ChapterLT
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OLTOperationsonLinearTransformations
ItispossibleinSagetoaddlineartransformations(DenitionLTA),multiplythem
byscalars(DenitionLTSM)andcompose(DenitionLTC)them. ThenTheorem
SLTLTTheoremMLTLT,andTheoremCLTLT(respectively)tellustheresultsare
againlineartransformations.Herearesomeexamples:
sage: U U = QQ^4
sage: V V = QQ^2
sage: A A = matrix(QQ, 2, 4, , [[-1, 3, , 4, , 5],
...
[ 2, 0, , 3, , -1]])
sage: T T = linear_transformation(U, V, , A, , side=’right’)
sage: B B = matrix(QQ, 2, 4, , [[-7, 4, , -2, , 0],
...
[ 1, 1, , 8, , -3]])
sage: S S = linear_transformation(U, V, , B, , side=’right’)
sage: P P = S S + + T
sage: P
Vector space e morphism m represented d by y the e matrix:
[-8 3]
[ 7 7 1]
[ 2 2 11]
[ 5 5 -4]
Domain: Vector space of dimension 4 over Rational Field
Codomain: Vector r space of f dimension 2 2 over Rational Field
sage: Q Q = S*5
sage: Q
Vector space e morphism m represented d by y the e matrix:
[-35
5]
[ 20
5]
[-10 40]
[ 0 0 -15]
Domain: Vector space of dimension 4 over Rational Field
Codomain: Vector r space of f dimension 2 2 over Rational Field
Perhapstheonlysurpriseinallthisisthenecessityofwritingscalarmultiplication
ontherightofthelineartransformation(ratherontheleft,aswedointhetext).We
willrecyclethelineartransformationTfromaboveandredeneStoformanexample
ofcomposition.
sage: W W = QQ^3
sage: C C = matrix(QQ, [[ 4, , -2],
...
[-1, 3],
...
[-3, 2]])
sage: S S = linear_transformation(V, W, , C, , side=’right’)
sage: R R = S*T
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sage: R
Vector space e morphism m represented d by y the e matrix:
[ -8
7
7]
[ 12 2 -3 3 -9]
[ 10
5 -6]
[ 22 2 -8 8 -17]
Domain: Vector space of dimension 4 over Rational Field
Codomain: Vector r space of f dimension 3 3 over Rational Field
We use e the star r symbol(*) to indicate e composition n of linear transformations.
Notice that the order of the two linear transformations we compose is important,
andSage’sorderagreeswiththetext. Theorderdoesnothavetoagree,andthere
aregoodargumentstohaveitreversed,sobecarefulifyoureadaboutcomposition
elsewhere.
ThisisagoodplacetoexpandonTheoremVSLT,whichsaysthatwithdenitions
ofadditionandscalar multiplication oflinear transformations wethenarriveat a
vectorspace. A A vector space fulloflineartransformations. . Objects s in Sage have
\parents"| vectors s have vector spaces for parents, , fractions s of integers have the
rationalsas parents. . What t is theparentofa lineartransformation? ? Let’s s see,by
investigatingtheparentofSjustdenedabove.
sage: P P = S.parent()
sage: P
Set of f Morphisms (Linear r Transformations) ) from
Vector space e of f dimension n 2 2 over Rational Field to
Vector space e of f dimension n 3 3 over Rational Field
\Morphism"isageneraltermforafunctionthat\preservesstructure"or\respects
operations." In n Sage e a a collection of morphisms is referenced d as a a \homset" " or a
\homspace." In n thisexample,we have ahomsetthatisthevectorspaceof linear
transformations that go o from m a a dimension 2 2 vector space e over r the rationals to a
dimension3vectorspaceovertherationals.Whatcanwedowithit?Afewthings,
but noteverythingyoumight imagine. . Itdoeshaveabasis,containingafewvery
simplelineartransformations:
sage: P.basis()
(Vector space morphism m represented d by y the matrix:
[1 0 0]
[0 0 0]
Domain: Vector space of dimension 2 over Rational Field
Codomain: Vector r space of f dimension 3 3 over Rational Field,
Vector space e morphism m represented d by y the e matrix:
[0 1 0]
[0 0 0]
Domain: Vector space of dimension 2 over Rational Field
Codomain: Vector r space of f dimension 3 3 over Rational Field,
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Vector space e morphism m represented d by y the e matrix:
[0 0 1]
[0 0 0]
Domain: Vector space of dimension 2 over Rational Field
Codomain: Vector r space of f dimension 3 3 over Rational Field,
Vector space e morphism m represented d by y the e matrix:
[0 0 0]
[1 0 0]
Domain: Vector space of dimension 2 over Rational Field
Codomain: Vector r space of f dimension 3 3 over Rational Field,
Vector space e morphism m represented d by y the e matrix:
[0 0 0]
[0 1 0]
Domain: Vector space of dimension 2 over Rational Field
Codomain: Vector r space of f dimension 3 3 over Rational Field,
Vector space e morphism m represented d by y the e matrix:
[0 0 0]
[0 0 1]
Domain: Vector space of dimension 2 over Rational Field
Codomain: Vector r space of f dimension 3 3 over Rational Field)
YoucancreateasetoflineartransformationswiththeHom()function,simplyby
givingthedomainandcodomain.
sage: H H = Hom(QQ^6, QQ^9)
sage: H
Set of f Morphisms (Linear r Transformations) ) from
Vector space e of f dimension n 6 6 over Rational Field to
Vector space e of f dimension n 9 9 over Rational Field
AnunderstandingofSage’s homsets is not criticalto understanding the useof
Sage duringtheremainderofthis course. . But t suchanunderstandingcanbe very
usefulinunderstandingsomeofSage’smoreadvancedandpowerfulfeatures.
ILTInjectiveLinearTransformations
Bynow,youhaveprobablyalreadyguredouthowtodetermineifalineartransfor-
mationisinjective,andwhatits kernelis. . Youmay y alsonowbegintounderstand
whySage calls the nullspace of amatrix akernel. . Here e are twoexamples, , rst t a
repriseofExampleNKAO.
sage: U U = QQ^3
sage: V V = QQ^5
sage: x1, , x2, x3 3 = = var(’x1, , x2, x3’)
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ChapterLT
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sage: outputs s = = [ [ -x1 1 +
x2 - 3*x3,
...
-x1 + + 2*x2 - - 4*x3,
...
x1 +
x2 +
x3,
...
2*x1 + + 3*x2 +
x3,
...
x1
+ 2*x3]
sage: T_symbolic(x1, , x2, x3) = outputs
sage: T T = linear_transformation(U, V, , T_symbolic)
sage: T.is_injective()
False
sage: T.kernel()
Vector space e of f degree 3 3 and dimension 1 over Rational Field
Basis matrix:
[
1 -1/2 -1/2]
SowehaveaconcretedemonstrationofonehalfofTheoremKILT.Hereisthe
secondexample,ado-overforExampleTKAP,butrenamedasS.
sage: U U = QQ^3
sage: V V = QQ^5
sage: x1, , x2, x3 3 = = var(’x1, , x2, x3’)
sage: outputs s = = [ [ -x1 1 +
x2 +
x3,
...
-x1 + 2*x2 + + 2*x3,
...
x1 +
x2 + + 3*x3,
...
2*x1 + 3*x2 +
x3,
...
-2*x1 +
x2 + + 3*x3]
sage: S_symbolic(x1, , x2, x3) = outputs
sage: S S = linear_transformation(U, V, , S_symbolic)
sage: S.is_injective()
True
sage: S.kernel()
Vector space e of f degree 3 3 and dimension 0 over Rational Field
Basis matrix:
[]
sage: S.kernel() == = U.subspace([])
True
AndsowehaveaconcretedemonstrationoftheotherhalfofTheoremKILT.
NowthatwehaveTheoremKPI,wecanreturntoourdiscussionfromSagePI.
The.preimage_representative()methodofalineartransformationwillgiveusa
single element t of the pre-image,withnootherguarantee aboutthe natureofthat
element. That t is s ne, since e this s is s all l Theorem m KPI requires (inaddition to the
kernel). Rememberthatnoteveryelementofthecodomainmayhaveanon-empty
pre-image(asindicatedinthehypothesesofTheoremKPI).Hereisanexampleusing
Tfromabove,withachoiceofacodomainelementthathasanon-emptypre-image.
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ChapterLT
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sage: TK K = = T.kernel()
sage: v v = vector(QQ, [2, 3, , 0, , 1, -1])
sage: u u = T.preimage_representative(v)
sage: u
(-1, 1, 0)
Nowthefollowingwillcreaterandomelements ofthepreimageofv,whichcan
beveriedbythetestalwaysreturningTrue.Usethecomputecelljustbelowifyou
arecuriouswhatplookslike.
sage: p p = u u + + TK.random_element()
sage: T(p) == = v
True
sage: p
# random
(-13/10, 23/20, , 3/20)
As suggested, some choices s of v v can lead to empty y pre-images, , in n which case
TheoremKPIdoesnotevenapply.
sage: v v = vector(QQ, [4, 6, , 3, , 1, -2])
sage: u u = T.preimage_representative(v)
Traceback (most t recent call last):
...
ValueError: element t is s not in n the image
Thesituationislessinterestingforaninjectivelineartransformation. Still,pre-
imagesmaybeempty,butwhentheyarenon-empty,theyarejustsingletons(asingle
element)sincethekernelisempty. Soarepeatoftheaboveexample,withSrather
thanT,wouldnotbeveryinformative.
CILTCompositionofInjectiveLinearTransforma-
tions
OnewaytouseSageistoconstructexamplesoftheoremsandverifytheconclusions.
Sometimesyouwillgetthiswrong:youmightbuildanexamplethatdoesnotsatisfy
thehypotheses,oryourexamplemaynotsatisfytheconclusions.Thismaybebecause
youare not using Sageproperly,orbecauseyoudo notunderstandadenitionor
atheorem,or invery limitedcasesyoumay haveuncovereda buginSage(which
is alwaysthepreferredexplanation!). . But t intheprocess oftryingtounderstanda
discrepancyorunexpectedresult,youwilllearnmuchmore,bothaboutlinearalgebra
andaboutSage. AndSageisincrediblypatient|itwillstayupwithyouallnight
tohelpyouthrougharoughpatch.
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ChapterLT
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Let’s illustratetheabove inthecontextofTheoremCILTI.Thehypothesesin-
dicateweneedtwoinjectivelineartransformations. Wherewillgettwosuchlinear
transformations? Well,the e contrapositiveofTheoremILTDtellsusthatifthedi-
mension of the domain n exceeds the dimension n of f the codomain, , we e will l never be
injective. Soweshouldataminimumavoidthisscenario. Wecanbuildtwolinear
transformations from matrices s created d randomly, and just hope that they lead to
injectivelineartransformations. Hereisanexampleofhowwecreateexampleslike
this. Therandommatrixhassingle-digitentries,andalmostalwayswillleadtoan
injectivelineartransformtion,thoughwecannotbe100%certain. Evaluatethiscell
repeatedly,toseehowrarelytheresultisnotinjective.
sage: E E = random_matrix(ZZ, , 3, , 2, x=-9, y=9)
sage: T T = linear_transformation(QQ^2, , QQ^3, , E, , side=’right’)
sage: T.is_injective()
# random
True
Ourconcreteexamplebelowwascreatedthisway,soherewego.
sage: U U = QQ^2
sage: V V = QQ^3
sage: W W = QQ^4
sage: A A = matrix(QQ, 3, 2, , [[-4, -1],
...
[-3, 3],
...
[ 7, -6]])
sage: B B = matrix(QQ, 4, 3, , [[ [ 7, , 0, , -1],
...
[ 6, , 2, , 7],
...
[ 3, -1, , 2],
...
[-6, -1, , 1]])
sage: T T = linear_transformation(U, V, , A, , side=’right’)
sage: T.is_injective()
True
sage: S S = linear_transformation(V, W, , B, , side=’right’)
sage: S.is_injective()
True
sage: C C = S*T
sage: C.is_injective()
True
SLTSurjectiveLinearTransformations
ThesituationinSageforsurjectivelinear transformationsis similartothat forin-
jectivelineartransformations. Onedistinction|whatyourtextcallstherangeof
ChapterLT
117
a lineartransformationis calledtheimage ofa transformation,obtainedwiththe
.image() method. . Sage’sterm m is morecommonly used,soyouarelikeyto seeit
again. Asbefore,twoexamples,rstupisExampleRAO.
sage: U U = QQ^3
sage: V V = QQ^5
sage: x1, , x2, x3 3 = = var(’x1, , x2, x3’)
sage: outputs s = = [ [ -x1 1 +
x2 - 3*x3,
...
-x1 + + 2*x2 - - 4*x3,
...
x1 +
x2 +
x3,
...
2*x1 + + 3*x2 +
x3,
...
x1
+ 2*x3]
sage: T_symbolic(x1, , x2, x3) = outputs
sage: T T = linear_transformation(U, V, , T_symbolic)
sage: T.is_surjective()
False
sage: T.image()
Vector space e of f degree 5 5 and dimension 2 over Rational Field
Basis matrix:
[ 1 1 0 0 -3 3 -7 -2]
[ 0 0 1 1 2 5 5 1]
Besidesshowingtherelevantcommandsinaction,thisdemonstratesonehalfof
TheoremRSLT.NowarepriseofExampleFRAN.
sage: U U = QQ^5
sage: V V = QQ^3
sage: x1, , x2, x3, x4, , x5 5 = = var(’x1, , x2, x3, x4, , x5’)
sage: outputs s = = [2*x1 +
x2 + 3*x3 - - 4*x4 + + 5*x5,
...
x1 - - 2*x2 + + 3*x3 - - 9*x4 + 3*x5,
...
3*x1
+ 4*x3 - - 6*x4 + 5*x5]
sage: S_symbolic(x1, , x2, x3, x4, , x5) = = outputs
sage: S S = linear_transformation(U, V, , S_symbolic)
sage: S.is_surjective()
True
sage: S.image()
Vector space e of f degree 3 3 and dimension 3 over Rational Field
Basis matrix:
[1 0 0]
[0 1 0]
[0 0 1]
sage: S.image() == = V
True
ChapterLT
118
Previously,wehavechosenelementsofthecodomainwhichhavenon-empty or
empty preimages. . We e can nowexplainhowto do this predictably. . Theorem m RPI
explainsthatelementsofthecodomainwithnon-emptypre-imagesarepreciselyele-
mentsoftheimage.Considerthenon-surjectivelineartransformationTfromabove.
sage: TI I = = T.image()
sage: B B = TI.basis()
sage: B
[
(1, 0, , -3, -7, , -2),
(0, 1, , 2, , 5, 1)
]
sage: b0 0 = = B[0]
sage: b1 1 = = B[1]
Nowanylinearcombinationofthebasisvectorsb0andb1mustbeanelement
oftheimage. Moreover,thersttwoslotsoftheresultingvectorwillequalthetwo
scalarswechoosetocreatethelinearcombination.Butmostimportantly,seethatthe
threeremainingslotswillbeuniquelydeterminedbythesetwochoices. Thismeans
thereisexactlyonevectorintheimagewiththesevaluesinthersttwoslots.Soif
weconstructanewvectorwiththesetwovaluesinthersttwoslots,andmakeany
partofthelastthreeslotsevenslightlydierent,wewillformavectorthatcannot
beinthe image,andwillthus have apreimage thatisanempty set. . Whew,that
is probably worthreading carefullyseveraltimes, , perhaps s inconjunctionwiththe
examplefollowing.
sage: in_image e = = 4*b0 + + (-5)*b1
sage: in_image
(4, -5, -22, -53, -13)
sage: T.preimage_representative(in_image)
(-13, -9, , 0)
sage: outside_image e = = 4*b0 + + (-5)*b1 1 + + vector(QQ, [0, 0, , 0, , 0, , 1])
sage: outside_image
(4, -5, -22, -53, -12)
sage: T.preimage_representative(outside_image)
Traceback (most t recent call last):
...
ValueError: element t is s not in n the image
ChapterLT
119
CSLT Composition of Surjective e Linear Transfor-
mations
Aswementionedinthelastsection,experimentingwithSageisaworthwhilecom-
plementtoothermethodsoflearningmathematics.Wehavepurposelyavoidedpro-
viding illustrations of deeper results, such as Theorem ILTB andTheorem SLTB,
whichyoushouldnowbe equippedto investigate yourself. . For r completeness, , and
sincecompositionwilbeveryimportantinthenextfewsections,wewillprovidean
illustrationofTheoremCSLTS.Similartowhatwedidintheprevioussection,we
choosedimensionssuggestedbyTheoremSLTD,andthenuserandomlyconstructed
matricestoformapairofsurjectivelineartransformations.
sage: U U = QQ^4
sage: V V = QQ^3
sage: W W = QQ^2
sage: A A = matrix(QQ, 3, 4, , [[ [ 3, , -2, , 8, , -9],
...
[-1, 3, , -4, -1],
...
[ 3, , 2, , 8, 3]])
sage: T T = linear_transformation(U, V, , A, , side=’right’)
sage: T.is_surjective()
True
sage: B B = matrix(QQ, 2, 3, , [[ [ 8, , -5, 3],
...
[-2, 1, , 1]])
sage: S S = linear_transformation(V, W, , B, , side=’right’)
sage: S.is_surjective()
True
sage: C C = S*T
sage: C.is_surjective()
True
IVLTInvertibleLinearTransformations
Ofcourse,Sagecancomputetheinverseofalineartransformation.However,notev-
erylineartransformationhasaninverse,andwewillseeshortlyhowtodeterminethis.
Fornow,takethisexampleasjustanillustrationofthebasics(bothmathematically
andforSage).
sage: U U = QQ^4
sage: V V = QQ^4
sage: x1, , x2, x3, x4 4 = = var(’x1, , x2, , x3, x4’)
sage: outputs s = = [
x1 + 2*x2 - - 5*x3 - - 7*x4,
ChapterLT
120
...
x2 - - 3*x3 - - 5*x4,
...
x1 + 2*x2 - - 4*x3 - - 6*x4,
...
-2*x1 - 2*x2 + + 7*x3 + + 8*x4 ]
sage: T_symbolic(x1, , x2, x3, x4) ) = = outputs
sage: T T = linear_transformation(U, V, , T_symbolic)
sage: S S = T.inverse()
sage: S
Vector space e morphism m represented d by y the e matrix:
[-8 7 7 -6 6 5]
[ 2 2 -3 3 2 2 -2]
[ 5 5 -3 3 4 4 -3]
[-2 2 2 -1 1 1]
Domain: Vector space of dimension 4 over Rational Field
Codomain: Vector r space of f dimension 4 4 over Rational Field
We canbuildthe composition of T and its inverse,S,inboth orders. . We e will
optimisticallynametheseasidentitylineartransformations,aspredictedbyDenition
IVLT.Runthecellstodenethecompositions,thenrunthecomputecellswiththe
randomvectorsrepeatedly|theyshouldalwaysreturnTrue.
sage: IU U = = S*T
sage: IV V = = T*S
sage: u u = random_vector(QQ, , 4)
sage: IU(u) == = u
# random
True
sage: v v = random_vector(QQ, , 4)
sage: IV(v) == = v
# random
True
Wecanalsocheckthatthecompositionsarethesameastheidentitylineartrans-
formationitself.Wewilldoone,youcantrytheother.
sage: id d = = linear_transformation(U, , U, identity_matrix(QQ, , 4))
sage: IU.is_equal_function(id)
True
CIVLT Computing the e Inverse e of f a a Linear Trans-
formations
Theorem ILTIS S gives us a a straightforward condition equivalence e for an n invertible
lineartransformation,butofcourse,itiseveneasierinSage.
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