课件mth017数学系线性代数algebra6rank

Algebra6:TheRankofTheRankodMatrix矩阵的秩，是矩阵 概念之一性方程组的研究中，矩阵“秩”这个概念有关键作用为此，首先介绍矩阵的子(行列)式和代 式的概念

矩阵的子(行列)式

A=

a22a2n

k行、k an1a22amn

kk阶子式 -3

A=

-5

det

-3

取第1、3和4行，第2、4和5 得到三阶子方阵和3 A=

-5

det

2 Therankofa

A=

2 an1a22amnAk阶子方阵是可逆矩阵。k+1阶以上的子方阵都是不可逆的，A的秩(rank)k.Ak阶子行列式不等于零，但所有k+1阶以上的子行列式都等于零。则矩阵的秩等于k：

1 9 A=

B=

r(B)=1 0r(A)=

C= 1

=2„

0

00112 2=00112

1=

1= 0r(C)=

0

-3

A= -5

=-14„

Thereexistsasub-determinantoforder4thatisnonzero.SothattherankofAisgreaterthanorequalto4.Intheotherhand,thedimensionofthismatrix is4â5.Sothat,therankofAnotgreaterthan4.ThereforetherankofAequals一个4行5列的矩阵，它的秩不会超过ThepropertiesofrankofLetAbeamatrixofdimensionmân,r(A)£

12

A=

22

r(B)

a25

B=

In

r(A)£r(A)£min{m,

42 Ifr(A)=m,thenthematrixiscalledrowfullrank(行满秩)：Ifr(A)=nthenthematrixiscalledrowcolumnrank(列满秩)r(AT)=r(nnThenecessaryandsufficientconditionfortheinvertibilityofmatrixAisdet(A)π0.初等变换不改变矩阵的秩r(AB)£min{r(任意梯形矩阵的秩，等于它的非零行数2020122100000000000000000A=000

,r(A)=B=

r(B)3

就是该矩阵的秩 对矩阵AEshelonExample:Findtherankoffollowing0010011111111110011001fi00A=00

- -

11

00

11

r

r1)fi

r(A)=

Example:FindtherankofmatrixA r12(-

r(-1)fi

- 4A=

r(-

- 4

7

- 4 0 0

r(A)=a11x1+a12

++a1nxn= x+ x++ 21 Matrixof

++amnxn=augmented

b1 bA=

A=

2

b m mn

m mDoesthesystemhaveHowmanysolutionsdothesystemWhataretheformulaofthegenerala11x1+a12

++a1nxn= x+

x++ =

21 Linear

++amnxn=①Thesystemhasnontrivialsolutionsifandonlyifr(A)<n②Supposethatr(A)<n,thentherearek=n-r(A)arbitraryconstantsintheformulaofgeneralsolutionsofthehomogeneoussystems.③Therearek=n-r(A)linearindependentvectorsinthesystemoffoundationalsolutions.方程通解表达式中含有k=n-r(A)基本解系中有k=n-r(A)Example：Solvethehomogeneouslinear x12x24x3+x4 2x+ +8x =

MatrixofA

r(-

1 fi

0r23

0rA)2n4有非平凡解k=n-r(A)=2个向量 x1+2x2+4x3+x4=

1 2x+ +8x+ = 系数矩阵 =

0r(1

-1/

x+2x- x= fi 3/

x1=-2x2+5

1 x+ x+

x4=

=-1x -2t+1t

1 52

5x2= x2= 5

=t1+t0

=t+t x

-1

10

2-1

4=

3

10

10x4

0 1

基础解系v1,Example：Solvethe x1+2x2+ 3x+

3 2x+5x+

Matrixof

A= 7 x1+

+4x3=

4 3

r(-

1

r(-1)fi

7 r(-1)

4

x1 =2 x 3 Examplex+

x0l (x,x,x)T=(0, matrixofl l1=l

1- 1+

l1时方程组有非零解l=1x1+x2+x3=

x+x+x=

-

+x3=

11 x1+x2+x3

x1+x2=-

4x=-

x3=

x1=-

x2=-2x

1 1- 3 2 2- 3

1

=

-t

=t-

=t3

=t

3=v 2x

2 2

3

1 齐次方程组定理的合理解释（用例题解释a11x1+a12x2+a13x3+a14

+a15x5= x+ x+ x+ x+ x=

a12a1521

a31x1+a32x2+a33x3+a34

+a35x5=

25 x+ x+ x+ x+ x=

41 x+ x+ x+

+a55x5=

51

+a64

+a65x5=若r(A)=3<

c15

x+

+cx+

x+ x=

25

11

c25 c

c22

+

+c25x5= 0

c33x3+c34x4+c35x5=0 0

0 0x4x5c11x1+c12

+c13x3=-c14x4-

+c23

- 于是方程组有无穷多组解.t1和t2 c15 25

c45

55 0

x+ x+cx+ x+cx=

11

c22

+c23x3+c24

+c25x5=

+c35x5= c44

+c45x5=c55x5=Non-HomogeneousLineara11x1+a12

++a1nxn=

x1

b1x x+ x++ =x

x

b21

2

b= 2Ax=

++amnxn=

xn

bmDoesthesystemhaveHowmanysolutionsdothesystemWhataretheformulaofthegeneralMatrixof Augmented

b1

b2A= 2n

A=

mnThesystemisconsistentifandonlyifr(A)=r(Thesystemhasuniquesolutionifandonlyifr(A)=r(A)=Thesystemhasinfinitesolutionsifandonlyifr(A)=r(A)<Thegeneralsolutioncontaink=n-r(A)arbitrary(1)线性方程组相容的充分必要条件是r(A)=r(A).(2)线性方程组有惟一解的充分必要条件是rA)rA)n.(3)rA)rAn.例3x1-7x2+14x3-8x4= x-4x+3x

x-3x+4x-

2x1-15x2-x3+

判断系数矩阵和增广矩阵的秩的关系将方程组化为等价方程组，方便求解3424r-2 34455 fi22 33

-2 2 2

244

0 00 0

306系列变换00

r(A)=r(A)=2<60600Thesystemoflinerequationsis33

-2 2 2

244

0 00 0

30600系列变 00

66000

r(A)=r(A)=2<Thesystemoflinerequationsis06 06

000 00Thesystemoflinerequationsx1-4x2+3x3-x4=- x1-4x2=-3x3+x4-xxxx2+

-x4=

2=6-

+

=

,

x1-4x2=-3t12x =6-t+t

+

-2x1=22-7t1+ =6-t+= =

x 6 2= -t+tx3 0 1 2 0 x4 0 Thegeneralsolutionofthelinear

x 6

2=

=x+tv+v 0

10

2

1 x 0 0 4

6

,v2=0 0 t1v1+t2v2isthegeneralsolutionofthehomogeneousandxisaparticularsolution.ofthenonhomogeneousa11x1+a12x2+a13x3+a14

+a15x5= x+ x+ x+ x+ x=21 a31x1+a32

+a33x3+a34x4+a35x5= x+ x+

x+

x+

=41

51

+a52

+a53x3+a54x4+a55x5=

+a63x3+a64x4+a65x5= a12a15

b1

b

a22a25

2

a61a62a65

6rA)rA3 aaa 2 00000000000000000

0 0 0a11x1+a12

+a13x3+a14x4+a15x5= ax+ax+ax+ax= a 33x3+a34x4+a35x5=ax5x6移至等式右端作为自由变量a11x1+a12

+a13x3=b1-a14x4-a15aa 22

+a23x3=b2-a24x4-a25 a33x3=b3-a34x4-a35x5t1x6a11x1+a12

+a13x3=b1-a14t1-aa 22

+a23x3=b2-a24t1-a a

=b-at-a

34

35继续求解，通解表达式中含有5-3=2个任意常数rA)rA) aaa 2

a 4

0 5个方程，5rA)4,rA) aaa b 2

b 4 00 00a11x1+a12

+a13x3+a14x4+a15x5= a22x2+a23x3+a24x4+a25x5= a33

+a34x4+a35x5= 0x1+0x2+0x3+0x4+0x5=r(A)<r(A)Theremustbeequationwhichis3x1-7x2+14x3-8x4= x-4x+3x

x-3x+4x-

2x1-15x2-x3+

33-224

24

4 4

33

-2 2 2

244

0 00 0

30600系列变 00

66000

r(A)=r(A)=2<Thesystemoflinerequationsis06 06

000 00Thesystemoflinerequationsx1-4x2+3x3-x4=- x1-4x2=-3x3+x4-xx 2+

-x4=

=6-x+

=

,

x1-4x2=-3t12x =6-t+t

+

-2x1=22-7t1+ =6-t+= =

x 6 2= -t+tx3 0 1 2 0 x4 0 Thegeneralsolutionofthelinear

x 6

2= -

=x+tv+vx 0 2

1 3 x4 0 6 wherex=

,v1=,v2=0

2x1-3x2+6x3-5x4=

x2-

+x4=1

A=

k-r(-1)fi

k= r(A)=r(

k-当k=7660111fi11000000

Afi

fi00

11

x1-3x3-x4=

x1=3x3+x4+xxx3=t1,x4=

-

+x4

= - x

4t-t

1 2=

=t+t

+

1

20 xt xt4

即nndimensionv1 u1

v1+u1

v u lv

+u v=2,u=2,lv= 2,u+v=

2,q= v u lv

+u

n n n n 向量的加法(addition)和数乘(scalarmultiplication)，(LinearSpace)

是一组n维向量，k,k ab=ka+ a a,a

的一个线性组合(linear如果向量b等于a1,a2,,am b可以由a1,a2,,am线性表示

-

5

=

=2

b=

-

=21-32 1,

=-

2

-6

0 2 ba1a2Example:Inthevectorspaceofn

e1= ,111 1 b1b Theneachvectorb=2inRn,theequation b bnb1 1 b 1

2=b +b ++b

b=be+be+b 1 2 n

1 b 1n

IfVisanonemptysubsetofRnthatsatisfiestheax˛Vwheneverx˛Vforanyscalarx+y˛Vwheneverx˛Vandy˛VThenVisasubspaceofRn什么是子空间设VR3中xOy平面上所有向量组成的集合，是R3的一个二维子空间。LR3中经过原点的一条直线，LR3LetC[a,b]denotthesetofallreal-valuedcontinuousfunctionson[a,b]:(f+g)(x)=f(x)+g(x),(af)(x)=af(ThenC[a,b]isavectorThesetofallpolynomialsisasubspaceofC[a,ThesetofalldifferentiablefunctionsisasuubspaceofLetv1,v2,,vmbevectorsinavectorspaceVAsuma1v1+a2v2++amvm,wherea1,a2,,amarescalars,iscalledalinearcombinnationofv1,v2,,vm.Thesetofalllinrarconbinationsofv1,v2,,vmiscalledthespanofv1,v2,,vm,thatdenotedbyspan(v1,v2,,vm).Thesetspan(v1,v2,,vm)isasubspaceofVThenullspaceofamatrixandthesolutionsofhomogenouslinearsystem

x1

a12a1n x+ x++

21

x=x2

a22a2n

x+ x++

n1

n

xn

an1an2anna11 a12

a1n

x 21+x

22++x

2n= 1 2

n

n1 n2 xa+xa++xa a11x1+a12

++a1mxm= x+ x++ =21

++anmxm=

y1x yx=2,y=2

x xnIfxandyaresolitionsofthehomogeneouethenax+byiealsosolutionofthissystem.(Inwhich,aandbarearbitraryscalars)

yny齐次线性方程组的解集合，就是系数矩阵的零空间(NullN(A)={x˛Rn|Ax=x= 平凡解(trivial

=00这个解空间完全由系数矩阵AN(A).N(A)是所有满足Ax=0的那些向量组成的集合.A的零空间.Linear(向量集的线性无关性Lineardependence(线性相关 或者成倍数关系：b=aaora=bb三个向量a和b c=aa+bb在 Rn中，称向量集{v,v,,v}线性相关 vm=a1v1+a2v2++am-1vm-Example： e=0, =1,e=

Thesetofvectorsislinear

Example：

3

Thissetofvectorsis

c=a-a=1,b=2,c=-

Example：在 e=,e=,,e=

Thesetofvectorsislinearindependent. Definition:Lineardependence(线性相关Thevectorsv1,v2,,vminavectorspaceVaresaidtolinearlydependentiftherearenontrialchoicesofscalarsforwhichthelinearcombination 在n Rn中，称向量集{v,v,,v} Thesesimpleconceptsprovidethekeytounderstandingthestructureofvectorspaces.DefinitionLinearindependence(线性无关如果它不是线性相关，则称它们是线性无关{v1v2,vm}是指其中任何一个向量都不能由其它向量线性表示Therigorousdefinitionoflinearlyindependence:Thevectorsv1,v2,,vminavectorspaceVaresaidtobeallthescalarsc1,c2,,cmmustequal0.因为方 矩阵和线性方程组的工具研究向量集的有关问题如何判断线性相关,还是线性无关a11 a12 a1n x 21+x

22++x

2n=

1 2

n

n1 n2 nn xa+xa++x = 又转化为系数矩阵的求秩(rank)Proposition1:Thenecessaryandsufficientconditionforthelinearlydependentisthelinearsystemhasnontrivialsolutions.Proposition2:Thenecessaryandsufficientconditionforthelinearlydependentisthattherankofthecoefficientmatrixlessthann.Key{v1v2,,vm}线性相关的充分必要条件是x1v1+x2v2++xmvm=将向量v1,v2,, 按列排成n·m矩阵VV vmVx1v1+x2v2++xmvm={v1v2,vm}r(V{v1v2,vm}r(VDeterminewhetherornotthesetofvectorsis a1=(1,1,1),a2

=

,a3

A= Convertitintoaechelon

0 fi

0r(-1),r(-1)fi

0 0 R(A)=3，所以这个向量集线性无关 Determinewhetherornotthatthesetofvectorsislinearlydependent. a T，a= T，a=

3

3

10

19

~

~

1

1

3 3 ~ ~

raaa= 19

1 0

Thereforethissetofvectorsislinearly a(1,11)Ta(025)Ta(13 解 A= r(-

fi

rA)23fi线性相关Supposethatthesetofvectors{,

}islinearly

aa 1 aLetb1=a1+a2,b2=a2+a3,b3=a1+a3 Showthat{b1,b1,b3}arealsolinear Onthecontrary,wesupposethat{b1,b2,b3}arelinearelythenthereexistconstantsk1,k2andk3thatisnottrivialsuch That k(

)+k(

)+k(+)=0.1a

a2 2a a3 3a (k1+k3)a1+(k1+k2)a2+(k2+k3)a3=0.(k+k)a+(k+k

k1+k3=0,k1+k2=0,k2+k3=0.k1+k3=Thuswehavethelinearsystem k1+k2=0k+k= Thecoefficient

=2„Thislinearsystemhasonlytrivialk1=k2=k3=0. Thus,{b1,b2,b3}arelinearely什么是齐次线性方程组的基础解系(基本解组)

x+ x++ 21

假设下列m,a ,,,a 如果这m这组向量向量是线性无关的Systemofsolutions).Example：Solvethelinear x1+2x2+4x3+x4=2x+4x+8x+ = 3x+6x+ = Solution：

r(-

1 fi 0 r(-

-1 1

5 r(-

3r23

10fi 10

0r21(-4) 00 00 Example：Solvethelinear x1+2x2+4x3+x4=2x+4x+8x+ = 3x+6x+ = Solution：

r(-

1 fi 0 r(-

-1 1

5 r(-

3r23

10fi 10

0r21(-4) 00 00 x1+2x2+4x3+x4=

112x+4x+8x+ =

3

Afi

+ =

0 0

1000

x=- +1x1+2x2-5x4=

fi x3x3x+

x=

=-

Letx2=t1,x4=t2

x=-2t+1 5Thesolutionsofthesystemis

= 10 x=-2t+1

5

=

a=1,b=

0

3t =t 10

-

Thesystemofx=x=

x 1

2=t +t

xtat x 10

2 3 3 x4 0

-

Theconstantst1andt2canbearbitrary x1+2x2+3x3-x4= x+ +4x+ Example: 2x+ +6x+ x1-2x2-3x3+4x4=Solution：Reducethecoefficientmatrixintoreduced

-1r(-1)

3

4 4 r13(-2)fi

fi 4

6

1

- 4

3

0

x1+2x3=

fi

x+1

原方程组化为 22 r(-

4

x1+2x3=

x+x+2 2

+2x4=

x2=-2x

x

1Let

=t

=-

x -

2=t 2

x=x=

x3

1

4 0 1 0

Thesystema11x1+a12

a11x1+a12

x+ x++

x+

x++ 21

21

KeyThesetofsolutionsofthehomogeneoussystemisaThedifferenceoftwosolutionsofnon-homogenneoussystemisasolutionofhomogeneoussystem.Thegeneralsolutionsisthe非齐次方程组通解=非齐次方程组一个特解+x=x0+c1x1++ RankofVectorSet(向量集的秩向量集的 alLinearlyindependent(2)那么这个子集就称为原向量集的一个极大线性无关组 allinearlyindependencesubset)LetV={v1,v2,,vm}beasetofvectors.U={vi1,vi2,,vik}beasubsetofVItissaidthatthesubsetU={,v,, isa allinearlyindependentsubset,ifandonlyif(1)vi1,v12,,vikarelinear(2)Alltheotherscanbelinearlyexpessusevi1,v12,,vik

v=0, =1,v=0,

v1,v2,v3is allinearlyindependentv1,v2,v4isalso allinearlyindependentLet{v1,v2,,vm}beasetof{v1,v2,v3}is allinearlyindependentProof:在向量集中任取4u1u2u3u4 只有平凡解[x Since{v1,v2,v3}is allinearlyindependentSothatwehaveu1=a11v1+a21v2+u4=a14v1+a24v2+a34v3 (a1