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Algebra6:TheRankofTheRankodMatrix矩陣的秩,是矩陣 概念之一性方程組的研究中,矩陣“秩”這個概念有關鍵作用為此,首先介紹矩陣的子(行列)式和代 式的概念
矩陣的子(行列)式
A=
a22a2n
k行、k an1a22amn
kk階子式 -3
A=
-5
det
-3
取第1、3和4行,第2、4和5 得到三階子方陣和3 A=
-5
det
2 Therankofa
A=
2 an1a22amnAk階子方陣是可逆矩陣。k+1階以上的子方陣都是不可逆的,A的秩(rank)k.Ak階子行列式不等于零,但所有k+1階以上的子行列式都等于零。則矩陣的秩等于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 is4a5.Sothat,therankofAnotgreaterthan4.ThereforetherankofAequals一個4行5列的矩陣,它的秩不會超過ThepropertiesofrankofLetAbeamatrixofdimensionman,r(A)£
12
A=
22
r(B)
a25
B=
In
r(A)£r(A)£min{m,
42 Ifr(A)=m,thenthematrixiscalledrowfullrank(行滿秩):Ifr(A)=nthenthematrixiscalledrowcolumnrank(列滿秩)r(AT)=r(nnThenecessaryandsufficientconditionfortheinvertibilityofmatrixAisdet(A)π0.初等變換不改變矩陣的秩r(AB)£min{r(任意梯形矩陣的秩,等于它的非零行數(shù)2020122100000000000000000A=000
,r(A)=B=
r(B)3
就是該矩陣的秩 對矩陣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.方程通解表達式中含有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個向量 x1+2x2+4x3+x4=
1 2x+ +8x+ = 系數(shù)矩陣 =
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
基礎解系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時方程組有非零解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 齊次方程組定理的合理解釋(用例題解釋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
- 于是方程組有無窮多組解.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)線性方程組相容的充分必要條件是r(A)=r(A).(2)線性方程組有惟一解的充分必要條件是rA)rA)n.(3)rA)rAn.例3x1-7x2+14x3-8x4= x-4x+3x
x-3x+4x-
2x1-15x2-x3+
判斷系數(shù)矩陣和增廣矩陣的秩的關系將方程組化為等價方程組,方便求解3424r-2 34455 fi22 33
-2 2 2
244
0 00 0
306系列變換00
r(A)=r(A)=2<60600Thesystemoflinerequationsis33
-2 2 2
244
0 00 0
30600系列變 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移至等式右端作為自由變量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繼續(xù)求解,通解表達式中含有5-3=2個任意常數(shù)rA)rA) aaa 2
a 4
0 5個方程,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系列變 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-當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)和數(shù)乘(scalarmultiplication),(LinearSpace)
是一組n維向量,k,k ab=ka+ a a,a
的一個線性組合(linear如果向量b等于a1,a2,,am b可以由a1,a2,,am線性表示
-
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什么是子空間設VR3中xOy平面上所有向量組成的集合,是R3的一個二維子空間。LR3中經(jīng)過原點的一條直線,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齊次線性方程組的解集合,就是系數(shù)矩陣的零空間(NullN(A)={x?Rn|Ax=x= 平凡解(trivial
=00這個解空間完全由系數(shù)矩陣AN(A).N(A)是所有滿足Ax=0的那些向量組成的集合.A的零空間.Linear(向量集的線性無關性Lineardependence(線性相關 或者成倍數(shù)關系:b=aaora=bb三個向量a和b c=aa+bb在 Rn中,稱向量集{v,v,,v}線性相關 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(線性相關Thevectorsv1,v2,,vminavectorspaceVaresaidtolinearlydependentiftherearenontrialchoicesofscalarsforwhichthelinearcombination 在n Rn中,稱向量集{v,v,,v} Thesesimpleconceptsprovidethekeytounderstandingthestructureofvectorspaces.DefinitionLinearindependence(線性無關如果它不是線性相關,則稱它們是線性無關{v1v2,vm}是指其中任何一個向量都不能由其它向量線性表示Therigorousdefinitionoflinearlyindependence:Thevectorsv1,v2,,vminavectorspaceVaresaidtobeallthescalarsc1,c2,,cmmustequal0.因為方 矩陣和線性方程組的工具研究向量集的有關問題如何判斷線性相關,還是線性無關a11 a12 a1n x 21+x
22++x
2n=
1 2
n
n1 n2 nn xa+xa++x = 又轉(zhuǎn)化為系數(shù)矩陣的求秩(rank)Proposition1:Thenecessaryandsufficientconditionforthelinearlydependentisthelinearsystemhasnontrivialsolutions.Proposition2:Thenecessaryandsufficientconditionforthelinearlydependentisthattherankofthecoefficientmatrixlessthann.Key{v1v2,,vm}線性相關的充分必要條件是x1v1+x2v2++xmvm=將向量v1,v2,, 按列排成n·m矩陣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,所以這個向量集線性無關 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線性相關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什么是齊次線性方程組的基礎解系(基本解組)
x+ x++ 21
假設下列m,a ,,,a 如果這m這組向量向量是線性無關的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
原方程組化為 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非齊次方程組通解=非齊次方程組一個特解+x=x0+c1x1++ RankofVectorSet(向量集的秩向量集的 alLinearlyindependent(2)那么這個子集就稱為原向量集的一個極大線性無關組 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
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