ChapterR
131
sage: d0 0 = = vector(QQ, [3, , 4])
sage: d1 1 = = vector(QQ, [2, , 3])
sage: W W = (QQ^2).subspace_with_basis([d0, , d1])
sage: T T = linear_transformation(U, V, , T_symbolic)
sage: S S = linear_transformation(V, W, , S_symbolic)
sage: (S*T).matrix(’right’)
[-321 218 297]
[ 456 -310 -422]
sage: S.matrix(side=’right’)*T.matrix(side=’right’)
[-321 218 297]
[ 456 -310 -422]
Extracredit:whatchangesdoyouneedtomakeifyoudroppedtheside=’right’
optiononthesethreematrixrepresentations?
LTRLinearTransformationRestrictions
TheoremKNSIandTheoremRCSIhavetwoofthemostsubtleproofswehaveseen
sofar.Theconclusionthattwovectorspacesareisomorphicisestablishedbyactually
constructinganisomorphismbetweenthevectorspaces. Tobuildtheisomorphism,
webeginwithafamilarobject,avectorrepresentationlineartransformation,butthe
hardworkisshowingthatwecan\restrict"thedomainandcodomainofthisfunction
andstillarriveatalegitimate(bijective)lineartransformation.Inaneorttomake
theproofs more concrete,we willwalk through anontrivialexample for Theorem
KNSI,andyoumighttrytodothesameforTheoremRCSI.(Anunderstandingof
thissubsectionisnotneededfortheremainder|itssecondpurposeistodemonstrate
someofthepowerfultoolsSageprovides.)
Herearethepieces. Webuildalinear r transformationwithtwodierent repre-
sentations,onewithrespecttostandardbases,theotherwithrespecttoless-obvious
bases.
sage: x1, , x2, x3, x4, , x5 5 = = var(’x1, , x2, x3, x4, , x5’)
sage: outputs s = = [ [ x1 1 - - x2 - - 5*x3 +
x4 +
x5,
...
x1
- 2*x3 -
x4 -
x5,
...
- x2 2 - - 3*x3 + 2*x4 + + 2*x5,
...
-x1 + + x2 2 + + 5*x3 -
x4 -
x5]
sage: T_symbolic(x1, , x2, x3, x4, , x5) = = outputs
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ChapterR
132
sage: b0 0 = = vector(QQ, [-1, , 6, 5, , 5, 1])
sage: b1 1 = = vector(QQ, [-1, , 5, 4, , 4, 1])
sage: b2 2 = = vector(QQ, [-2, , 4, 3, , 2, 5])
sage: b3 3 = = vector(QQ, [ [ 1, , -1, , 0, , 1, , -5])
sage: b4 4 = = vector(QQ, [ [ 3, , -7, , -6, -5, -4])
sage: U U = (QQ^5).subspace_with_basis([b0, , b1, , b2, b3, b4])
sage: c0 0 = = vector(QQ, [1, , 1, 1, , -3])
sage: c1 1 = = vector(QQ, [-2, 3, , -6, -7])
sage: c2 2 = = vector(QQ, [0, , -1, 1, , 2])
sage: c3 3 = = vector(QQ, [-1, 3, , -4, -7])
sage: V V = (QQ^4).subspace_with_basis([c0, , c1, , c2, c3])
sage: T_plain n = = linear_transformation(QQ^5, , QQ^4, T_symbolic)
sage: T_fancy y = = linear_transformation(
U,
V, T_symbolic)
Nowwecomputethekernelofthelineartransformationusingthe\plain"version,
andthenullspaceofamatrixrepresentationcomingfromthe\fancy"version.
sage: K K = T_plain.kernel()
sage: K
Vector space e of f degree 5 5 and dimension 3 over Rational Field
Basis matrix:
[
1
0 2/7
0 3/7]
[
0
1 -1/7
0 2/7]
[
0
0
0
1
-1]
sage: MK K = = T_fancy.matrix(side=’right’).right_kernel()
sage: MK
Vector space e of f degree 5 5 and dimension 3 over Rational Field
Basis matrix:
[
1
0
0 -97/203 164/203]
[
0
1
0 -10/29
19/29]
[
0
0
1 129/203 100/203]
Soquiteobviously,thekernelofthelineartransformationisquitedierentlooking
fromthenullspaceofthematrixrepresentation. Thoughitisnoaccidentthatthey
have the same dimension. . Nowwe e build the necessary vector representation, , and
usetwoSagecommandsto \restrict"thefunctiontoasmallerdomain(thekernel
ofthelineartransformation)andasmallercodomain(thenullspaceofthematrix
representationrelativetononstandardbases).
sage: rhoB = = linear_transformation(U, , QQ^5, , (QQ^5).basis())
sage: rho_restrict t = rhoB.restrict_domain(K).restrict_codomain(MK)
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ChapterR
133
sage: rho_restrict
Vector space e morphism m represented d by y the e matrix:
[ 33/7 -37/7 7 -11/7]
[-13/7 22/7 7 -12/7]
[
-4
5
6]
Domain: Vector space of degree 5 5 and dimension 3 3 over Rational l Field
Basis matrix:
[
1
0 2/7
0 3/7]
[
0
1 -1/7
0 2/7]
[
0
0
0
1
-1]
Codomain: Vector r space of f degree e 5 5 and dimension 3 over Rational Field
Basis matrix:
[
1
0
0 -97/203 164/203]
[
0
1
0 -10/29
19/29]
[
0
0
1 129/203 100/203]
Therstsuccessisthattherestrictionwasevencreated.Sagewouldrecognizeif
theoriginallineartransformationevercarriedaninputfromtherestricteddomainto
anoutputthatwasnotcontainedintheproposedcodomain,andwouldhaveraised
anerrorinthatevent.Phew! Guaranteeingthissuccesswastherstbigstepinthe
proofofTheoremKNSI.Noticethatthematrixrepresentationoftherestrictionisa
33matrix,sincetherestrictionrunsbetweenadomainandcodomainthateachhave
dimension3. Thesetwovectorspaces(thedomainandcodomainoftherestriction)
havedimension 3butstillcontain vectors with5 entries intheir un-coordinatized
versions.
Thenexttwostepsoftheproofshowthattherestrictionisinjective(easyinthe
proof)andsurjective(hardintheproof). InSage,hereisthesecondsuccess,
sage: rho_restrict.is_injective()
True
sage: rho_restrict.is_surjective()
True
Veriedas a bijection, , rho_restrict t qualies as anisomorphism betweenthe
lineartransformationkernel,K,andthe matrixrepresentationnullspace,MK.Only
anexample,butstillverynice. Yourturn|canyoucreateavercationofTheorem
RCSI(forthisexample,orsomeothernontrivialexampleyoumightcreateyourself)?
NME9NonsingularMatrixEquivalences,Round9
Ournalfactaboutnonsingularmatricesexpressesthecorrespondencebetweenin-
vertiblematricesandinvertiblelineartransformations. AsaSagedemonstration,we
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ChapterR
134
willbeginwithaninvertiblelineartransformationandexaminetwomatrixrepresen-
tations. Wewillcreatethelineartransformationwithnonstandardbasesandthen
computeitsrepresentationrelativetostandardbases.
sage: x1, , x2, x3, x4 4 = = var(’x1, , x2, , x3, x4’)
sage: outputs s = = [ [ x1
- 2*x3 - - 4*x4,
...
x2 -
x3 - - 5*x4,
...
-x1 - - 2*x2 + 2*x3 + + 7*x4,
...
-
x2
+
x4]
sage: T_symbolic(x1, , x2, x3, x4) ) = = outputs
sage: b0 0 = = vector(QQ, [ [ 1, , -2, , -1, , 8])
sage: b1 1 = = vector(QQ, [ [ 0, , 1, 0, , -2])
sage: b2 2 = = vector(QQ, [-1, -2, , 2, , -5])
sage: b3 3 = = vector(QQ, [-1, -3, , 2, , -2])
sage: U U = (QQ^4).subspace_with_basis([b0, , b1, , b2, b3])
sage: c0 0 = = vector(QQ, [ [ 3, , -1, , 4, , -8])
sage: c1 1 = = vector(QQ, [ [ 1, , 0, 1, , -1])
sage: c2 2 = = vector(QQ, [ [ 0, , 2, , -1, , 6])
sage: c3 3 = = vector(QQ, [-1, , 2, , -2, , 8])
sage: V V = (QQ^4).subspace_with_basis([c0, , c1, , c2, c3])
sage: T T = linear_transformation(U, V, , T_symbolic)
sage: T
Vector space e morphism m represented d by y the e matrix:
[ 131 1 -56 6 -321 1 366]
[ -37
17
89 -102]
[ -61
25 153 3 -173]
[ -7
-1
24 -25]
Domain: Vector space of degree 4 4 and dimension 4 4 over Rational l Field
User basis s matrix:
[ 1 1 -2 2 -1 1 8]
[ 0 0 1 1 0 0 -2]
[-1 -2 2 2 2 -5]
[-1 -3 3 2 2 -2]
Codomain: Vector r space of f degree e 4 4 and dimension 4 over Rational Field
User basis s matrix:
[ 3 3 -1 1 4 4 -8]
[ 1 1 0 0 1 1 -1]
[ 0 0 2 2 -1 1 6]
[-1 2 2 -2 2 8]
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ChapterR
135
sage: T.is_invertible()
True
sage: T.matrix(side=’right’).is_invertible()
True
sage: (T^-1).matrix(side=’right’) ) == = (T.matrix(side=’right’))^-1
True
WecanconvertTtoanewrepresentationusingstandardbasesforQQ^4bycom-
putingimagesofthestandardbasis.
sage: images s = [T(u) for u u in n (QQ^4).basis()]
sage: T_standard = = linear_transformation(QQ^4, , QQ^4, images)
sage: T_standard
Vector space e morphism m represented d by y the e matrix:
[ 1 1 0 0 -1 1 0]
[ 0 0 1 1 -2 2 -1]
[-2 -1 1 2 0]
[-4 -5 5 7 1]
Domain: Vector space of dimension 4 over Rational Field
Codomain: Vector r space of f dimension 4 4 over Rational Field
sage: T_standard.matrix(side=’right’).is_invertible()
True
UnderstandthatanymatrixrepresentationofT_symbolicwillhaveaninvertible
matrixrepresentation,nomatterwhichbasesareused.Ifyoulookatthematrixrep-
resentationofT_standardandthedenitionofT_symbolictheconstructionofthis
examplewillbetransparent,especiallyifyouknowtherandommatrixconstructor,
sage: A A = = random_matrix(QQ, , 4, , algorithm=’unimodular’, , upper_bound=9)
sage: A
# random
[-1 -1 1 2 2 1]
[ 1 1 1 1 -1 1 0]
[-2 -1 1 5 5 6]
[ 1 1 0 0 -4 4 -5]
ENDOEndomorphisms
An endomorphism is s an n \operation-preserving" function (a \morphism") whose
domainandcodomainareequal. Sagetakesthisdenitiononestepfurtherforlinear
transformationsandrequiresthatthedomainandcodomainhavethesamebases(ei-
theradefaultechelonizedbasisorthesameuserbasis).Whenalineartransformation
meetsthisextrarequirement,severalnaturalmethodsbecomeavailable.
Principally, we e can compute theeigenvalues providedby Denition EELT.We
alsogetanaturalnotionofacharacteristicpolynomial.
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ChapterR
136
sage: x1, , x2, x3, x4 4 = = var(’x1, , x2, , x3, x4’)
sage: outputs s = = [ [ 4*x1 1 + + 2*x2 -
x3 + + 8*x4,
...
3*x1 - 5*x2 - - 9*x3
,
...
6*x2 + + 7*x3 + + 6*x4,
...
-3*x1 + 2*x2 + + 5*x3 - - 3*x4]
sage: T_symbolic(x1, , x2, x3, x4) ) = = outputs
sage: T T = linear_transformation(QQ^4, , QQ^4, , T_symbolic)
sage: T.eigenvalues()
[3, -2, 1, , 1]
sage: cp p = = T.characteristic_polynomial()
sage: cp
x^4 - - 3*x^3 - - 3*x^2 + + 11*x - - 6
sage: cp.factor()
(x - 3) * (x + 2) * (x - 1)^2
Nowthequestionofeigenvaluesbeingelementsofthesetofscalarsusedforthe
vectorspacebecomesevenmoreobvious.Ifwedeneanendomorphismonavector
spacewhosescalarsaretherationalnumbers,shouldwe\allow"irrationalorcomplex
eigenvalues? Youwillnowrecognizeouruseofthecomplexnumbersinthetextfor
thegrossconveniencethatitis.
CBMChange-of-BasisMatrix
Tocreateachange-of-basismatrix,itisenoughtoconstructanidentitylineartrans-
formationrelative toa domainandcodomainwiththespecieduserbases,which
is simply astraightapplicationofDenitionCBM.Herewegowithtwo arbitrary
bases.
sage: b0 0 = = vector(QQ, [-5, 8, , 0, , 4])
sage: b1 1 = = vector(QQ, [-3, 9, , -2, 4])
sage: b2 2 = = vector(QQ, [-1, 4, , -1, 2])
sage: b3 3 = = vector(QQ, [-1, 2, , 0, , 1])
sage: B B = [b0, b1, b2, b3]
sage: U U = (QQ^4).subspace_with_basis(B)
sage: c0 0 = = vector(QQ, [ [ 0, , 2, , -7, , 5])
sage: c1 1 = = vector(QQ, [-1, , 2, , -1, , 4])
sage: c2 2 = = vector(QQ, [ [ 1, , -3, , 5, , -7])
sage: c3 3 = = vector(QQ, [ [ 1, , 1, , -8, , 3])
sage: C C = [c0, c1, c2, c3]
sage: V V = (QQ^4).subspace_with_basis(C)
ChapterR
137
sage: x1, , x2, x3, x4 4 = = var(’x1, , x2, , x3, x4’)
sage: id_symbolic(x1, x2, , x3, x4) = [x1, x2, x3, x4]
sage: S S = linear_transformation(U, V, , id_symbolic)
sage: CB B = = S.matrix(side=’right’)
sage: CB
[ 36 6 25
8
7]
[ 27 7 34 4 15
7]
[ 35 5 35 5 14
8]
[-13 -4
0 -2]
sage: S.is_invertible()
True
We candemonstrate that CB is indeed the change-of-basis matrix from B toC,
convertingvectorrepresentationsrelativetoBintovectorrepresentationsrelativeto
C.Wechooseanarbitraryvector,x,toexperimentwith(youcouldexperimentwith
other possibilities). . Weusethe e Sageconveniences tocreate vector representations
relativetothetwobases,andthenverifyTheoremCBM.Recognizethatx,uandv
areallthesamevector.
sage: x x = vector(QQ, [-45, , 62, , 171, 85])
sage: u u = U.coordinate_vector(x)
sage: u
(-103, -108, , 45, , 839)
sage: v v = V.coordinate_vector(x)
sage: v
(-175, 95, -43, 93)
sage: v v == = CB*u
True
Wecanalsoverifytheconstructionabovebybuildingthechange-of-basismatrix
directly(i.e.,withoutconstructingalineartransformation).
sage: cols = = [V.coordinate_vector(u) ) for r u u in n U.basis()]
sage: M M = column_matrix(cols)
sage: M
[ 36 6 25
8
7]
[ 27 7 34 4 15
7]
[ 35 5 35 5 14
8]
[-13 -4
0 -2]
ChapterR
138
MRCBMatrixRepresentationandChange-of-Basis
InSage MRwe builttwomatrix representations ofone linear transformation,rel-
ative to o two o dierent pairs s of f bases. . We e now w understand how these e two matrix
representations are related |Theorem MRCB gives s the e precise relationshipwith
change-of-basis matrices,oneconvertingvectorrepresentations onthe domain,the
other converting g vector representations on the e codomain. . Here e is the demonstra-
tion. WeuseMTastheprex x ofnamesformatrixrepresentations,CBastheprex
forchange-of-basismatrices,andnumeralstodistinguishthetwodomain-codomain
pairs.
sage: x1, , x2, x3, x4 4 = = var(’x1, , x2, , x3, x4’)
sage: outputs s = = [3*x1 + + 7*x2 + x3 - 4*x4,
...
2*x1 + + 5*x2 + + x3 3 - - 3*x4,
...
-x1 - - 2*x2
+
x4]
sage: T_symbolic(x1, , x2, x3, x4) ) = = outputs
sage: U U = QQ^4
sage: V V = QQ^3
sage: b0 0 = = vector(QQ, [ [ 1, , 1, , -1, 0])
sage: b1 1 = = vector(QQ, [-1, 0, , -2, 7])
sage: b2 2 = = vector(QQ, [ [ 0, , 1, , -2, 4])
sage: b3 3 = = vector(QQ, [-2, 0, , -1, 6])
sage: B B = [b0, b1, b2, b3]
sage: c0 0 = = vector(QQ, [ [ 1, , 6, , -6])
sage: c1 1 = = vector(QQ, [ [ 0, , 1, , -1])
sage: c2 2 = = vector(QQ, [-2, -3, , 4])
sage: C C = [c0, c1, c2]
sage: d0 0 = = vector(QQ, [ [ 1, , -3, , 2, , -1])
sage: d1 1 = = vector(QQ, [ [ 0, , 1, 0, , 1])
sage: d2 2 = = vector(QQ, [-1, , 2, , -1, -1])
sage: d3 3 = = vector(QQ, [ [ 2, , -8, , 4, , -3])
sage: D D = [d0, d1, d2, d3]
sage: e0 0 = = vector(QQ, [ [ 0, , 1, , -3])
sage: e1 1 = = vector(QQ, [-1, , 2, , -1])
sage: e2 2 = = vector(QQ, [ [ 2, , -4, , 3])
sage: E E = [e0, e1, e2]
ChapterR
139
sage: U1 1 = = U.subspace_with_basis(B)
sage: V1 1 = = V.subspace_with_basis(C)
sage: T1 1 = = linear_transformation(U1, V1, , T_symbolic)
sage: MTBC =T1.matrix(side=’right’)
sage: MTBC
[ 15 5 -67 -25 -61]
[-75 326 120 298]
[ 3 3 -17 7 -7 7 -15]
sage: U2 2 = = U.subspace_with_basis(D)
sage: V2 2 = = V.subspace_with_basis(E)
sage: T2 2 = = linear_transformation(U2, V2, , T_symbolic)
sage: MTDE = = T2.matrix(side=’right’)
sage: MTDE
[ -32
8
38 -91]
[-148
37 178 8 -422]
[ -80
20
96 -228]
Thisisasfaraswecouldgoback inSectionMR.Thesetwomatricesrepresent
thesamelineartransformation(namelyT_symbolic),butthequestionnowis\how
are these representations related?" " We e need two change-of-basis matrices. . Notice
thatwithdierentdimensionsforthedomainandcodomain,wegetsquarematrices
ofdierentsizes.
sage: identity4(x1, , x2, x3, , x4) = = [x1, x2, , x3, x4]
sage: CU U = = linear_transformation(U2, U1, , identity4)
sage: CBDB = = CU.matrix(side=’right’)
sage: CBDB
[ 6 6 7 7 -8 8 1]
[ 5 5 1 1 -5 5 9]
[-9 -6 6 10 0 -9]
[ 0 0 3 3 -1 1 -5]
sage: identity3(x1, , x2, x3) ) = = [x1, x2, x3]
sage: CV V = = linear_transformation(V1, V2, , identity3)
sage: CBCE = = CV.matrix(side=’right’)
sage: CBCE
[ 8
1 -7]
[ 33
4 -28]
[ 17
2 -15]
Finally,hereisTheoremMRCB,relatingthethetwomatrixrepresentationsvia
thechange-of-basismatrices.
sage: MTDE == = CBCE * * MTBC C * * CBDB
True
ChapterR
140
We canwalk throughthis theoremjust abit more carefully, , step-by-step. . We
will compute three matrix-vector products, , using three e vector representations, , to
demonstratetheequalityabove. Toprepare,wechoosethevectorxarbitrarily,and
wecomputeitsvaluewhenevalutedbyT_symbolic,andthenverifythevectorand
matrixrepresentationsrelativetoDandE.
sage: T_symbolic(34, , -61, 55, 18)
(-342, -236, , 106)
sage: x x = vector(QQ, [34, -61, 55, 18])
sage: u_D D = = U2.coordinate_vector(x)
sage: u_D
(25, 24, -13, -2)
sage: v_E E = = V2.coordinate_vector(vector(QQ, , [-342, -236, 106]))
sage: v_E
(-920, -4282, , -2312)
sage: v_E E == MTDE*u_D
True
Sofarthisisnotreallynew,wehavejustveriedtherepresentationMTDEinthe
caseofoneinputvector(x),butnowwewillusethealternateversionofthismatrix
representation,CBCE * * MTBC * * CBDB,insteps.
First,converttheinputvectorfromarepresentationrelativetoDtoarepresen-
tationrelativetoB.
sage: u_B B = = CBDB*u_D
sage: u_B
(420, 196, -481, 95)
Nowapplythematrixrepresentation,whichexpects\input"coordinatizedrelative
toBandproduces\output"coordinatizedrelativetoC.
sage: v_C C = = MTBC*u_B
sage: v_C
(-602, 2986, , -130)
NowconverttheoutputvectorfromarepresentationrelativetoCtoarepresen-
tationrelativetoE.
sage: v_E E = = CBCE*v_C
sage: v_E
(-920, -4282, , -2312)
Itisnosurprise thatthis versionofv_Eequalsthepreviousone,sincewehave
checked theequality y of thematrices s earlier. . But t it may beinstructiveto see the
inputconvertedbychange-of-basismatricesbeforeandafterbeinghitbythelinear
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