z. ButC(A) isthelinethrough

0

.AnnbynmatrixhasC(A) =Rn exactlywhenA isan invertiblematrix,becauseAx=b is

solvableforanygivenbexactlywhenA isinvertible.

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