CHAP. 7 Linear .-...L l I L A .Matrices, Determinants. Linear ....\Ic:',.on"l~
Rx == f is inconsistent: No solution is possible. Therefore the system Ax == b is inconsistent as well. See Example 4, where r == 2< m == 3 and ==
13
== 12.If the system is consistent (eitherr == m,orr< m andall the numberslr+l,lr+2, ...
.i«
are zero), then there are solutions.
solution. If the system is consistent and r == n, there is exactly one
u'U' ... ..,... ''U'.I..I..which can be found back substitution. See 2, wherer == n == 3
andm == 4.
many solutions. To obtain any of these choose values of Then solve the rth for xr terms of those then the (r - forXr-l, and so on up the line. See
Gauss elimination is reasonable in time and demand.
We shall consider those in Sec. 20.1 in the on numeric linear _ . Section 7.4 fundamental of linear such as linear moenenuence and rank of a matrix. These in turn be used in Sec. 7.5 to characterize the
hAI'1It:l'{71Arof linear in terms of and of solutions.
o
15. relation. definition, an eauivatence relationon a set is a relation three conditions:
(named asmdtcatcd)
(i) Each element A of the set is to itself or its
Solve the linear system matrix. Show details. then A is equivalent to C (Transitivity).
Show that row equivalence of matrices satisfies these three conditions. Hint. Show that for each of the three elementary row these conditions hold.
SEC. 7.3 Linear of "''-1,..a0. '" """1 I,,;). Gauss Elimination
-the analog of Kirchhoff's Current Law, find -the traffic flow (cars per hour) in the net of one-way streets (in the directions indicated by the arrows) shown in the figure. Is the solution unique?
Models of markets. Determine the equilibrium solution (D1 = Sl'D2 = S2) of the two-commodity marketwith linear model CD,S, P = demand, supply, price; index 1= first commodity, index 2 = second commodity)
PROJECT. Matrices. The idea is that elementarv operations can be accomplished matrix munmncanon. If A is anm Xnmatrix on which we want to do an elementary operation, then there is a matrix E such that EA is the new matrix after the
VIJ'...,.LU\.,.LVJU,.Such anEis called an matrix.
This idea can be for instance, in the design
of it is
prefer-able to do row rather than
multiplication E.)
(a) Show that the following are elementary matrices, for Rows 2 and 3, for adding - 5 times the first row to the third, and for the fourth row 8.
23. + ~
+ means finding integerXl, X 2, X 3, X 4
such that the numbers of atoms of carbon (C), hydrogen (H), and oxygen are the same on both sides of this reaction, in which propane and give carbon dioxide and water. Find the smallest
1200
Net of one-way streets 600
1000
Problem 20
Wheatstone bridge
Traffic Methods of electrical circuit analysis have applications to other fields. For instance, applying 18.
16.. CAS PROJECT.. Gauss Elimination and Back Substitution..Write a program for Gauss elimination and back substitution (a) that does not include pivoting and that does include pivoting. Apply the programs to Probs. 11-14 and to some larger systems of your choice.
In Probs. 17-19, using Kirchhoff's laws (see Example 2) and showing the details, find the currents:
17..
20..Wheatstone Show that if in
the figure, then I = O. of the
instrument which I is This bridge is a method for Rx .R1,R2 ,R3 are known.
is variable. To get makeI = 0 R3 .Then calculate
CHAP. 7 Linear Matrices, Determinants. LinearSvsterns
Apply E1 ,E2 , to a vector and to a 4X 3 matrix of your choice. Find B = lI...:J:-{JiLJ")'liU I.cA.
the general 4 X2 matrix. Is
(b) Conclude that are obtained doing the corresponding eiementarv operations on the 4 X 4
unit matrix. Prove that is obtained from by an elementary row operation, then
M=EA,
where is obtained from the n X n unit matrix by the same rowoneratt.on.
Since our next characterize the behavior of linear in terms
of existence and of solutions we have to introduce new
fundamental linear <:ll llrAh,r'<:l10 0j~nr'Ant"that will aid us in so. Foremost among
these are and the mind that these
lIr1l1-·11!'Y1l'""lf"all"'{Tlinked with Gauss elimination method and how
Given any set ofm vectorsa(1), ... , aCm) the same number of0.n.lnB1'""Ir.nonTCI
COlmtnnatJ.onof these vectors is an.o'V1n1l""Cl.0 011r.n of form
a
where C1, C2, ... , Cm are any scalars. Now consider the v'4'....UI,.,Jl'J.I..1.
this vectorannllnt"-.r.rIl
If this is the
a(1), ... , aCm) are said to form a
holds if we choose all zero, because then it becomes of scalars for which then our vectors tnaenenaent set or, more we call them if also holds with scalars not all zero, we call these vectors This means that we can express at least one of the vectors as a linear combination of the other vectors. For if holds say,
C1
"*
0, we can solve fora(1):a(1) ==
+ ... +
where kj ==may be zero. Or even all of linear
uenenuent, then we can
ifa(1) ==
if a set of vectors is
rid of at least one or more of the vectors until we
a IlIrIlt::lln1l""Iumdenenuent set. This set is then the smallest essential" set with
which we can work. we cannot express any of the vectors, of this set, in terms of the others.
向量空間 線性獨立
線性獨立：
c1, c2, c3...cm全為0 線性相依：
c1, c2, c3...cm不全為0 線性組合
相依性：向量彼此間有特定的關係存在
a1,a2,a3...am可當 作基底(basis)
SEC. 7.4 Linear tnoeoenoence. Rank of a Matrix. Vector
The three vectors
a(1)= [3 0
a(2)= [-6 42
a(3)= [21 -21
2 2J
24 54J
o
-15J are linearly dependent because6a(1)
-Although this is easily checked by vector arithmetic (do it!), it is not so easy to discover. However, a systematic method for finding out about linear independence and dependence follows below.
The first two of the three vectors are linearly independent becausecla(1)+c2a(2)= 0 impliesC2= 0 (from the second components) and thenCl= 0 (from any other component ofa(1).
further discussion will show that the rank of a matrix is an11'Yl1·nn.1I~1"1JI1I'"\1"
1""\1l""",,1""\'::Jl1l"'i"1 OCl of matrices and linear C',,,~rpnlC'
The matrix
=
l~:
0 22J
(2) 42 24 54
-21 0 -15
has rank 2, because Example 1 shows that the first two row vectors are linearly independent, whereas all three row vectors are linearly dependent.
Note further that rank A=0 if and only if A= O. This follows directly from the definition.
can be obtained from if
row-euutvatent to a matrix ctemcntarv row 'U'!-""""..L ... "..L'U'Juu-.>.
Now the maximum number of row vectors of a matrix does not if we the order of rows or a row a nonzero c or take a linear combination of a row to another row. This shows that rank is
in1liTQlri€1lntunder row /"'<1""\01l""""lf"lI,n.n,C'·
Hence we can determine the rank of a matrix the matrix to row-echelon form, as was done in Sec. 7.3. Once the matrix is in row-echelon we count the number of nonzero rows, which is the rank of the matrix.
如何找出係數呢？
到底a1,a2,a3誰是基底？
寫成矩陣型式
列運算後所得之列梯形矩陣，
非全為0的列數即為Rank
經過列運算後之矩陣，與原矩陣形成列等效矩 陣，彼此間有相同的Rank數
CHAP. 7 Linear Vectors, Determinants. Linear svstems
For the matrix in Example 2 we obtain successively
A=
1-:
420 242 5421
L 21 -21 0 - 15....J
l
003 -21420 -14282 -29582J
3-+l
300 4200 28205: J
3The last matrix is in row-echelon form and has two nonzero rows. Hence rank A= 2, as before.
... '<Tr'"""""'"f'...I""''' 1-3 illustrate the"tAlIAlUl1110"useful theorem
the matrix ==
p == 3,n == 3, and the rank of
Consider p vectors that each have n components. Then these vectors are
inaepenaeni
if
the matrix with these vectors as row vectors, has rank p.these vectors are
if
that matrix has rank less than p.Further111('1'nAlr1"r::ll"11"..,...·"..,,11"'.0.,'...1-.. ,.-,." will result from the basic
The rank r a matrix A column vectors
HenceAand its transpose
the maximum number
have the same rank.
In this we write "rows" and "columns" for row and column vectors. Let A be an m X nmatrix of rank A == r.Then definition A has r
rows which we denote v(1), ... , vCr) of their in and all the rows
3(1), ... , 3Cm) of are linear combinations of say,
(3)
3Cm) == Cm1V(1)
+
Cm2v(2)+
進行列運算
R12(2) = R1 × (2) + R2
R13(-7) = R1 × (-7) + R3
R23(1/2) = R2 × (1/2) + R3
不全為0的列數=2
，故Rank A = 2 (1) 第1、2列向量線性獨立，可當 作基底
(2) 兩個基底，故可組成2維向量空間
Rank (A) = Rank (AT) 行運算後所得之行梯形矩陣，
非全為0的行數即為Rank Rank (A)=p，則x(1), x(2),…x(p)線性獨立 Rank (A)<p，則x(1), x(2),…x(p)線性相依
SEC. 7.4 LinearInrilonon/....or"lro Rank of a Matrix. Vector
These are vector a r n l n ....n.r\C\ for rows. To switch to corumns, we write (3) in terms of components asn such with k== 1,"',n,
and collectn.n.1MY'\~,.n."Y\a"Y\11-C\in columns. we can write as
== Vlk
+ ... +
Vrkwherek == 1,·",n. Now the vector on the left is the kth column vector of A. We see that each of thesencolumns is a linear combination of the same columns on the Hence A cannot have more columns than rows, whose number is rank Now rows of are columns of the . For our conclusion is that cannot have more columns than rows, so that cannot have more
mdependent rows than columns. the number of columns
of A must be r, the rank of A. This the
The matrix in (2) has rank 2. From Example 3 we see that the first two row vectors are linearly independent and by "working backward" we can verify that Row 3= 6 Row 1 -
i
Row 2. Similarly, the first two columns are linearly independent, and by reducing the last matrix in Example 3 by columns we find thatColumn 3=
i
Column 1+i
Column 2'.n.1"Y1lhll-nllYl,(1rTheorems 2 and 3 we obtain
and Column 4 =
i
Column 1+~ Column 2.Consider p vectors each n componerus. If n <p, then these vectors are
The matrix A with thosep vectors as row vectors hasp rows andn<p r-.n.lllrY1lnc· hence Theorem 3 it has rank A ~ n <p, which linear Theorem 2.
The related are of interest in linear In the
context a clarification of essential of matrices and their role in connection with linear I;;.'\r';;:tpnl I;;.
(見補充資料)
例題3變成行向量作運算
向量數目(可視為方程式的數目) 向量分量數目(可視為未知數的數目)
向量空間 (見補充資料)
7 Linear Matrices, Determinants. Linear ...."11""''1-,, ...
Consider a set V of vectors where each vector has the same number of for any two vectors a and inV,we have that all their linear combinations aa
+ f3
any real are also elements ofV,and if, andthe laws and in Sec. as well as any vectors a, c in V
then V is a vector space. Note that here we wrote laws and of Sec. 7.1 lowercase b,C, which is our notation for vectors. More on vector spaces in Sec. 7.9.
The maximum number of vectors inVis called the U1
Vand is denoted dimV.Here we assume the dimension to be infinite dimension will be defined in Sec. 7.9.
A set in V of a maximum number of vectors
V is called a basis for V. In other any set of vectors V forms basis for V. That means, if we add one or more vector to that set, the set will
be also the of Sec. 7.4 on linear and
rI,::u'''H::llnrL:::II1,,.,,a of the number of vectors of a basis forV dim V.
The set of all linear combinations of vectors aCl), ... , with the same number of is called the of these vectors. a span is a vector space. If
o.4 ....j.-"-v.-..L''U'..L..L~the vectors aCl), ... , are then form a basis
for that vector space.
This then leads to another definition of basis. A set of vectors is a basis for a vector spaceVif the vectors in the set are
Vcan be as a linear combination of the vectors also say that the set of vectors the vector space V.
a of a vector spaceVwe mean a..LJL'U'.LJL"-'JL..llJLIIJ'" Y
that forms a vector space with to the two ....JLj'- .. "-'U'JL ....-'""-''U'IIJ"-'JL ...JL,JJL·....J \U,'U...i.L.A.V..LA.
..L..L.i.IL,JI...LL.LIJ.L.L",-,U,I,.LVJ,A.Jdefined for the vectors of V.
The span of the three vectors in Example 1 is a vector space of dimension 2. A basis of this vector space consists of any two of those three vectors, for instance,a(l), a(2),ora(l), a(3),etc.
We further note the
The vector space has dimension n.
all vectors with components (n realVJIJ1VWI101l"C'
A basis of n vectors is aCl) == [1 0 1 0
aCn) ==
o
].For a matrix we call the span of the row vectors the of the span of the column vectors of A is called thecotumn
Theorem 3 shows that a matrix A has as many rows as columns. the definition of their number is the dimension of the row space or the column space of A. This proves
The row space and the column space to rankA.
matrixA have the same aimension.
基底
SEC. 7.4 LinearInrilonon,l""'Iov""ro Rank of a Matrix. Vector
for a matrix A the solution set of the nomogeneous vector space, called the space of and its dimension is called the the next section we motivate and prove the basic relation
Ax == 0 is a of In
Find the rank. Find a basis for the row space. Find a basis for the column space. Hint. Row-reduce the matrix and its transpose. (You may omit obvious factors from the vectors
of these rank = rank
= rank AB. (Note the a counterexample.
IfAis not square, either the row vectors or the column vectors ofAare
If the row vectors of a square matrix are so are the column vectors, andr»A"'",,p>'rc p>II"
Give that the rank of a of
matrices cannot exceed the rank of either factor.
Show the-tAIIA'nrlno-' large matrices of low rank mdependent ofn.
7. 0 -1 omit one after another until you get a linearly
set.
Are the sets of vectors Show the details of your work.
25.