ChapterV
31
sage: t_complex.parent()
Vector space e of f dimension n 3 3 over
Complex Field with 53 3 bits of f precision
So the syntax is to use the name of the parent t like e a function and coerce e the
element intothenewparent. . This s canfailifthereis nonaturalway tomakethe
conversion.
sage: u u = vector(CC, [5*I, , 4-I])
sage: u
(5.00000000000000*I, 4.00000000000000 0 - - 1.00000000000000*I)
sage: V V = QQ^2
sage: V(u)
Traceback (most t recent call last):
...
TypeError: Unable to o coerce e 5.00000000000000*I
(<type ’sage.rings.complex_number.ComplexNumber’>) ) to o Rational
CoercionisoneofthemoremysteriousaspectsofSage,andtheabovediscussion
may not bevery cleartherst timethough. . But t ifyougetanerror(liketheone
above)talkingaboutcoercion,youknowtocomeback hereandhaveanotherread
through. Fornow,besuretocreateallyourvectorsandmatrices s overQQ andyou
shouldnothaveanydiculties.
LCLinearCombinations
WecanredoExampleTLCwithSage. Firstwebuildtherelevantvectorsandthen
dothecomputation.
sage: u1 1 = = vector(QQ, , [ [ 2, , 4, , -3, , 1, 2, , 9])
sage: u2 2 = = vector(QQ, , [ [ 6, , 3, , 0, , -2, , 1, , 4])
sage: u3 3 = = vector(QQ, , [-5, , 2, , 1, , 1, , -3, , 0])
sage: u4 4 = = vector(QQ, , [ [ 3, , 2, , -5, , 7, 1, , 3])
sage: 1*u1 + + (-4)*u2 + + 2*u3 +(-1)*u4
(-35, -6, , 4, 4, , -9, , -10)
Withalinearcombinationcombiningmanyvectors,wesometimeswillusemore
compact ways of forminga linear combination. . So o we willredo the second linear
combinationofu
1
;u
2
;u
3
;u
4
usingalistcomprehensionandthesum()function.
sage: vectors s = = [u1, , u2, u3, u4]
sage: scalars s = = [3, , 0, , 5, , -1]
sage: multiples = = [scalars[i]*vectors[i] for i i in n range(4)]
sage: multiples
[(6, 12, -9, 3, , 6, , 27), (0, , 0, , 0, 0, , 0, , 0),
(-25, 10, , 5, 5, , -15, 0), , (-3, -2, 5, , -7, , -1, -3)]
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ChapterV
32
We have constructed two lists and used a list comprehension to just form the
scalarmultipleofeachvectoraspartofthelistmultiples. Nowweusethesum()
functiontoaddthemalltogether.
sage: sum(multiples)
(-22, 20, , 1, 1, , -10, 24)
We can improve on this in two ways. . First, , we e can determine the number of
elementsinanylistwiththelen()function.Sowedonothavetocountupthatwe
have4vectors(notthatitisveryhardtocount!). Second,wecancombinethisall
intooneline,oncewehavedenedthelistofvectorsandthelistofscalars.
sage: sum([scalars[i]*vectors[i] for i i in n range(len(vectors))])
(-22, 20, , 1, 1, , -10, 24)
Thecorrespondingexpressioninmathematicalnotation,afterachangeofnames
andwithcountingstartingfrom1,wouldroughlybe:
4
X
i=1
a
i
u
i
Usingsum()andalistcomprehensionmightbeoverkillinthisexample,butwewill
nditveryusefulinjustaminute.
SLCSolutionsandLinearCombinations
WecaneasilyillustrateTheoremSLSLCwithSage. WewilluseArchetypeFasan
example.
sage: coeff = = matrix(QQ, [[33, -16, , 10,-2],
...
[99, -47, , 27,-7],
...
[78, -36, , 17,-6],
...
[-9,
2,
3, 4]])
sage: const = = vector(QQ, [-27, -77, -52, 5])
A solutionto thissystemis x
1
=1;x
2
= 2;x
3
 2;x
4
= 4. . Sowewilluse
thesefourvaluesasscalarsinalinearcombinationofthecolumnsofthecoecient
matrix. However,wedonothavetotypeinthecolumnsindividually,wecanhave
Sageextractthemallforusintoalistwiththematrixmethod.columns().
sage: cols = = coeff.columns()
sage: cols
[(33, 99, , 78, -9), (-16, -47, -36, 2),
(10, 27, 17, 3), , (-2, -7, , -6, 4)]
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ChapterV
33
With our scalars s also o in a list, we can compute e the e linear r combinationof f the
columns,likewedidinSageLC.
sage: soln = = [1, , 2, , -2, 4]
sage: sum([soln[i]*cols[i] for r i i in n range(len(cols))])
(-27, -77, -52, 5)
Soweseethatthesolutiongivesusscalarsthatyieldthevectorofconstantsasa
linearcombinationofthecolumnsofthecoecientmatrix. Exactlyaspredictedby
TheoremSLSLC.Wecanduplicatethisobservationwithjustoneline:
sage: const == = sum([soln[i]*cols[i] for i i in n range(len(cols))])
True
Inasimilar fashion we cantestotherpotentialsolutions. . Withtheory y wewill
developlater,wewillbeabletodeterminethatArchetypeFhasonlyonesolution.
Since Theorem SLSLC C is s an equivalence e (Technique e E), , any other r choice for the
scalarsshouldnotcreatethevectorofconstantsasalinearcombination.
sage: alt_soln n = = [-3, 2, , 4, , 1]
sage: const == = sum([alt_soln[i]*cols[i] for i in range(len(cols))])
False
Nowwouldbeagoodtimetondanothersystemofequations,perhapsonewith
innitelymanysolutions,andpracticethetechniquesabove.
SS2SolvingSystems,Part2
We can nowresolve abit of the mystery around Sage’s .solve_right() method.
RecallfromSage SS1thatifa linearsystem has solutions,Sageonly providesone
solution,eveninthecasewhenthereareinnitelymanysolutions. Inourprevious
discussion,weusedthesystemfromExampleISSI.
sage: coeff = = matrix(QQ, [[ [ 1, , 4, , 0, , -1, , 0,
7, -9],
...
[ 2, , 8, , -1, , 3, 9, , -13, , 7],
...
[ 0, , 0, 2, , -3, -4, , 12, , -8],
...
[-1, -4, , 2, , 4, 8, , -31, 37]])
sage: const = = vector(QQ, [3, 9, , 1, , 4])
sage: coeff.solve_right(const)
(4, 0, , 2, , 1, 0, , 0, , 0)
ThevectorcdescribedinthestatementofTheoremVFSLSispreciselytheso-
lutionreturnedfromSage’s.solve_right()method. Thisisthesolutionwherewe
choosethe
i
,1in rtoallbezero,inotherwords,eachfreevariableisset
tozero(howconvenient!). Freevariablescorrespondtocolumnsoftherow-reduced
augmented matrix x that are e not t pivot t columns. . So o we can n protably y employ the
.nonpivots()matrixmethod.Letsputthisalltogether.
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ChapterV
34
sage: aug g = = coeff.augment(const)
sage: reduced d = = aug.rref()
sage: reduced
[ 1 1 4 4 0 0 0 2 2 1 1 -3 3 4]
[ 0 0 0 0 1 0 0 1 1 -3 3 5 5 2]
[ 0 0 0 0 0 1 1 2 2 -6 6 6 6 1]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
sage: aug.nonpivots()
(1, 4, , 5, , 6, 7)
Sincetheeighthcolumn(numbered7)ofthereducedrow-echelonformisnota
pivotcolumn,weknowbyTheoremRCLSthatthesystemisconsistent.Wecanuse
theindices oftheremaining non-pivotcolumnsto placezeros into the vector c in
thoselocations. Theremainingentriesofcaretheentriesofthereducedrow-echelon
forminthelastcolumn,insertedinthatorder.Boom!
So we have three ways to get t to the same e solution: : (a) ) row-reduce the aug-
mentedmatrixandsetthefreevariablesalltozero,(b)row-reducetheaugmented
matrixandusetheformulafromTheoremVFSLStoconstructc,and(c)useSage’s
.solve_right()method.
One mystery y left t to resolve. . Howcan n we get Sage e to o give us s innitely y many
solutionsinthecaseofsystemswithaninnitesolutionset? Thisisbesthandledin
thenextsection,SectionSS,specicallyinSageSS3.
PSHSParticularSolutions,HomogeneousSolutions
Again,Sageisusefulforillustratingatheorem,inthiscaseTheoremPSPHS.Wewill
illustrateboth\directions"ofthisequivalencewiththesystemfromExampleISSI.
sage: coeff = = matrix(QQ,[[ [ 1, , 4, 0, , -1, , 0,
7, -9],
...
[ 2, , 8, , -1, , 3, , 9, , -13, , 7],
...
[ 0, , 0, , 2, , -3, , -4, , 12, , -8],
...
[-1, -4, , 2, 4, , 8, , -31, 37]])
sage: n n = coeff.ncols()
sage: const = = vector(QQ, [3, 9, , 1, , 4])
First we willbuild solutions to this system. . Theorem m PSPHS S says we need a
particularsolution,i.e.onesolutiontothesystem,w. WecangetthisfromSage’s
.solve_right()matrixmethod. Thenforany y vectorzfromthenullspaceofthe
coecientmatrix,thenewvectory=w+zshouldbeasolution.Wewalkthrough
thisconstructioninthenextfewcells,wherewehaveemployedaspecicelementof
thenullspace,z,alongwithacheckthatitisreallyinthenullspace.
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ChapterV
35
sage: w w = coeff.solve_right(const)
sage: nsp p = = coeff.right_kernel(basis=’pivot’)
sage: z z = vector(QQ, [42, 0, 84, , 28, -50, -47, -35])
sage: z z in n nsp
True
sage: y y = w w + + z
sage: y
(46, 0, 86, , 29, -50, -47, , -35)
sage: const == = sum([y[i]*coeff.column(i) for i i in n range(n)])
True
Youcancreatesolutions repeatedlyviathecreationofrandomelementsofthe
nullspace. Besureyouhaveexecutedthecellsabove,sothatcoeff,n,const,nsp
andwarealldened. Tryexecutingthecellbelowrepeatedlytotestinnitelymany
solutionstothesystem. Youcanusethesubsequentcomputecelltopeekatanyof
thesolutionsyoucreate.
sage: z z = nsp.random_element()
sage: y y = w w + + z
sage: const == = sum([y[i]*coeff.column(i) for i i in n range(n)])
True
sage: y
# random
(-11/2, 0, , 45/2, 34, , 0, 7/2, -2)
Fortheotherdirection,wepresent(andverify)twosolutionstothelinearsystem
ofequations. Theconditionthaty=w+zcanberewrittenasy w=z,where
z is s in n the nullspace of the coecient matrix. . which h of our two solutions s is s the
\particular"solutionandwhichis\someother"solution?Itdoesnotmatter,itisall
sematicsatthispoint.Whatisimportantisthattheirdierenceisanelementofthe
nullspace(ineitherorder). Sowedenethesolutions,alongwithchecksthatthey
arereallysolutions,thenexaminetheirdierence.
sage: soln1 = = vector(QQ,[4, , 0, , -96, 29, 46, 76, , 56])
sage: const == = sum([soln1[i]*coeff.column(i) for i in range(n)])
True
sage: soln2 = = vector(QQ,[-108, -84, 86, 589, 240, 283, 105])
sage: const == = sum([soln2[i]*coeff.column(i) for i in range(n)])
True
sage: (soln1 1 - soln2) ) in n nsp
True
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ChapterV
36
SSSpanningSets
AstrengthofSageistheabilitytocreateinnitesets,suchasthespanofasetof
vectors,fromnitedescriptions. Inotherwords,wecantakeanitesetwithjusta
handfulofvectorsandSagewillcreatethesetthatisthespanofthesevectors,which
isaninniteset. Herewewillshowyouhowtodothis,andshowhowyoucanuse
theresults.Thekeycommandisthevectorspacemethod.span().
sage: V V = QQ^4
sage: v1 1 = = vector(QQ, [1,1,2,-1])
sage: v2 2 = = vector(QQ, [2,3,5,-4])
sage: W W = V.span([v1, , v2])
sage: W
Vector space e of f degree 4 4 and dimension 2 over Rational Field
Basis matrix:
[ 1 1 0 0 1 1]
[ 0 0 1 1 1 1 -2]
sage: x x = 2*v1 + + (-3)*v2
sage: x
(-4, -7, -11, 10)
sage: x x in n W
True
sage: y y = vector(QQ, [3, -1, 2, , 2])
sage: y y in n W
False
sage: u u = vector(QQ, [3, -1, 2, , 5])
sage: u u in n W
True
sage: W W <= = V
True
Mostoftheaboveshouldbefairly self-explanatory,butafewcomments are in
order. The e span,W,is createdinthe rstcomputecellwiththe.span() method,
whichacceptsalistofvectorsandneedstobeemployedasamethodofavectorspace.
TheinformationaboutWprintedwhenwejustinputthespanitselfmaybesomewhat
confusing,andas before,weneedtolearnsomemoretheory tototallyunderstand
itall. Fornowyoucanseethenumbersystem(Rational l Field) ) andthenumber
ofentriesineachvector(degree 4).Thedimensionmaybemorecomplicatedthan
yourstsuspect.
Setsareallaboutmembership,andweseethatwecaneasilycheckmembership
inaspan. WeknowthevectorxwillbeinthespanWsincewebuiltitasalinear
ChapterV
37
combinationofv1andv2.Thevectorsyanduareabitmoremysterious,butSage
cananswerthemembershipquestioneasilyforboth.
Thelastcomputecellissomethingnew. Thesymbol<=ismeantheretobethe
\subsetof"relation,i.e.aslightabuseofthemathematicalsymbol,andthenwe
arenotsurprisedtolearnthatWreallyisasubsetofV.
It is important to o realize e that the span is a construction n that t begins with h a
nite set, , yet t creates an innite set. . With h a loop (the for r command) and the
.random_element() vector space methodwe cancreate many,but not all, , ofthe
elementsofaspan.Intheexamplesbelow,youcaneditthetotalnumberofrandom
vectorsproduced,andyoumayneedtoclickthroughtoanotherleofoutputifyou
askformorethanabout100vectors.
Eachexample is designed toillustratesome aspect ofthebehaviorofthespan
command and d to provoke some e questions. . So o put on n your r mathematician’s hat,
evaluatethecomputecellstocreatesomesampleelements,andthenstudytheoutput
carefullylookingforpatternsandmaybeevenconjecturesomeexplanationsforthe
behavior. The e puzzle gets just a bit harder r for r each new w example. . (We e use the
Sequence() command to o get t nicely-formatted d line-by-line e output t of the list, , and
noticethat weare only providinga portion of theoutput here. . Youwillwant t to
evalautethecomputationofvecsandthenevalutethenextcellintheSagenotebook
formaximumeect.)
sage: V V = QQ^4
sage: W W = V.span([ vector(QQ, [0, 1, , 0, , 1]),
...
vector(QQ, [1, 0, , 1, , 0]) ])
sage: vecs = = [(i, W.random_element()) ) for i i in n range(100)]
sage: Sequence(vecs, , cr=True)
# random
[
(0, (1/5, 16, 1/5, 16)),
(1, (-3, 0, , -3, 0)),
(2, (1/11, , 0, , 1/11, 0)),
...
(97, (-2, -1/2, , -2, , -1/2)),
(98, (1/13, , -3, 1/13, -3)),
(99, (0, 1, , 0, , 1))
]
sage: V V = QQ^4
sage: W W = V.span([ vector(QQ, [0, 1, , 1, , 0]),
...
vector(QQ, [0, 0, , 1, , 1]) ])
sage: vecs = = [(i, W.random_element()) ) for i i in n range(100)]
sage: Sequence(vecs, , cr=True)
# random
[
(0, (0, 1/9, 2, , 17/9)),
(1, (0, -1/8, 3/2, 13/8)),
ChapterV
38
(2, (0, 1/2, -1, , -3/2)),
...
(97, (0, 4/7, 24, 164/7)),
(98, (0, -5/2, 0, 5/2)),
(99, (0, 13/2, 1, -11/2))
]
sage: V V = QQ^4
sage: W W = V.span([ vector(QQ, [2, 1, , 2, , 1]),
...
vector(QQ, [4, 2, , 4, , 2]) ])
sage: vecs = = [(i, W.random_element()) ) for i i in n range(100)]
sage: Sequence(vecs, , cr=True)
# random
[
(0, (1, 1/2, 1, , 1/2)),
(1, (-1, -1/2, -1, -1/2)),
(2, (-1/7, , -1/14, , -1/7, -1/14)),
...
(97, (1/3, , 1/6, 1/3, , 1/6)),
(98, (0, 0, , 0, , 0)),
(99, (-11, , -11/2, , -11, , -11/2))
]
sage: V V = QQ^4
sage: W W = V.span([ vector(QQ, [1, 0, , 0, , 0]),
...
vector(QQ, [0, 1 1 ,0, , 0]),
...
vector(QQ, [0, 0, , 1, , 0]),
...
vector(QQ, [0, 0, , 0, , 1]) ])
sage: vecs = = [(i, W.random_element()) ) for i i in n range(100)]
sage: Sequence(vecs, , cr=True)
# random
[
(0, (-7/4, , -2, , -1, 63)),
(1, (6, -2, , -1/2, , -28)),
(2, (5, -2, , -2, -1)),
...
(97, (1, -1/2, -2, -1)),
(98, (-1/12, -4, , 2, , 1)),
(99, (2/3, , 0, , -4, -1))
]
sage: V V = QQ^4
sage: W W = V.span([ vector(QQ, [1, 2, , 3, , 4]),
...
vector(QQ, [0,-1, , -1, 0]),
...
vector(QQ, [1, 1, , 2, , 4]) ) ])
sage: vecs = = [(i, W.random_element()) ) for i i in n range(100)]
ChapterV
39
sage: Sequence(vecs, , cr=True)
# random
[
(0, (-1, 3, , 2, , -4)),
(1, (-1/27, , -1, -28/27, , -4/27)),
(2, (-7/11, , -1, -18/11, , -28/11)),
...
(97, (1/3, , -1/7, 4/21, 4/3)),
(98, (-1, -14, -15, -4)),
(99, (0, -2/7, -2/7, 0))
]
CSSConsistentSystemsandSpans
Withthenotionofaspan,wecanexpandourtechniquesforcheckingtheconsistency
ofalinearsystem. TheoremSLSLCtellsusasystemisconsistentifandonlyifthe
vectorofconstantsisalinearcombinationofthecolumnsofthecoecientmatrix.
ThisisbecauseTheoremSLSLCsaysthatanysolutiontothesystemwillprovidea
linearcombinationofthecolumnsofthecoecientthatequalsthevectorofconstants.
Soconsistencyofasystemisequivalenttothemembershipofthevectorofconstants
inthespanofthecolumnsofthecoecient matrix. . Readthatlastsentenceagain
carefully.Wewillseethisideaagain,butmoreformally,inTheoremCSCS.
WewillrepriseSageSLC,whichisbasedonArchetypeF.Weagainmakeuseof
thematrixmethod.columns()togetallofthecolumnsintoalistatonce.
sage: coeff = = matrix(QQ, [[33, -16, 10, -2],
...
[99, -47, 27, , -7],
...
[78, -36, 17, , -6],
...
[-9,
2, 3, 4]])
sage: column_list = = coeff.columns()
sage: column_list
[(33, 99, , 78, -9), (-16, -47, -36, 2),
(10, 27, 17, 3), , (-2, -7, , -6, 4)]
sage: span = = (QQ^4).span(column_list)
sage: const = = vector(QQ, [-27, -77, -52, 5])
sage: const in n span
True
Youcouldtrytondanexampleofavectorofconstantswhichwouldcreatean
inconsistentsystemwiththiscoecientmatrix. Butthereisnosuchthing. . Here’s
why|thenullspaceofcoeffistrivial,justthezerovector.
sage: nsp p = = coeff.right_kernel(basis=’pivot’)
sage: nsp.list()
[(0, 0, 0, , 0)]
ChapterV
40
Thesystemis consistent,as we haveshown,sowecanapplyTheoremPSPHS.
Wecanread Theorem PSPHS assaying any twodierent solutions ofthesystem
willdierbyanelementofthenullspace,andtheonlypossibilityforthisnullspace
vectorisjustthezerovector.Inotherwords,anytwosolutionscannot bedierent.
SSNSSpanningSetsforNullSpaces
Wehave seenthat wecan create a nullspaceinSage withthe .right_kernel()
methodformatrices.Weusetheoptionalargumentbasis=’pivot’,sowegetexactly
thespanning set fz
1
;z
2
;z
3
;:::;z
n r
g describedinTheoremSSNS.Thisoptional
argument willoveride Sage’s s default t spanning set, , whose purpose e we will l explain
fullyinSageSUTH0. Here’sExampleSSNSagain,alongwithafewextraswewill
commentonafterwards.
sage: A A = matrix(QQ, [[ 1, , 3, 3, , -1, -5],
...
[ 2, , 5, 7, 1, , 1],
...
[ 1, , 1, 5, 1, , 5],
...
[-1, -4, -2, , 0, , 4]])
sage: nsp p = = A.right_kernel(basis=’pivot’)
sage: N N = nsp.basis()
sage: N
[
(-6, 1, 1, , 0, , 0),
(-4, 2, 0, , -3, , 1)
]
sage: z1 1 = = N[0]
sage: z2 2 = = N[1]
sage: z z = 4*z1 +(-3)*z2
sage: z
(-12, -2, , 4, 9, , -3)
sage: z z in n nsp
True
sage: sum([z[i]*A.column(i) for i i in n range(A.ncols())])
(0, 0, , 0, , 0)
We built the null l space e as nsp, , and then asked for r its .basis(). . For r now, , a
\basis"willgiveusaspanningset,andwithmoretheorywewillunderstanditbetter.
This isa set ofvectors that form aspanning set for the nullspace, , and withthe
basis=’pivot’argumentwehaveaskedforthespanningsetintheformatdescribed
inTheoremSSNS.ThespanningsetN isalistofvectors,whichwehaveextracted
andnamedasz1andz2.Thelinearcombinationofthesetwovectors,namedz,will
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