Advanced Linear Algebra [Lecture notes]

44 Pages·2017·0.306 MB·English

by  Rachel Quinlan

Advanced Linear Algebra MA500-1: Lecture Notes
Semester 1 2015-2016
DrRachelQuinlan
SchoolofMathematics,StatisticsandAppliedMathematics,NUIGalway
March14,2017
Contents
1 Threewaystothinkaboutamatrix 2
1.1 Lineartransformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1.1 Interpretingamatrixasalineartransformation . . . . . . . . . . . . . . . . 2
1.1.2 Interpretingalineartransformationasamatrix . . . . . . . . . . . . . . . . 3
1.1.3 ChangeofBasis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.1.4 Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.1.5 Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.2 BilinearForms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.2.1 Symmetricandalternatingforms . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.2.2 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.3 MatricesandGraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2 SpectralProperties 24
2.1 Thedeterminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.2 Thespectrumofamatrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.3 Positivematrices-theFrobenius-PerronTheorem . . . . . . . . . . . . . . . . . . . 31
2.4 SupplementtoChapter2: evenandoddpermutations. . . . . . . . . . . . . . . . . 38
1
Chapter 1
Three ways to think about a matrix
Matricesareubiquitousinmathematicsandarisecentrallyinmanyareasthatarenotnecessarily
closelyrelatedinanobviousway.Matricesarenaturallyequippedwithlotsofalgebraicstructure
(forexamplethesetofn×nmatricesoveraringRorfieldFisitselfaringwithlotsofinteresting
properties). Whichaspectsofthisextensivealgebraicstructureareofinterestcandependaloton
thecontext. Therearemanywaysofthinkingaboutwhatamatrixis,anditisoftenhelpful,even
necessary, to have access to more than one of them. We discuss three different viewpoints (and
thealgebraicconsiderationsthataccompanythem)inthischapter.
1.1 Linear transformations
ForafieldFandpositiveintegern,wewillwriteFnforthevectorspaceconsistingofallcolumn
vectorsoflengthnwithentriesinF,andMn(F)forthesetofalln×nmatriceswithentriesinF.
1.1.1 Interpretingamatrixasalineartransformation
IfA∈Mn(F)andv∈Fn,wecan“multiply”AbyvtogetanotherelementofFn.
   
−1 1 2 1
Example1.1.1. InM3(Q),writeA= −12 8 6 . InQ3,writev= 2 . Then
12 −7 −3 3
      
−1 1 2 1 −1(1)+1(2)+2(3) 7
Av= −12 8 6  2 = −12(1)+8(2)+6(3) = 22 .
12 −7 −3 3 12(1)−7(2)−3(3) −11
Example 1.1.1 demonstrates the process of matrix-vector multiplication. Although this is al-
readyfamiliaritisworthwhiletoconsiderwhatisgoingoninslightlymoredetail. Foramatrix
Aandcolumnvectorv, youcancalculatetheproductAvonlyifthenumberofcolumnsofAis
the same as the number of entries in v. What we are doing when we calculate Av is taking the
linearcombinationofthecolumnsofAthatisdeterminedbytheentriesofv. Thismeansthatif
 
a
1
v= .. andAhasncolumns,thenthecolumnvectorAvisgivenby
 . 
a
n
     
Col1 Col2 Coln
a1 of +a2 of +···+an of .
A A A
Exercise1.1.2. Ifindoubt,confirmthisfromExample1.1.1.
2
ThuseverymatrixA∈Mn×n(F)definesafunctionTA :Fn →Fnby
TA(v)=Av.
Thisfunctionisalineartransformation,whichmeansthat
• Forallu,v∈Fn,TA(u+v)=TA(u)+TA(v)(sinceA(u+v)=Au+Av),and
• Forallv∈Fn andα∈F,TA(αu)=αTA(u). (Thefieldelementαisreferredtoasascalarin
thiscontext).
Ingeneralalineartransformationisafunctionthatrespectsadditionandscalarmultiplication(in
acontextwherethatmakessense).
NOTES
1. Moregenerally,ap×nmatrixA(prows,ncolumns)maybethoughtofasalineartrans-
formationTA fromFn toFp, viamatrix-vectormultiplication. Ifv ∈ Fn, thenTA(v) = Av
isthelinearcombinationofthecolumnsofAinwhichthecoefficientofColumniistheith
entryofv.
2. If A ∈ Mp×n(F) then Column i of A is the image under TA of the vector ei, which has
entry1intheithpositionandzeroinallotherpositions. Sodifferentmatricescorrespond
todifferentlineartransformations.
3. If T : Fn → Fp is any linear transformation, let A be the matrix in Mp×n(F) whose ith
columnisT(ei). ThenT(v)=Avforallvectorsv∈Fn.
Exercise1.1.3. ProvethestatementinItem3above.
SowecanthinkofthematrixspaceMp×n(F)asbeingthesetoflineartransformationsfromFn
toFp. Thepresentationgivenhereinvolvesachoicetodiscussmatricesmultiplied(ontheright)
bycolumnvectors. Itcouldequallywellbepresentedasintermsofmatricesbeingmultipliedon
theleftbyrowvectors.
Welet(Fp)T denotethetransposeofFp,i.e. thespaceofrowvectorsoflengthpwithentriesin
F. Thenap×nmatrixAdescribesalineartransformationfrom(Fp)T to(Fn)T via
v→vA.
NotethatvA,whichbelongstoFn,isthelinearcombinationoftherowsofAinwhichthecoeffi-
cientofRowiisentryiofv. Soa(row-)vector-matrixproductgivesalinearcombinationofthe
rowsofthematrix,withcoefficientsgivenbythevectorentries.
1.1.2 Interpretingalineartransformationasamatrix
Suppose now that V and W are vector spaces of finite dimensions n and p respectively over F,
andletf : V → W bealineartransformation. LetB = {b1,...,bn}andC = {c1,...,ck}bebases
for V and W respectively. Write Mf,B,C for the p×n matrix whose ith column contains the C-
coordinatesoftheelementf(bi)ofW.
Theorem 1.1.4. If v ∈ V, then the C-coordinates of f(v) are the entries of the matrix-vector product
Mf,B,C[v],where[v]isthevectorinFnwhoseentriesaretheB-coordinatesofv.
 
a
1
Proof. Writev=a1b1+···+anbn,so[v]= ... . Then
a
n
f(v)=a1f(b1)+a2f(b2)+···+anf(bn).
3
SincetheC-coordinatesoff(bi)arewrittenintoColumniofMf,B,C,itfollowsthattheC-coordinates
off(v)aretheentriesof
     
Col1 Col2 Coln
a1 of +a2 of +···+an of =Mf,B,C[v].
Mf,B,C Mf,B,C Mf,B,C
Example 1.1.5. Let V = Q [x], the space of polynomials in x of degree at most 4 over Q, with basis
4
B = {1,x,x2,x3,x4}. Let W = Q [x], the space of polynomials of degree at most 3 over Q, with basis
3
C = 1,x,x2,x3. Let D : V → W be the differential operator, which maps a polynomial to its derivative.
ThenDisalineartransformationand
 
0 1 0 0 0
 0 0 2 0 0 
MD,B,C = 0 0 0 3 0 .
0 0 0 0 4
Ifwewanttousethistocalculatethederivativeofx4−3x3+2x2−x,wecancalculatethematrix-vector
product
 
  0  
0 1 0 0 0 −1
 −1 
 0 0 2 0 0    4 
  2 = .
 0 0 0 3 0    −9 
 −3 
0 0 0 0 4 4
1
SoD(x4−3x3+2x2−x)=4x3−9x2+4x−1.
Exercise1.1.6. Intheaboveexample,supposeweusedthebasisC(cid:48) ={1,1+x,1+x+x2,1+x+x2+x3}
forW insteadofC. Howwouldthematrixchange?
The point here is that given a linear transformation f between two vector spaces of finite
dimension,thechoiceofabasisforeachspaceallowsusconsiderfasamatrix. Itisnotexactly
truetosaythateverytransformationfcorrespondstoamatrixinsomeobjectiveway,becauseit
isnotonlyfbutalsothechoiceoftwobasesthatdeterminethematrix.
Question1.1.7. Supposethatf:V →Wisalineartransformationbetweendifferentvectorspaces. How
dothematricesthatrepresentfwithrespecttodifferentbasesresembleeachother?
Inordertoanswerthisquestionweneedsomematrixmachinery.
1.1.3 ChangeofBasis
LetV beaF-vectorspaceofdimensionnandsupposethatB={b1,...,bn}andB(cid:48) ={v1,...,vn}
arebasesofV. Theneachbi canbewritteninauniquewayasalinearcombinationofv1,...,vn.
Forj=1,...,nwrite
(cid:88)n
bj = aijvi, aij ∈F.
i=1
Let P denote the n×n matrix whose entry in the (i,j) position is aij. Possibly a better way to
thinkaboutthematrixP isthatColumnjofP isthevectorwhoseentriesaretheB(cid:48)-coordinates
ofbj. SothecolumnsofPexpresstheelementsofthebasisBintermsoftheirB(cid:48)-coordinates.
(cid:80)
Lemma1.1.8. Letx∈V,andsupposethatx= ni=1cibi,sothattheB-coordinatesofxarec1,...,cn.
c
1
ThentheB(cid:48)-coordinatesofxaregivenbytheentriesofthematrix-vectorproductP ... .
c
n
4
Proof. Thematrix-vectorproductis
     
Col1 Col2 Coln
c1 of +c2 of +···+cn of .
P P P
SinceColumnjofPexpressesbj intermsofitsB(cid:48)-coordinates,theentriesofthisproductarethe
coordinatesofc1b1+···+cnbn =xintermsofB(cid:48).
Definition1.1.9. InviewofLemma1.1.8,werefertoPasthechangeofbasismatrixfromBtoB(cid:48). Its
columnsaretheelementsofBexpressedintermsofB(cid:48).
Remark In the context of Section 1.1.2, you can consider P to be the matrix M ,B,B(cid:48), where
id
id:V →V istheidentitymapping. (ThankstoBenforthisniceobservation).
NowletQbethechangeofbasismatrixfromB(cid:48) toB,definedequivalently-thecolumnsof
QaretheB-columnrepresentationsoftheelementsv1,...,vn ofB(cid:48). Asabove,wecanpassfrom
theB(cid:48)-columnrepresentationofanyelementofV toitsB-columnrepresentationbymultiplying
ontheleftbyQ.
NowletcbeanyvectorinFn. ThencistheB-columnrepresentationofsomeelementvofV,
andtheB(cid:48) columnrepresentationofvisPc. ButthentheB-columnrepresentationofvisgiven
byQ(Pc)=QPc. HoweverthismustbeequaltocThus
QPc=cforallc∈Fn.
Hence QP = In, the n×n identity matrix. Similarly PQ = In, so P and Q are inverses of each
other.Inparticular,everychangeofbasismatrixisinvertible,anditsinverseisthereversechange
ofbasismatrix.
On the other hand, every invertible n×n matrix determines a change of basis in Fn. To see
thisletP ∈Mn(F)beinvertibleandletQbeitsinverse. ThismeansthatPQ=QP =In.
We focus on the product QP = In. This means that e1, the first column of In, is the linear
combination of the columns of Q whose coefficients are the entries of Column 1 of P. Similarly
e2,...,en are linear combinations of the columns of Q with coefficients given by the entries of
Columns 2,...,n of P. Then in particular all of the standard basis vectors of Fn belong to the
span of the Columns of Q, and so the columns of Q form a spanning set of Fn. Since Fn has
dimension n it cannot be spanned by fewer than n columns, and so the columns of Q form a
minimal spanning set of Fn, hence they must be linearly independent. Thus the columns of the
invertiblematrixQformabasisBQofFn.
Moreover,forj = 1,...,n,theentriesofColumn1ofP arethecoordinatesofei withrespect
to the basis BQ. If v is any vector in Fn, then the coordinates of v with respect to BQ are given
by the entries of Pv (or equivalently Q−1v). So P or Q−1 is the change of basis matrix from the
standardbasisofFntoBQ.
Remarks
1. Asabove,wecouldusethefactthatPQ=IntoshowthatthecolumnsofPformabasisfor
Fn,andbythinkingoftherowsofPQ(orQP)aslinearcombinationsoftherowsofQ(or
P),wecanshowthattherowsofQ(orP)formabasisfor(Fn)T.
2. The above argument shows that if A ∈ Mn(F) has a right inverse (i.e. there exists B ∈
Mn(F) with AB = In, then the columns of A form a basis of Fn. It is true that if A has
a right inverse then it also has a left inverse and these coincide, but we are not quite in a
positiontoprovethatyet. Wewillsoon.
WearenowinapositiontoanswerQuestion1.1.7.
SupposethatVandWareF-vectorspacesofdimensionspandnrespectivelyandthatf:V →
W isalineartransformation. LetBandB(cid:48) betwobasesofV andletCandC(cid:48) betwobasesofW.
5
LetthechangeofbasismatricesfromB(cid:48)toBandfromCtoC(cid:48)bedenotedbyPandQrespectively
(soPisanonsingularn×nmatrixandQisanonsingularp×pmatrix). Then
Mf,B(cid:48),C(cid:48) =QMf,B,CP.
Explanation: SupposethatcistheB(cid:48)-columnrepresentationofsomeelementvofV. Wewantto
knowwhatmatrixshouldmultiplycontheleftinordertogivetheC(cid:48)-columnrepresentationof
f(v). MultiplyingcbyPgivesustheB-columnrepresentationofv,multiplyingthatbythep×n
matrix Mf,B,C gives us the C-column representation of f(v), and multiplying that by Q gives us
theC(cid:48)-columnrepresentationoff(v). Thus,overallwehave
[f(v)]C(cid:48) =QMf,B,CP[v]B(cid:48).
Thismotivatesthefollowingdefinition.
Definition1.1.10. LetAandBbematricesinMp×n(F). ThenAandBaresaidtobeequivalentifthere
existnonsingularmatricesP ∈M (F)andQ∈M (F)forwhich
n p
B=QAP.
Iftwop×nmatricesareequivalent,ifmeansthattheyrepresentthesamelineartransforma-
tionfromFntoFp,possiblywithrespecttodifferentbasesforbothspaces.
1.1.4 Similarity
We now specialize the discussion to the case where V = W. Our set up now is that we have a
single vector space V of dimension n, and a linear transformation f : V → V. If B is a basis of
V,wewriteMf,B forthematrixthatwascalledMf,B,B inSection??. SupposethatB(cid:48) isanother
basis of V, and let P be the change of basis matrix from B(cid:48) to B (so the columns of P are the B-
representationsoftheelementsofB(cid:48). ThenP−1 isthechangeofbasismatrixfromBtoB(cid:48) (and
itscolumnsaretheB(cid:48)-representationsoftheelementsofB). Then
Mf,B(cid:48) =P−1Mf,BP.
Definition1.1.11. SupposethatAandBarematricesinMn(F). ThenAandBaresaidtobesimilarif
thereexistsaninvertiblematrixP ∈M (F)forwhich
n
B=P−1AP.
If two matrices are similar, it means that they describe the same linear transformation, with
respecttodifferentbases.
Givenalineartransformationf:V →V,itisreasonabletoaskwhetherthereissomebasisof
V withrespecttowhichitsmatrixhasaparticularyniceform. Thenicestformthatyoucanhope
forisadiagonalform,inwhichallentriesawayfromthemaindiagonal(fromupperlefttolower
right)arezeros. IfthereisabasisB={b1,...,bn}ofV forwhich
 
λ 0 ... 0
1
 0 λ2 ... 0 
Mf,B =diag(λ1,...,λn)= ... ... ... ,
0 0 ... λn
itmeansthatf(bi)=λi(bi)fori=1,...,n. Thismeansexactlythateachbiisaneigenvectoroff.
Definition1.1.12. LetV beaF-vectorspaceandletf : V → V bealineartransformation. Anon-zero
element v of V is called an eigenvector of f if f(v) = λv for some λ ∈ F. In this case λ is called the
eigenvalueofftowhichvcorresponds.
6
So if v is an eigenvector of f, it means that f(v) is just a scalar multiple of v. The mapping
f : V → V is called diagonalizable (or diagonable) if there exists a basis of V with respect to
whichthematrixoffisdiagonal. ThismeansthatthereexistsabasisofV consistingentirelyof
eigenvaluesoff.
The“matrix”versionsof(someof)thesedefinitionsaregivenbelow.Therearemanyconcepts
andstatementsinlinearalgebrathatcanbeexpressedeitherintermsofmatricesorintermsof
lineartransformations.
Definition 1.1.13. Let A ∈ Mn(F). A non-zero column vector v ∈ Fn is a right eigenvector of A if
Av = λvforsomeλ ∈ F. Anon-zerorowvectorw ∈ Fn isalefteigenvectorofAifwA = λwforsme
λinF. IneachcasethescalarλisthecorrespondingeigenvalueofA.
Ifisnotalwaystrueforalineartransformationf : V → V orforasquarematrixA ∈ Mn(F)
(cid:18) (cid:19)
1 1
that there exists a basis of V (or Fn) consisting of eigenvectors. For example let A =
0 1
andinvestigaterighteigenvectorsofA. Supposethat
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
1 1 x x x+y = λx
=λ =⇒
0 1 y y y = λy.
Foraneigenvectorxandycannotbothbezero. Fromthesecondequation,eithery=0orλ=1.
Howeverify=0thenλ=1anywayfromthefirstequation,sowemusthaveλ=1,and1isthe
onlyeigenvalueofA. Now
x+y=x=⇒y=0,
and any vector of the form (cid:0)x(cid:1) is an eigenvector of A corresponding to the eigenvalue 1. These
0
aretheonlyeigenvectorsofAandtheyareallscalarmultiplesof(cid:0)1(cid:1),theyforma1-dimensional
0
subspaceofF2. SoF2 doesnothaveabasisconsistingofeigenvectorsofA,andAisnotsimilar
toadiagonalmatrix.
Wedohavethefollowingtheorem,butitsconverseisnottrue(sincewecanobviouslyeasily
writedownexamplesofdiagonalmatricesthathaverepeatedeigenvalues).
Theorem1.1.14. SupposethatA∈Mn(F)hasndistincteigenvaluesλ1,...,λn,andthatv1,...,vnare
respectiveeigenvectorsofA. ThenB={v1,...,vn}isabasisofFn.
Proof. WeshowthatthesetBislinearlyindependent. Supposeit’snot,andseekacontradiction.
Then k be the least index for which {v1,v2,...,vk} is linearly dependent. Then k (cid:54) n since B
is linearly dependent, and k (cid:62) 2 since v1 is not the zero vector. Then v1,...,vk−1 are linearly
independentandvk isalinearcombinationofthese,sothereexista1,...,ak−1 ∈ F,notallzero,
with
vk =a1v1+···+ak−1vk−1. (1.1)
Multiplying1.1ontheleftbyAgives
λkvk =a1λ1v1+a2λ2v2+···+ak−1λk−1vk−1,
andmultiplying1.1bythescalarλkgives
λkvk =a1λkv1+a2λkv2+···+ak−1λk−1vk−1.
Equatingtherighthandsidesofthesetwoexpressionsforλkvkgives
a1(λk−λ1)+a2(λk−λ2)+···+ak−1(λk−λk−1)vk−1. (1.2)
Since the eigenvalues λ1,...,λk are distinct, the field elements λk −λi in 1.2 above are all non-
zero. Furthermoretheai inthisexpressionarenotallzero,so1.2isalineardependencerelation
among v1,...,vk−1. This contradicts the choice of k as the least index for which {v1,...,vk} has
sucharelation.
7
A matrix A ∈ Mn(F) is diagonalizable (over F) if there exists a basis B of Fn consisting of
eigenvectors of A. In this case P−1AP is diagonal, where P is matrix whose columns are the
elements of B (this is the change of basis matrix from B to the standard basis). So a matrix is
diagonalizableifitissimilartoadiagonalmatrix.
There is an issue here that might not be immediately obvious. We demonstrate it with an
example.
(cid:18) (cid:19)
0 −1
Example1.1.15. LetA= inM (R). ToinvestigateeigenvectorsofAwecanwrite
1 0 2
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
0 −1 x x −y = λx
=λ =⇒
1 0 y y x = λy
Now substituting the second equation into the first gives −y = λx = λ(λy) = λ2y. Taking
y=0isnotanoptionasthiswouldforcex=0also,andthezerovectorcannotbeaneigenvector.
So from −y = λ2y we must conclude λ2 = −1. There is no real number λ with this property,
so A has no eigenvalues in R and has no eigenvectors with real entries. However, if we allow
ourselves to consider complex eigenvalues, we can try λ = i and λ = −i. An eigenvector (cid:0)x(cid:1)
1 2 y
correspondingtoλ mustsatisfyx=iy,forexample(cid:0)i(cid:1),andaneigenvectorcorrespondingtoλ
1 1 2
mustsatisfyx=−iy,forexample(cid:0)−i(cid:1).Now(cid:0)i(cid:1)and(cid:0)−i(cid:1)formabasisofC2,soAisdiagonalizable
1 1 1
ifweconsideritasanelementofM (C),andinthiscase
2
(cid:18) (cid:19) (cid:18) (cid:19)
i 0 i −i
P−1AP = , whereP = . (Checkthis!)
0 −i 1 1
However,AisnotdiagonalizablewithinM (R).
2
GivenamatrixA ∈ Mn(F)(oralineartransformationf : V → V ofan-dimensionalF-vector
space V) we can ask about the existence of “nice” bases for describing A, or of “nice” matrices
thataresimilartoA. TheeigenvaluesofAmayormaynotbelongtothefieldF. Wecanconsider
twocases:
• If we only want to consider similarity within Mn(F), we can try to identify the “nicest”
matrix of the form P−1AP, where P ∈ GL(n,F). This leads to the theory of the rational
canonicalform.
• If F¯ is a field that has F has a subfield and contains all the eigenvalues of F, then we can
consider A to be an element of Mn(F¯) and consider similarity over F¯. Then we would
be looking for the “nicest” matrix of the form P−1AP where the entries of the invertible
matrixP(andofP−1AP)belongtoF¯ butnotnecessarilytoF. Thisleadstothetheoryofthe
Jordancanonicalform.TheJordancanonicalformofthematrixinExample1.1.15aboveisthe
(cid:18) (cid:19)
i 0
diagonalmatrix .
0 −i
Wewillrevisittheconceptofsimilaritylaterinthecourse.
1.1.5 Rank
Letf:V →W bealineartransformationofF-vectorspaces. Thekernelandimageoffaredefined
by
kerf={x∈V :f(x)=0}; (cid:61)f=f(x):x∈V.
Lemma1.1.16. ThekernelandimageoffaresubspacesofV andW respectively.
Theproofisleftasanexercise.
Definition1.1.17. Thedimensionof(cid:61)fiscalledtherankoff.
8
LetA∈Mp×n(F). ThecolumnspaceofAisthesubspaceofFpthatisspannedbythecolumns
of A. The dimension of this space is called the column rank of A. Since linear combinations of
thecolumns of A areprecisely equalto matrix-vectorproducts ofthe form Av with v ∈ Fn, the
columnspaceofAisthesetofallsuchvectors. Thisisalsotheimageofthelineartransformation
TA :Fn →Fpdefinedasleft-multiplicationbyA,anditsdimensionistherankofTA. Thus
ThecolumnrankofAistherankofthelineartransformationTA.
Analagously, we can define the row rank of A to be the dimension of the subspace of (Fn)T
spanned by the rows of A. This space is called the row space of A and it is the image of the
lineartransformationfromLA :(Fp)T →(Fn)T definedforarowvectorw∈(Fp)T by
LA(w)=wA.
WhatpreciselyistheconnectionbetweenTAandLAisnotaparticularlyeasyquestiontoanswer
butwewillcomebacktoitinSection1.2. Fornowwecanprovethesurprisingandnon-obvious
factthattheyhavethesamerank. Manydifferentproofsofthistheoremcanbefoundinbooks,
notallofthemofferalotofinsightintowhythestatementistrue-manyrelyonrowandcolumn
operationsthatreducethematrixtoanechelonform.
Theorem1.1.18. LetA∈M (F). ThentherowrankandcolumnrankofAareequal.
p×n
Proof. WriterfortherowrankofAandcforthecolumnrank. Wewanttoshowthatr = c. Let
v1,...,vc becolumnvectorsinFn thatformabasisforthecolumnspaceofA,andletCA bethe
p×c matrix that has v1,...,vc as its columns. Then every column of A can be expressed as a
linearcombinationofv1,...,vc inauniqueway,anditfollowsthatthereexistsac×nmatrixR
forwhich
(cid:124)C(cid:123)(cid:122)A(cid:125)(cid:124)(cid:123)R(cid:122)(cid:125)=A.
p×c c×n
The first column in R contains the coordinates of Column 1 of A in terms of v1,...,vc, and so
on. Looking at the same product the other way round, we see that Row 1 of A is the linear
combination of the rows of R whose coefficients are the entries of Row 1 of CA and so on; each
rowofAisalinearcombinationofthecrowsofR. ThusthecrowsofRspantherowspaceofA,
andtherowrankofAisatmostc,sor(cid:54)c.
We use essentially the same argument to show that c (cid:54) r and hence that r = c. The row
rankofAisr,hencethereexistsaspanningsetw1,...,wr fortherowspaceofA. LetRA bethe
r×nmatrixwhoserowsarew1,...,wr. TheneveryrowofAhasauniqueexpressionasalinear
combinationoftherowsofRA,sothereexistsap×rmatrixCforwhich
(cid:124)(cid:123)C(cid:122)(cid:125)(cid:124)R(cid:123)A(cid:122)(cid:125)=A.
p×r r×n
TheentriesinRowiofCarethecoefficientsintheexpressionforRowiofAasalinearcombina-
tionoftherowsofRA. ButnoweverycolumnofAisalinearcombinationofthercolumnsofC,
andhencethedimensionofthecolumnspaceofAisatmostr,soc(cid:54)r.
Sincec(cid:54)randr(cid:54)c,weconcludethatc=randthattherowrankandcolumnrankofAare
equal.
InviewofTheorem1.1.18,wedonotneedtodistinguishbetweentherowrankandcolumn
rankofamatrix,andwecanjustrefertoitsrank. IfAisap×nmatrix,thenrank(A)isaninteger
between0andmin(p,n). Hereareafewremarksaboutrank.
1. Amatrixhasrank0ifandonlyifitisthezeromatrix.
2. A matrix has rank 1 if and only if it is not the zero matrix and all of its non-zero rows are
scalarmultiplesofeachother(sameforcolumns).
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