线性代数英文讲义

合集下载

线性代数英文讲义

Chapter 1 Matrices and Systems of EquationsLinear systems arise in applications to such areas as engineering, physics, electronics, business, economics, sociology（社会学）, ecology （生态学）, demography(人口统计学), and genetics（遗传学）, etc. §1. Systems of Linear EquationsNew words and phrases in this section:Linear equation 线性方程Linear system，System of linear equations 线性方程组Unknown 未知量Consistent 相容的Consistence 相容性Inconsistent不相容的Inconsistence 不相容性Solution 解Solution set 解集Equivalent 等价的Equivalence 等价性Equivalent system 等价方程组Strict triangular system 严格上三角方程组Strict triangular form 严格上三角形式Back Substitution 回代法Matrix 矩阵Coefficient matrix 系数矩阵Augmented matrix 增广矩阵Pivot element 主元Pivotal row 主行Echelon form 阶梯形1.1 DefinitionsA linear equation (线性方程) in n unknowns（未知量）is1122...n na x a x a x b+++=A linear system of m equations in n unknowns is11112211211222221122...... .........n n n n m m m n n m a x a x a x b a x a x a x b a x a x a x b+++=⎧⎪+++=⎪⎨⎪⎪+++=⎩ This is called a m x n (read as m by n) system.A solution to an m x n system is an ordered n-tuple of numbers （n 元数组）12(,,...,)n x x x that satisfies all the equations.A system is said to be inconsistent （不相容的） if the system has no solutions.A system is said to be consistent （相容的）if the system has at least one solution.The set of all solutions to a linear system is called the solution set（解集）of the linear system.1.2 Geometric Interpretations of 2x2 Systems11112212112222a x a xb a x a x b +=⎧⎨+=⎩ Each equation can be represented graphically as a line in the plane. The ordered pair 12(,)x x will be a solution if and only if it lies on bothlines.In the plane, the possible relative positions are(1) two lines intersect at exactly a point; (The solution set has exactly one element)(2)two lines are parallel; (The solution set is empty)(3)two lines coincide. (The solution set has infinitely manyelements)The situation is the same for mxn systems. An mxn system may not be consistent. If it is consistent, it must either have exactly one solution or infinitely many solutions. These are only possibilities.Of more immediate concerns is the problem of finding all solutions to a given system.1.3 Equivalent systemsTwo systems of equations involving the same variables are said to be equivalent（等价的，同解的）if they have the same solution set.To find the solution set of a system, we usually use operations to reduce the original system to a simpler equivalent system.It is clear that the following three operations do not change the solution set of a system.(1)Interchange the order in which two equations of a system arewritten;(2)Multiply through one equation of a system by a nonzero realnumber;(3)Add a multiple of one equation to another equation. (subtracta multiple of one equation from another one)Remark: The three operations above are very important in dealing with linear systems. They coincide with the three row operations of matrices. Ask a student about the proof.1.4 n x n systemsIf an nxn system has exactly one solution, then operation 1 and 3 can be used to obtain an equivalent “strictly triangular system ”A system is said to be in strict triangular form (严格三角形) if in the k-th equation the coefficients of the first k-1 variables are all zero and the coefficient ofkx is nonzero. (k=1, 2, …,n)An example of a system in strict triangular form:123233331 2 24x x x x x x ++=⎧⎪-=⎨⎪=⎩Any nxn strictly triangular system can be solved by back substitution (回代法).(Note: A phrase: “substitute 3 for x ” == “replace x by 3”)In general, given a system of linear equations in n unknowns, we will use operation I and III to try to obtain an equivalent system that is strictly triangular.We can associate with a linear system an mxn array of numbers whose entries are coefficient of theix ’s. we will refer to this array as thecoefficient matrix (系数矩阵) of the system.111212122212.....................n nm m m n a a a a a a a a a ⎛⎫⎪ ⎪ ⎪ ⎪⎝⎭A matrix （矩阵） is a rectangular array of numbersIf we attach to the coefficient matrix an additional column whose entries are the numbers on the right-hand side of the system, we obtain the new matrix11121121222212n n s m m m na a ab a a a b b a a a ⎛⎫ ⎪ ⎪ ⎪⎝⎭We refer to this new matrix as the augmented matrix （增广矩阵） of a linear system.The system can be solved by performing operations on the augmented matrix. i x ’s are placeholders that can be omitted until the endof computation.Corresponding to the three operations used to obtain equivalent systems, the following row operation may be applied to the augmented matrix.1.5 Elementary row operationsThere are three elementary row operations:(1)Interchange two rows;(2)Multiply a row by a nonzero number;(3)Replace a row by its sum with a multiple of another row.Remark: The importance of these three operations is that they do not change the solution set of a linear system and may reduce a linear system to a simpler form.An example is given here to illustrate how to perform row operations on a matrix.★Example:The procedure for applying the three elementary row operations:Step 1: Choose a pivot element (主元)(nonzero) from among the entries in the first column. The row containing the pivotnumber is called a pivotal row（主行）. We interchange therows (if necessary) so that the pivotal row is the new firstrow.Multiples of the pivotal row are then subtracted form each of the remaining n-1 rows so as to obtain 0’s in the firstentries of rows 2 through n.Step2: Choose a pivot element from the nonzero entries in column 2, rows 2 through n of the matrix. The row containing thepivot element is then interchanged with the second row ( ifnecessary) of the matrix and is used as the new pivotal row.Multiples of the pivotal row are then subtracted form eachof the remaining n-2 rows so as to eliminate all entries belowthe pivot element in the second column.Step 3: The same procedure is repeated for columns 3 through n-1.Note that at the second step, row 1 and column 1 remain unchanged, at the third step, the first two rows and first two columns remain unchanged, and so on.At each step, the overall dimensions of the system are effectively reduced by 1. (The number of equations and the number of unknowns all decrease by 1.)If the elimination process can be carried out as described, we will arrive at an equivalent strictly triangular system after n-1 steps.However, the procedure will break down if all possible choices for a pivot element are all zero. When this happens, the alternative is to reduce the system to certain special echelon form（梯形矩阵）. AssignmentStudents should be able to do all problems.Hand-in problems are: # 7--#11§2. Row Echelon FormNew words and phrases:Row echelon form 行阶梯形Reduced echelon form 简化阶梯形 Lead variable 首变量 Free variable 自由变量Gaussian elimination 高斯消元Gaussian-Jordan reduction. 高斯-若当消元 Overdetermined system 超定方程组 Underdetermined systemHomogeneous system 齐次方程组 Trivial solution 平凡解2.1 Examples and DefinitionIn this section, we discuss how to use elementary row operations to solve mxn systems.Use an example to illustrate the idea.★ Example : Example 1 on page 13. Consider a system represented by the augmented matrix111111110011220031001131112241⎛⎫ ⎪--- ⎪ ⎪-- ⎪- ⎪ ⎪⎝⎭ 111111001120002253001131001130⎛⎫⎪ ⎪ ⎪ ⎪- ⎪ ⎪⎝⎭………..(The details will given in class)We see that at this stage the reduction to strict triangular form breaks down. Since our goal is to simplify the system as much as possible, we move over to the third column. From the example above, we see that the coefficient matrix that we end up with is not in strict triangular form,it is in staircase or echelon form (梯形矩阵).111111001120000013000004003⎛⎫ ⎪ ⎪ ⎪ ⎪- ⎪ ⎪-⎝⎭The equations represented by the last two rows are:12345345512=0 2=3 0=4 03x x x x x x x x x ++++=⎧⎪++⎪⎪⎨⎪-⎪=-⎪⎩Since there are no 5-tuples that could possibly satisfy these equations, the system is inconsistent.Change the system above to a consistent system.111111110011220031001133112244⎛⎫ ⎪--- ⎪ ⎪-- ⎪ ⎪ ⎪⎝⎭ 111111001120000013000000000⎛⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭The last two equations of the reduced system will be satisfied for any 5-tuple. Thus the solution set will be the set of all 5-tuples satisfying the first 3 equations.The variables corresponding to the first nonzero element in each row of the augment matrix will be referred to as lead variable .（首变量） The remaining variables corresponding to the columns skipped in the reduction process will be referred to as free variables （自由变量）.If we transfer the free variables over to the right-hand side in the above system, then we obtain the system:1352435451 2 3x x x x x x x x x ++=--⎧⎪+=-⎨⎪=⎩which is strictly triangular in the unknown 1x 3x 5x . Thus for each pairof values assigned to 2xand4x , there will be a unique solution.★Definition: A matrix is said to be in row echelon form (i) If the first nonzero entry in each nonzero row is 1.(ii)If row k does not consist entirely of zeros, the number of leading zero entries in row k+1 is greater than the number of leading zero entries in row k.(iii) If there are rows whose entries are all zero, they are below therows having nonzero entries.★Definition : The process of using row operations I, II and III to transform a linear system into one whose augmented matrix is in row echelon form is called Gaussian elimination （高斯消元法）.Note that row operation II is necessary in order to scale the rows so that the lead coefficients are all 1.It is clear that if the row echelon form of the augmented matrix contains a row of the form (), the system is inconsistent.000|1Otherwise, the system will be consistent.If the system is consistent and the nonzero rows of the row echelon form of the matrix form a strictly triangular system (the number of nonzero rows<the number of unknowns), the system will have a unique solution. If the number of nonzero rows<the number of unknowns, then the system has infinitely many solutions. (There must be at least one free variable. We can assign the free variables arbitrary values and solve for the lead variables.)2.2 Overdetermined SystemsA linear system is said to be overdetermined if there are more equations than unknowns.2.3 Underdetermined SystemsA system of m linear equations in n unknowns is said to be underdetermined if there are fewer equations than unknowns (m<n). It is impossible for an underdetermined system to have only one solution.In the case where the row echelon form of a consistent system has free variables, it is convenient to continue the elimination process until all the entries above each lead 1 have been eliminated. The resulting reduced matrix is said to be in reduced row echelon form. For instance,111111001120000013000000000⎛⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭ 110004001106000013000000000⎛⎫⎪- ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭Put the free variables on the right-hand side, it follows that12345463x x x x x =-=--=Thus for any real numbersαandβ, the 5-tuple()463ααββ---is a solution.Thus all ordered 5-tuple of the form ()463ααββ--- aresolutions to the system.2.4 Reduced Row Echelon Form★Definition : A matrix is said to be in reduced row echelon form if :(i)the matrix is in row echelon form.(ii) The first nonzero entry in each row is the only nonzero entry in its column.The process of using elementary row operations to transform a matrix into reduced echelon form is called Gaussian-Jordan reduction.The procedure for solving a linear system:(i) Write down the augmented matrix associated to the system; (ii) Perform elementary row operations to reduce the augmented matrix into a row echelon form;(iii) If the system if consistent, reduce the row echelon form into areduced row echelon form. (iv) Write the solution in an n-tuple formRemark: Make sure that the students know the difference between the row echelon form and the reduced echelon form.Example 6 on page 18: Use Gauss-Jordan reduction to solve the system:1234123412343030220x x x x x x x x x x x x -+-+=⎧⎪+--=⎨⎪---=⎩The details of the solution will be given in class.2.5 Homogeneous SystemsA system of linear equations is said to be homogeneous if theconstants on the right-hand side are all zero.Homogeneous systems are always consistent since it has a trivial solution. If a homogeneous system has a unique solution, it must be the trivial solution.In the case that m<n (an underdetermined system), there will always free variables and, consequently, additional nontrivial solution.Theorem 1.2.1 An mxn homogeneous system of linear equations has a nontrivial solution if m<n.Proof A homogeneous system is always consistent. The row echelon form of the augmented matrix can have at most m nonzero rows. Thus there are at most m lead variables. There must be some free variable. The free variables can be assigned arbitrary values. For each assignment of values to the free variables, there is a solution to the system.AssignmentStudents should be able to do all problems except 17, 18, 20.Hand-in problems are 9, 10, 16,Select one problem from 14 and 19.§3. Matrix AlgebraNew words and phrases:Algebra 代数Scalar 数量，标量Scalar multiplication 数乘 Real number 实数 Complex number 复数 V ector 向量Row vector 行向量 Column vector 列向量Euclidean n-space n 维欧氏空间 Linear combination 线性组合 Zero matrix 零矩阵Identity matrix 单位矩阵 Diagonal matrix 对角矩阵 Triangular matrix 三角矩阵Upper triangular matrix 上三角矩阵 Lower triangular matrix 下三角矩阵 Transpose of a matrix 矩阵的转置(Multiplicative ) Inverse of a matrix 矩阵的逆 Singular matrix 奇异矩阵 Singularity 奇异性Nonsingular matrix 非奇异矩阵 Nonsingularity 非奇异性The term scalar (标量，数量) is referred to as a real number (实数) or a complex number （复数）. Matrix notationAn mxn matrix, a rectangular array of mn numbers.111212122212.....................n nm m m n a a a a a a a a a ⎛⎫⎪ ⎪ ⎪ ⎪⎝⎭()ij A a =3.1 VectorsMatrices that have only one row or one column are of special interest since they are used to represent solutions to linear systems.We will refer to an ordered n-tuple of real numbers as a vector （向量）.If an n-tuple is represented in terms of a 1xn matrix, then we will refer to it as a row vector . Alternatively, if the n-tuple is represented by an nx1 matrix, then we will refer to it as a column vector . In this course, we represent a vector as a column vector.The set of all nx1 matrices of real number is called Euclidean n-space (n 维欧氏空间) and is usually denoted by nR.Given a mxn matrix A, it is often necessary to refer to a particular row or column. The matrix A can be represented in terms of either its column vectors or its row vectors.12(a ,a ,,a )n A = ora (1,:)a(2,:)a(,:)A m ⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭3.2 EqualityFor two matrices to be equal, they must have the same dimensions and their corresponding entries must agree★Definition : Two mxn matrices A and B are said to be equal ifij ij a b =for each ordered pair (i, j)3.3 Scalar MultiplicationIf A is a matrix,αis a scalar, thenαA is the mxn matrix formed by multiplying each of the entries of A byα.★Definition : If A is an mxn matrix, αis a scalar, thenαA is themxn matrix whose (i, j) is ij a αfor each ordered pair (i, j) .3.4 Matrix AdditionTwo matrices with the same dimensions can be added by adding their corresponding entries.★Definition : If A and B are both mxn matrices, then the sum A+B is the mxn matrix whose (i,j) entry isij ija b + for each ordered pair (i, j).An mxn zero matrix (零矩阵) is a matrix whose entries are all zero. It acts as an additive identity on the set of all mxn matrices.A+O=O+A=AThe additive of A is (-1)A since A+(-1)A=O=(-1)A+A.A-B=A+(-1)B-A=(-1)A3.5 Matrix Multiplication and Linear Systems3.5.1 MotivationsRepresent a linear system as a matrix equationWe have yet to defined the most important operation, the multiplications of two matrices. A 1x1 system can be writtena xb =A scalar can be treated as a 1x1 matrix. Our goal is to generalize the equation above so that we can represent an mxn system by a single equation.A X B=Case 1: 1xn systems 1122... n n a x a x a x b +++=If we set()12n A a a a =and12n x x X x ⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭, and define1122...n n AX a x a x a x =+++Then the equation can be written as A X b =。

麻省理工上课讲义线性代数[第1集]An overview of key ideas

Vectors
What do you do with vectors? Take combinations. We can multiply vectors by scalars, add, and subtract. Given vectors u, v and w we can form the linear combination x1 u + x2 v + x3 w = b. An example in R3 would be: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 u = ⎣ −1 ⎦ , v = ⎣ 1 ⎦ , w = ⎣ 0 ⎦ . −1 1 0 The collection of all multiples of u forms a line through the origin. The collec tion of all multiples of v forms another line. The collection of all combinations of u and v forms a plane. Taking all combinations of some vectors creates a subspace. We could continue like this, or we can use a matrix to add in all multiples of w.
Subspaces
Geometrically, the columns of C lie in the same plane (they are dependent; the columns of A are independent). There are many vectors in R3 which do not lie in that plane. Those vectors cannot be written as a linear combination of the columns of C and so correspond to values of b for which C x = b has no solu tion x. The linear combinations of the columns of C form a two dimensional subspace of R3 . This plane of combinations of u, v and w can be described as “all vectors C x”. But we know that the vectors b for which C x = b satisfy the condition b1 + b2 + b3 = 0. So the plane of all combinations of u and v consists of all vectors whose components sum to 0. If we take all combinations of: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 u = ⎣ −1 ⎦ , v = ⎣ 1 ⎦ , and w = ⎣ 0 ⎦ 0 −1 1 we get the entire space R3 ; the equation Ax = b has a solution for every b in R3 . We say that u, v and w form a basis for R3 . A basis for Rn is a collection of n independent vectors in Rn . Equivalently, a basis is a collection of n vectors whose combinations cover the whole space. Or, a collection of vectors forms a basis whenever a matrix which has those vectors as its columns is invertible. A vector space is a collection of vectors that is closed under linear combina tions. A subspace is a vector space inside another vector space; a plane through the origin in R3 is an example of a subspace. A subspace could be equal to the space it’s contained in; the smallest subspace contains only the zero vector. The subspaces of R3 are: 3

线性代数英文课件：ch1_3 Cofactor Expansion

a11 a12 a13 行列式给定后，某元素
D a21 a22
a23
的余子式、代数余子式和该元素本身无关，只
a31 a32 a33 和行列式的其他元素有关
M 23
a11 a31
a12 a32
is the complement minor of a23
A23
(1)23
a11 a31
a12 a32 is the cofactor of a23
3 1 1 2
5 1
3
4 8 r2 r1 r4 5r1
0
4
6
D
2 0 1 1
2 0 1 1
1 5 3 3 16 0 2 7
8 (1)12 1 2
4 1
6
c1 c3
2cTec2l2heemreeni1st i6onnclyolou4nmenn2o,nsz2oero
1 weexp0anse th1e 0
【 Corollary(推论)】
ai1 Aj1 ai2 Aj2 ain Ajn 0 when i j;
a1i A1 j a2i A2 j ani Anj 0 when i j;
For example,
2 1 3
A1 2 1 ,
23 4
then A11 2 A12 A13 0.
Math. Dept., Wuhan University of Technology
Proof: For determinant
a11 a12
a1n
The cofactor of a j1 ,a j2 , ,a jn
a21 a22
a2n
are Aj1 , Aj2 , , Ajn
DD1 ai1 ai 2

线性代数英文讲义

Definition
A matrix is said to be in reduced row echelon form if: ⅰ. The matrix is in row echelon form. ⅱ. The first nonzero entry in each row is the only nonzero entry in its column.
n×n Systems Definition
A system is said to be in strict triangular form if in the kth equation the coefficients of the first k-1 variables are all zero and the coefficient of xk is nonzero (k=1, …,n).
1×n matrix
column vector
x1 x2 X x n
n×1 matrix
Definition
Two m×n matrices A and B are said to be equal if aij=bij for each i and j.
1 1
Matrix Multiplication and Linear Systems
Case 1 One equation in Several Unknows
If we let A (a1 a2 an ) and
Example
x1 x2 1 (a ) x1 x2 3 x 2 x 2 2 1 x1 x2 x3 x4 x5 2 (b) x1 x2 x3 2 x4 2 x5 3 x x x 2 x 3x 2 4 5 1 2 3

线性代数英文课件：ch3-1 Elementary Operations

(注：增广矩阵化为最简形时，线性方程组的解亦求出)
下列哪些矩阵是行最简形？
1 0 3 3
1 0
0
1
0
2
?
0 0 1 1
0
1
0 0
0
0
0
0
0
0
0 3
0
2
?
1 1
2
0
1 0
0
1
0 0
0
0
0 3
0
2
?
1 1
0
1Leabharlann 1 0 0 301
0
2
0 0 0 0
√
0
0
0
0
1 0 0 3
I. ri rj : Interchange row i and row j.
II. kri (or k ri ) : Multiply the ith row by a nonzero
.scalar k.
III. ri krj : Add k times the jth row to the ith row (i j)
0
0
Echelon Form Matrix
Definition 4 A matrix is said to be in row echelon
form(行阶梯型矩阵), or simply an echelon matrix, if:
I. The zero rows, if any, are below all nonzero rows and
1. Elementary Operations and Gaussian Elimination Method
The Cramer’s Rule:前提：

线性代数讲义

线性代数（Linear Algebra ）引言（Introduction ）1. 数学数学（數學、mathematics ）在我国古代叫算（筭、祘）术，后来叫算学或数学；直到1939年6月，为了划一才确定统一用数学．“数学是研究现实世界的量的关系和空间形式的科学”，分为代数、几何等．2. 代数代数(algebra)分为古典代数和近世代数．古典代数(ancient algebra)基本上就是方程论，以方程的解法为中心．如：一元一次方程 )0(≠=a b ax 的解为b a x 1-=；一元二次方程 )0(02≠=++a c bx ax 的解为)2/()4(22,1a ac b b x -±-=；一元三、四次方程也有类似的求根公式（16世纪）；但是，一元n 次方程当n ≥5时却无一般的“求根公式”（参见数学史或近代数）；根式求解条件的探究导致群概念的引入，这最早出现在Lagrange 1770年和1771年的著作中；1799年Ruffini 给出“证明”（群论思想）；Abel 进一步给出严格的证明，开辟了近世代数方程论的道路（1824年和1826年），包括群论和方程的超越函数解法；Galois 引入代换群彻底解决了代数方程根式可解的条件，开辟了代数学的一个崭新的领域——群论．从而使代数的研究对象转向研究代数结构本身，此即近世代数．近世代数(modern algebra)又称抽象代数（abstract algebra ）包括代数数论、超复数系、线性代数、群论、环论、域论、格论、李（Lie ）群、李代数、代数几何、代数拓扑，等等． 3. 线性代数如果保持一元一次方程中未知量的指数（一次的）不变，而增加未知量及方程的个数，即得到线性（一次）方程组．先看下面三个例子：例1 （《孙子算经》卷下第31题）“今有雉兔同笼，上有三十五头，下有九十四足．问雉、兔各几何？答曰：雉二十三，兔一十二．”解法1 设雉、兔分别为x ，y ，则由⎩⎨⎧=+=+944235y x y x 解得⎩⎨⎧==1223y x .解法2 ⎪⎪⎭⎫ ⎝⎛9435足头⎪⎪⎭⎫⎝⎛−−→−⎪⎪⎭⎫ ⎝⎛−−→−⎪⎪⎭⎫ ⎝⎛−−→−122312354735兔雉兔头半足头再作差作差半其足 . 解法 3 请兔子全“起立”后，雉兔总“足”数为70235＝⨯，从而得兔“手”数为94－70＝24，于是兔子数为24÷2＝12，雉数为35－12＝23 .注：后两种解法心算即可．例2 某厂用四种原料生产五种产品，各产品的原料成份及各原料的用量为表1所示，求每种产品的产量（千克）．表1 各产品的原料成份（％）及各原料的用量（千克）解设A,B,C,D,E 五种产品的产量分别为X i （i ＝1，2，3，4，5），则问题归结为求解方程组 ⎪⎪⎩⎪⎪⎨⎧=++++=++++=++++=++++6001.06.01.02.01.07807.01.03.01.02.02501.02.02.06.04.01001.01.04.01.03.054321543215432154321X X X X X X X X X X X X X X X X X X X X这是一个含五个未知量、四个方程的方程组．例3 某商店经营四类商品，四个季度的销售额及利润额如表2所示．求每类商品的年平均利润率（％）．表2 各类商品四个季度的销售额及利润额（单位：元）解设四类商品A,B,C,D 的利润率分别为X i （i ＝1，2，3，4），则问题归结为解下面含四个未知量、四个方程的方程组 ⎪⎪⎩⎪⎪⎨⎧=+++=+++=+++=+++955005002503009075040030016085800500100200806003002002504321432143214321X X X X X X X X X X X X X X X X .现实中的很多问题，往往归结为求解含多个未知量（数）的一次方程组，称为线性方程组，其一般形式为 ⎪⎪⎩⎪⎪⎨⎧=+++=+++=+++mn mn m m n n n n b x a x a x a b x a x a x a b x a x a x a 22112222212111212111 .此类线性方程组可能有解，也可能无解；有解时可能只有一组解，也可能有多组甚至无穷多组解，如⑴⎩⎨⎧=-=+226132121x x x x 有唯一解⎩⎨⎧==03/121x x ; ⑵⎩⎨⎧=-=-226132121x x x x 有无穷多解⎩⎨⎧-==1321c x cx （其中c 为任意常数） ; ⑶⎩⎨⎧=-=-426132121x x x x 无解 .那么，如何判定一个给定的线性方程组有没有解？如果有解，究竟有多少组解（一组、多组、无穷多组）？这些解又怎样求（表示）出来？如果无解，又怎么办？因为无解的方程组如果是某一有解的实际问题的数学抽象，此时又如何（用这一线性方程组来）描述它所表示的实际问题的解（“广义解”）？这就要求我们研究解决线性方程组有解的判定条件、解法、解的结构与解的表示以及“广义解”等问题，这些都是线性代数所要解决的问题．线性代数( Linear algebra )是从线性方程组、行列式和矩阵等理论中产生出来的，是代数各分支中应用最广泛的分支．在历史上首先应归功于英国的J.J.Sylvester 、A.Cayley 、美国的Peirce 父子和L.E.Dickson 等人的工作．主要内容：行列式、矩阵、线性方程组、向量空间、相似矩阵及二次型等；主要方法：初等变换法、降阶法、分块法、标准形法、特征值法等．下面我们将分别介绍．当然我们这里所介绍的只是线性代数中最基本的内容，还有很多内容(如矩阵论或矩阵分析等)要等到我们进一步深造时再学；而且线性代数本身也是在不断发展的．参考书[1] 线性代数(第三版、第四版)，同济大学数学教研室编，高等教育出版社． [2] 线性代数（居余马等编）、线性代数与解析几何（俞正光等编）、线性代数辅导（胡金徳等编），清华大学出版社． [3] 线性代数（陈龙玄等编）、线性代数（李炯生等编），中国科技大学出版社． [4] 线性代数解题方法技巧归纳，毛纲源编，华中理工大学出版社． [5] 线性代数方法导引，屠伯埙编，复旦大学出版社． [6] Linear Algebra(UTM)，By L.Smith ，Springer-V erlag ．．．．第一讲行列式 ( Determinant )教学目的与要求：了解n 阶行列式的概念，掌握行列式的性质和二、三阶行列式的计算方法，会应用行列式的性质简化n 阶行列式的计算，会用Cramer 法则解线性方程组．重点：n 阶行列式的概念、性质与计算§1 二、三阶行列式（复习与总结）一、2阶行列例1 求下列二元一次方程组的解：(1) ⎩⎨⎧=+=+②①9442352121x x x x ；（2）⎩⎨⎧=+=+②① 22221211212111b x a x a b x a x a ……(1)（其中)021122211≠-=a a a a D ．解 (1) )1(4-⨯+⨯②①得,2346211=⇒=x x1)2(⨯+-⨯②①得1224222=⇒=x x ．(2) )(1222a a -⨯+⨯②①得121222111)(x b a a b D Dx ⇒-===D 1/D ，1121)(a a ⨯+-⨯②①得=⇒-==221121122)(x a b b a D Dx D 2/D ．为使⑴的解表示简单，Leibniz 于18世纪初引入2阶行列式的定义如下：定义设有4个元素（数）排成的两行（row ）、两列（column ）的22211211a a a a ，称为一个2阶行列式，其值为a 11a 22－a 12a 21，即2112221122211211a a a a a a a a -=．如例1（2）中的D=22211211a a a a 称为方程组⑴的系数行列式，而2221211a b a b D =，2211112b a b a D =；（1）中的24942351,46494135,2421121======D D D ．例２计算2315－=D ．解 1331252315＝）（－＝－⨯-⨯=D ．例３设132λλ=D ，问λ为何值时，（1）D = 0，（2）D ≠0？解因D =λ2－3λ＝λ（λ－3），故（1）当λ＝0或3时，D = 0；（2）当λ≠0，3时，D ≠0．例4 设1221--=k k D ，则D ≠0的充要条件是（）答：k ≠－1，3．(因D =(k －1)2－4＝（k ＋1）（k －3），故D ≠0的充要条件是k ≠－1，3) 例5 如果1222112110==a a a a D ，则下列（）是⎩⎨⎧=+-=+-0022221211212111b x a x a b x a x a 的解．(A)22111122221211b a b a x a b a b x ==，; (B)22111122221211b a b a x a b a b x =-=，;(C)22111122221211b a b a x a b a b x ----=----=，； (D)22111122221211b a b a x a b a b x ---=-----=，．答：（）(因原方程组即⎩⎨⎧-=-+-=-+22221211212111)()(b x a x a b x a x a 的系数行列式1022211211-=-=--=D a a a a D ，2221212221211a b a b a b a b D =----=，2211112211112b a b a b a b a D -=--=)二、3阶行列式例６求解下列三元一次方程组：（１） ⎪⎩⎪⎨⎧=++=++=++③②①333323213123232221211313212111b x a x a x a b x a x a x a b x a x a x a (2)（其中)0322311332112312213322113312312332211≠---++=a a a a a a a a a a a a a a a a a a D ；（２） ⎪⎩⎪⎨⎧-=++-=++=-③②① 232131232132131x x x x x x x x .解（1）记3332232211a a a a A =，3331232112a a a a A -=，3231222113a a a a A =, 3332131221a a a a A -=，3331131122a a a a A =,3231121123a a a a A -=,2322131231a a a a A =,2321131132a a a a A -=,2221121133a a a a A =，则： ①×A 11＋②×A 21＋③×A 31得D X 1=D 1（=b 1A 11+b 2A 21+b 3A 31）, X 1=D 1/D ， ①×A 12＋②×A 22＋③×A 32得D X 2=D 2（=b 1A 12+b 2A 22+b 3A 32）, X 2=D 2 /D ， ①×A 13＋②×A 23＋③×A 33得D X 3=D 3（=b 1A 13+b 2A 23+b 3A 33）, x 3=D 3/D ；(2) D=1+0-6-4+0-9=-18，23120A 61320A 81331A 312111＝－＝，＝－＝，＝－＝--, ，＝－－＝，＝－－－＝，＝－－＝53121A 31221A 71231A 322212--11101A 33201A 53211A 332313＝＝，＝－－＝，＝－＝-,①×A 11＋②×A 21＋③×A 31得－18x 1＝－18 ⇒x 1＝1， ①×A 12＋②×A 22＋③×A 32得－18x 2＝0 ⇒x 2＝0， ①×A 13＋②×A 23＋③×A 33得－18x 3＝0 ⇒x 3＝0．定义设有9个元素（数）排成的3行、3列的333231232221131211a a a a a a a a a 称为一个三阶行列式，其值为322311332112312213322113312312332211a a a a a a a a a a a a a a a a a a ---++．如例6中的D 即称为方程组的系数行列式．2、3阶行列式的值（代数和）可用沙路法（或对角线法则）来记忆：211222112112221122211211a a a a a a a a a a a a -=+=,322311332112312213322113312312332211333231232221131211a a a a a a a a a a a a a a a a a a a a a a a a a a a +++++==322311332112312213322113312312332211a a a a a a a a a a a a a a a a a ---++α;或在图333231232221131211333231232221131211a a a a a a a a a a a a a a a a a a 上操作．例7 计算 61504321-=D ．解58051642)1(03043)1(5260105164210343152601-=⋅⋅-⋅⋅--⋅⋅-⋅⋅+-⋅⋅+⋅⋅=-+-++-+=D例8 （1）010000=-=a bb aD 的充要条件是（）答：022=+b a ．（因为22b a D +=）（2）0114011>=a a D 的充要条件是（），其中R a ∈．答：1012>>-a a 或．（因为12-=a D ）（3）01110212=-=k kD 的充要条件是（）（A ）k ＝2；（B ）k ＝－2；（C ）k ＝0；（D ）k ＝3．答：（B ）或（D ）．（因为)3)(2(64222-+=--=---=k k k k k k D ）例９计算下列行列式的值（1）749651823=D ；（2）768452913'=D解（1）201721436032108105749651823-=---++==D ；（2）201147236010832105768452913'-=---++==D ．三、3阶行列式的性质 (由定义易验证，对2阶也成立且验证更易)性质1 D T =D . 其中D T 为将D 的行与列互换后所得的行列式，即如果333231232221131211a a a a a a a a a D =，则332313322212312111a a a a a a a a a D T =； D T 有时也记为D ˊ，称为行列式D 的转置行列式．此性质说明在（二、三阶）行列式中行、列等位．因此凡对行（或列）成立的性质对列（或行）也成立．性质2 交换两行(或列)使行列式仅变号，即有333231232221131211333231131211232221a a a a a a a a a a a a a a a a a a -=等．对换第i 行（列）与第j 行（列）记为（r i ，r j ）（（c i ，c j ））．推论两行（或列）相同的行列式值为0，即有0232221131211131211=a a a a a a a a a 等．性质3 行列式中某一行（或列）的公因子可以提到行列式外面来，即有333231232221131211333231232221131211a a a a a a a a a k a a a ka ka ka a a a =等．推论1 用数k 乘以某行列式相当于用k 乘以该行列的某一行（或列）．以k 数乘以第i 行（列）记为)(i ic k r k ⋅⋅．推论2 某一行（或列）全为0的行列式的值为0．推论3 有两行（或列）成比例的行列式的值为0．如0333231131211131211333231131211131211==a a a a a a a a a k a a a ka ka ka a a a 性质4 若行列式的某一行（或列）的每个元素都是两个元素之和，则此行列式可按此行（或列）分拆成两个行列式之和．如=+++=333231232322222121131211a a a cbc b c b a a a D D 1＋D 2，其中3332312322211312111a a a b b b a a a D =，3332312322211312112a a a c c c a a a D =．性质5 将某一行（或列）各元素的同一数倍加于另一行（或列）相应的元素上去，不改变行列式的值，即有333231232221131211333231132312221121131211a a a a a a a a a a a a la a la a la a a a a =+++等．将第j 行（列）的l 倍加到第i 行（列）记为 r i ＋⋅l r j （ c i ＋⋅l c j ）．注：性质2、3和5中的变换：对换两行（或列）、以非零常数乘某行（或列）和把某行（或列）的常数倍加到另一行（或列）上去，分别称为第一、第二和第三类初等行（或列）变换（详见第二讲§5）．性质6（按行（或列）展开定理）（1）∑====31333231232221131211)3,2,1(j ij ij i A a a a a a a a a a a D ，即333332323131232322222121131312121111A a A a A a A a A a A a A a A a A a D ++=++=++=；（2）∑==31i ij ijA aD （j=1,2,3）, 即313121211111A a A a A a D ++=333323231313323222221212A a A a A a A a A a A a ++=++=（其中A ij 如例6所示：ij ji ij M A +-=)1(，M ij 是将Ｄ中a ij 所在的第i 行和第j 列全划掉余下的二阶行列式，称为a ij 在D 中的余子式，而A ij 称为a ij 在D 中的代数余子式．）例10 计算下列行列式的值（1）151413---=kk D ；（2）12121-=k k kD ．解（1）)3)(1(1430114004315140132321++=-+-=-+++---=k k k k k k r r r r k k D ；；（2）)2(222020021121211312k kk k kkr r r r kk kD --=--=-----=；．性质7（代数余子式的性质）（1）D A a ik j kj ij δ=∑=31（其中⎩⎨⎧≠=ki 0ki 1，＝，δik 为Kronecker 记号．当i ＝k 时即为性质6（1）；当i ≠k ，如i ＝1，k ＝2时，0A 231322122111=+A a A a a ＋等）．（2）D A ajl i il ijδ=∑=31（当j ＝l 时即为性质6（2）；当j ≠l , 如j ＝1，l ＝2时， 0A 323122211211=+A a A a a ＋等）．例11 求132311201--=D 的值，并验证性质7．解 D 的23120A 61320A 81331A 312111＝－＝，＝－＝，＝－＝-- ，＝－－＝，＝－－－＝，＝－－＝53121A 31221A 71231A 322212--,11101A 33201A 53211A 332313＝＝，＝－－＝，＝－＝-(1) 按第1列展开得=⋅-⋅+⋅=312111211A A A D 1×(-8)+1×(-6)-2×2=-18；(2) 023)6(1)8(0310312111=⋅+-⋅+-⋅=⋅+⋅+⋅A A A ；其余类似．四、Cramer 法则1．一般情形由例1和例6即得定理（Cramer 法则）（1）二元一次线性方程组 ⎩⎨⎧=+=+②① 22221211212111b x a x a b x a x a …（1）当其系数行列式D=22211211a a a a ≠0时有唯一解D D x j j =（j ＝1，2）；（2）三元一次线性方程组⎪⎩⎪⎨⎧=++=++=++③②①333323213123232221211313212111b x a x a x a b x a x a x a b x a x a x a …（2）当其系数行列式D=333231232221131211a a a a a a a a a ≠0时有唯一解 D D x j j =（j ＝1，2，3）．例12(例6(2)的解法2 ) 18132311201＝－--=D ，D 1＝D ＝－18， 01223112112=---=D ，⎪⎩⎪⎨⎧⇒--0x 0x 1x 023211111D 3322113＝＝＝＝＝＝＝＝D D D D D D ．注：两种解法本质是一样的，只不过解法2是直接用Cramer 法则的结果（公式），而原解法是把消元（或Cramer 法则的证明）过程再写一遍．2．齐次情形推论奇次线性方程组 ⎪⎩⎪⎨⎧=++=++=++000333232131323222121313212111x a x a x a x a x a x a x a x a x a （2ˊ）当其系数行列式D ≠0时只有零解（ｘ1＝ｘ2＝ｘ3＝0）．以后将证明此推论的逆也成立，于是有命题（1）奇次线性方程组（2ˊ）只有零解⇔D ≠0；（2）奇次线性方程组（2ˊ）有非零解⇔D ＝0．例13 λ取何值时，奇次线性方程组⎪⎩⎪⎨⎧==-+=++00023321321x x x x x x x λλλ，有非零解？解因为)1(0011212-=-=λλλλλD ，故当λ＝0或±1时，该方程组有非零解．例14 如果方程组⎪⎩⎪⎨⎧=--=+=-+05403z y kx o z y z ky x 有非零解，则（）．（A ） k ＝0；（B ）k ＝1：（C ）k ＝－1；（D ）k ＝－3．答：（C ，D ）．(由例10（1）)3)(1(1514013++=---=k k kk D 即得) ．例15 当（）时，奇次线性方程组⎪⎩⎪⎨⎧=+-=++=+02020z y kx z ky x z kx 仅有零解；（A ） k ＝0；（B ）k ＝－1；（C ）k ＝2；（D ）k ＝－2．答：（A ，B ，D ）．(由例10（2）)2(2121210k kk kD --=-=即得) ．§2 全排列及其逆序列问题：行列式可否归纳定义 212112221111222112112)1()1(a a a a a a a a D ⋅-⋅+⋅-⋅==++，当n ≥2时，n n nn ijn A a A a A a a D 1112121111+++==⨯ ，其中j j j M A 111)1(+-=，M 1j 为a 1j 在D n 中的余子式（n －1阶行列式）？一、全排列例1 用1、2、3三个数字可以组成多少个没有重复数字的三位数？请写出．解共6个，分别为123，132，213，231，312，321．把n 个不同的元素（不妨设为1，2，…，n ）排成一列，叫做这n 个元素的一个（n 级）全排列，n 个不同元素的全排列的种数为P n ＝n!，如P 3＝3!＝6，P 4＝4!＝24，P 5＝5!=120等．记S n 为1，2，…，n 的所有n 级排列所组成的集合，即S n ＝{（j 1j 2…j n ）| (j 1j 2…j n )为n 级排列}，则|S n |= n!．二、逆序与逆序数1. 标准排列：对n 个不同的元素，先规定一个标准次序（如对1，2，…，n ，规定从小到大的次序为标准次序），从而得到一个标准排列．对1，2，…，n ，今后规定其标准排列为自然排列1 2 … (n －1) n ．2.逆序与逆序数在一个n 级排列中，当两个元素a 和b 的先后次序与标准顺序不同时，则说a 和b 形成一个逆序；一个排列中所有逆序的总数叫该排列的逆序数．排列p 1p 2…p n 的逆序数记为 t （p 1p 2…p n ）．逆序数为奇（或偶）数的排列称为奇（或偶）排列．例2 （1）2个2级排列12和21，一个为奇排列（21），一个为偶排列（12）．（2）3级排列的逆序数表（6个3级排列中奇、偶排列各3个）三、逆序数的求法不妨设n 个元素为1，2，…，n ，其标准排列为自然排列1 2 … n ，设p 1p 2…p n 为1，2，…，n 的一个排列，记t i =t(p i )为排列p 1p 2…p n 中p i 左（前）面的比p i 大的元素的个数，s i ＝s （p i ）为排列p 1p 2…p n 中p i 右（后）面的比p i 小的元素的个数，简记t(p 1p 2…p n )为t ，则(1)t(1)t(2) t(n)t(n) t(2)t(1)t t t t n 21+++=+++=+++= ；(2)s(1)s(2) s(n)s(n) s(2)s(1)s s s n 21+++=+++=+++= t ．注：显然0(1)t(n)t 1====s s n ， t(1 2 … n)=0．例３求（1）t=t(32514)；（2）t(7632451)；（3）t(2 3 … n 1)；（4）t(n (n-1) … 2 1) ．解（1）因t 1=t(3)=0，t 2=t(2)=1，t 3=t(5)=0，t 4=t(1)=3，t 5=t(4)=1⇒ t =5（奇），或因s 1=s(3)=2，s 2=s(2)=1，s 3=s(5)=2，s 4=s(1)=0，s 5=s(4)=0⇒ t =5．（2）t(7632451)=0+1+2+3+2+2+6=16，或=6+5+2+1+1+1+0=16（偶）．（3）t(2 3 … n 1)=0+0+…+0+(n-1)= n-1，或=1+1+…+1+0= n-1．（4）t(n … 2 1)=0+1+…+(n-2)+(n-1)=2)1(-n n ，或=(n-1)+(n-2)+…+1+0=2)1(-n n ．如62341) 2 3 t(4=⋅=，102451) 2 3 4 t(5=⋅=，152561) 2 3 4 5 t(6=⋅=，… 再看6个3级排列的逆序数：t(123)=0，t(132)=0+0+1=1，或=0+1+0=1；t(213)=0+1+0=1，或=1+0+0=1；t(231)=0+0+2=2，或=1+1+0=2；t(312)=0+1+1=2，或=2+0+0=2；t(321)=0+1+2=3，或=2+1+0=3．四、对换及其性质１．对换：在一个排列中，互换某两个元素（如i ，j ）的位置，而其余的元素不动，叫做对该排列的一次对换，记为（i ，j ）；互换相邻两个元素的对换叫做相邻对换．例４（321）0)123(,3)321(),123(3,1(==−−→−t t ）；（7632451）−−→−)1,6(（7132456），t(7632451)=16（偶）， t(7132456)=0+1+1+2+1+1+1=7（奇）．２．性质性质1 一次对换改变排列的奇偶性．证明（1）相邻对换改变排列的奇偶性：设（…a b …）−−→−),(b a （…b a …）因对换（a ，b ）只改变了a 和b 之间的逆序：当a<b 时，经对换后逆序数增加1；当a>b 时，经对换后逆序数减少1．而a 或b 与其他元素，以及其他元素之间的逆序数经对换后都没有改变，故相邻对换改变排列的奇偶性．（2）任一对换可由奇数次相邻对换而得到，从而改变奇偶性：（…ac 1c 2…c s b …）−−→−),(1c a(…c 1ac 2…c s b …) −−→−−−→−),(),(2sc a c a （…c 1c 2…c s ab …）−−→−),(b a （…c 1c 2…c s ba …）−−→−),(b c s (…c 1c 2…bc s a …)−−→−−→−),(2b c ( …c 1bc 2…c s a …)−−→−),(1b c (bc 1c 2…c s a …)，共2s+1次．例４（1）对换（7632451）−−→−)1,6(（7132456）可由9次相邻对换得到，相应的逆序变化为：（16=）t(7632451)= t(7362451)+1= t(7326451)+2= t(7324651)+3= t(7324561)+４ = t(7324516)+5=( t(7324156)+6= t(7321456)+7= t(7312456)+8= ) t(7132456)+9 (=7+9)．（2）(7=) t(7132456)= t(1324567)+6= t(1234567)+7 (=0+7)．（3）(16=) t(7632451)= t(6324517)+6= t(3245167)+11= t(3241567)+12= t(3214567)+13=t(2134567)+15=t(1234567)+16(=0+16)．性质2 （1）任一排列（p 1p 2…p n ）总可经有限次（相邻）对换成标准排列，且所作对换的次数k 与该排列有相同的奇偶性，即k 与t (p 1p 2…p n )奇偶性相同；（2）任一排列（p 1p 2…p n ）都可由标准排列1 2 … n 经有限次（相邻）对换而得到，且所作对换的次数k 与该排列有相同的奇偶性，即k 与t (p 1p 2…p n )奇偶性相同．（3）S n 中的任意两个n 级排列均可经有限次（相邻）对换而互相得到；且若这两个排列的奇偶性相（或不）同，则所作对换的次数为偶（或奇）数．证（1）对排列的阶n 归纳．当n=1时显然成立．假设结论对n －1已经成立，则对n ： ①若排列为p 1 … p n-1 n ，由归纳假设n －1级排列p 1 … p n-1可经有限次对换成为标准排列1 2 … （n －1），且所作对换的次数与t(p 1…p n-1)有相同的奇偶性，从而p 1…p n-1 n 经上述对换即成为标准排列1 2 …(n －1) n ，且所作对换的次数的次数与t(p 1…p n-1 n)＝t(p 1…p n-1)有相同的奇偶性．②若排列为（p 1…p i-1 n p i+1…p n ），则可经n －i 次相邻对换成为（p 1…p i-1p i+1…p n n ），且t （p 1…p i-1 n p i+1…p n ）=t （p 1…p i-1p i+1…p n n ）＋（n －i ），而由①得p 1…p i-1p i+1…p n n 可经有限次对换成为1 2 …(n －1) n ，且所作对换的次数m 与t （p 1…p i-1p i+1…p n n ）有相同的奇偶性，于是p 1…p i-1 n p i+1…p n 可经m ＋n －i 次对换成为1 2 … n ，且所作对换的次数k =m+n-i 的奇偶性与t （p 1…p i-1 n p i+1…p n ）即t （p 1…p i-1p i+1…p n n ）+n －i 的奇偶性相同．图示如下：（p 1…p i-1 n p i+1…p n ）−−→−-次i n （p 1…p i-1p i+1…p n n ）−−→−次m （1 2 …(n －1) n ）； ③所作对换次数与原排列有相同的奇偶性还可如下证明：设排列p 1…p n 经k 次对换成为标准排列，则t(p 1…p n )经k 次改变奇偶性后成为0 (=t(1 2 … n))，从而k 与t(p 1…p n )奇偶性相同（对k 为奇、偶数分别说明）．（2）将（1）中的变（对）换全倒过来便得．（3）由（1）和（2）：−→−次k n p p p )(21 （1 2 … n ）)21h n q q q （次−→−即得．性质3 n ！个n 级排列中奇偶排列各为）2(2!≥n n ．证因映射ϕ：{n 级奇排列}−−→−），（21{n 级偶排列}为一一对应，即得．如 {(21)}−−→−），（21{(12)} ； {(132), (213), (321) }−−→−），（21{(231), (123), (312)} .§3 n 阶行列式的概念一、二、三阶行列式的结构规律１. 二、三阶行列式定义式的结构（1）2112221122211211a a a a a a a a -=中的两项可统一表示为212121)()1(j j j j t a a -，其中（j 1j 2）取遍所有（2个）2级排列（12），（21）．（2）33⨯ija ＝322311332112312213322113312312332211a a a a a a a a a a a a a a a a a ---++α中的6项可统一表示为321321321)()1(j j j j j j t a a a -，（j 1j 2j 3）取遍所有（6个）3级排列（123），（231），（312），（321），（213），（132）．２. 二、三阶行列式的共同规律设n ＝2或3，则：(1) n 阶行列式为由n 2个数得到一个数的函数；(2) n 阶行列式为n ！项的代数和，每项为n 个元素的乘积，而这n 个元素是取自n 阶行列式中的不同的行、不同的列；(3) n 阶行列式中每项正负号的确定：当项中各乘积因子的第一个（行）下标为标准排列时，其第二个（列）下标为奇（偶）排列的项带负（正）号． 3. 二、三阶行列式的简单统一表达式（1）∑∈-=221221)(211)(22211211)1(s j j j j j j t a a a a a a ，其中)}1,2(),2,1{(2=S ；（2）∑∈-=332132321)(3211)(333231232221131211)1(s j j j j j j j j j t a a a a a a a a a a a a ，S 3＝{（j 1j 2j 3）| (j 1j 2j 3)为3级排列}={（123），（231），（312），（321），（213），（132）}．二、n 阶行列式的定义1．定义设有n 2个元素排成的n 行、n 列的nnn n nn a a a a a a a a a 212222111211称为一个n 阶行列式，其值为∑∈-nn n n s j j j nj j j j j j t a a a )(211)(21221)1( ．上述n 阶行列式可简记为nn ij a ⨯或det n (ij a )．注：⑴当n ＝2，3时，与前面定义一致；当n ＝1时，1111a a =（注意别与绝对值混淆）． ⑵当n ≥4时，“沙路法”不再成立（或不再那样简单），见例1(4)．例１（1）n 21n21λλλλ０λ０λ=（（主）对角线（形）行列式），n nn a a a a ＝⨯0，11101＝， |0|n ×n = 0；（2）nn nnn n a a a a a a a a a 221121222111＝（下三角形行列式）；（3）nn nn nna a a a a a a a a2211222112110=上三角形行列式）(；（4）)1(0112211111111212)1(112121n n nn n n n n n nn nn n n n na a a a a a a a a a a a a a a------=-=，n n n λλλλλλ212)1(21)1(0--=ｎ（次对角形行列式）；如,abcd 0dcb a 0= ,abcde 0ed cb a= ;abcdef 0fed c b a 0-=（5）abcd abcd abcd d c b a t -=-=-=3)3142()1()1(000000000000；（6）111111)1(11001001011010)4123(-=-=⋅⋅⋅-=３）（t （因第3行和第1列均只有一个非零元素，因此非零项必取含21a 32a 的，从而另两个乘积因子11j a 和44j a 只能分别取14a 和43a 才能使该项不为0，于是得结果）；（7）∑∑∈∈-⋅=-=34324324324432432432)(432)(11S )j j (1j 43211)1(44434241443332312423222111)1()1(000S j j j j j j j j j t j j j j j j t a a a a a a a a a a a a a a a a a a a a a })243,324,432,423,342,234{}3234{(344434234333234232211）（）（）（）（）（）（＝级排列的=⋅=S a a a a a a a a a a ；类似有nnn n nnn n na a a a a a a a a a a a2222112122221110⋅=，特别地，00002122221=nnn n na a a a a a，一般地，nnnr nr r r rrr r nnr n nrn n r r r r r r rrr r a a a a a a a a a a a a a a a a a a a a11111111111111111111000++++++++++⋅=，简记为C A CB O A ⋅=;（8）当n ≥2时，n n n n n nn n a a a a a a a a a a 12212)2)(1(1221)1(0000000000000-------=．（9）当n ≥2时，02!2!)1(1111111111)()(2121=-=-==∑⨯n n n n j j j j j j t nn ．（10）nn ijj n nj j j j j j j j t nn ji j i a b a b a b a b ba n n n ⨯---⨯-=-=≠∑)())(()1()0(2211212211)( ．2．等价定义定理1 n i i i i i i i i i t nn ijn n n a a a a D 21)()(212121)1(∑-==⨯（记为D 1）．证 ⎢⎢⎢⎣⎡−−−−−−−−→−⎥⎥⎥⎦⎤)n 12)12()(2121i i j j 212121i j j i 21n nj j j n n i i i j j j a a a n i i i a a a nn 列标（）行标（列标行标）＝，＝经若干次对换（因n n n nj j j j j j j j j t nn ija a a a 21212121)()()1(D ∑-=⨯＝由对换的性质2知对D 中任一项n n nj j j j j j t a a a 212121)()1(-总有且仅有D 1中的某一项n i i i j j j t n n a a a 21)(2121)1(-与之对应并相等；反之，对D 1中任一项n i i i i i i t n n a a a 21)(2121)1(-，也总有且仅有D 中的某一项n n nj j j j j j t a a a 212121)()1(-与之对应并相等，如Ｄ4中))1(()1())1(()1(42342113334134221)2413(42342113342342113)3142(a a a a a a a a a a a a a a a a t t -=-=-=-；于是D 与D 1中的项可以一一对应并相等，从而D ＝D 1．定理2 n n j i j i j i J t I t nn ija a a a 2211)()()1(∑+⨯-=，其中t(I)=t(i 1i 2…i n )，t(J)=t(j 1j 2…j n )，∑为对所有n 级排列(i 1i 2…i n )求和（此时(j 1j 2…j n )为某一固定的n 级排列），或为对所有n 级排列(j 1j 2…j n )求和（此时(i 1i 2…i n )为某一固定的n 级排列）．证用对换的性质2（3），与定理1类似证明即可．再看例1（3），a b c da b c d a c b d a b c d b d a c d c b a t t t -=-=-=-=-=++41)3412()1324()2413()1()1(,)1(000000000000又．注：此例中i 1＝2，i 4＝4，i 3＝1，i 4＝3；j 1＝3，j 2＝1，j 3＝4，j 4＝2；j i1＝j 2＝1，j i2＝j 4＝2，j i3＝j 1＝3，j i4＝j 3＝4；i j1＝i 3＝1，i j2＝i 1＝2，i j3＝i 4＝3，i j4＝i 2＝4．§4 行列式的性质一、性质设nnn n n na a a a a a a a a D212222111211=，记D T（或D ˊ）nnn nn n a a a a a a a a a212221212111=，称为D 的转置（行列式），由§3定理1立即得：性质1 D T＝D , 即任一行列式与其转置的值相等．此性质说明：行列的行与列具有同等的地位，凡对行（或列）成立的性质对列（或行）也成立．性质2 互换行列式的两行（或两列），行列式仅变号．证设k ia a a a D in i kn k111=，欲证D 1＝－D ，只需证D 1和D 的定义式中的一般项互为相反数即可．事实上，D 1中的一般项为n k i 1n k i 1nj ij kj j 1)j j j j (t a a a a )1( -n i k 1n i k 1nj kj j i j 1)j j j j (t a a a a )1( --=恰为D 中一般项的相反数；故得证．推论两行（或列）完全相同得行列式值为零．性质3 行列式某一行（或列）的公因子可以提到行列式外面来，kD a a ka ka a a nnn in i n=11111．证明与性质2的证明类似，考虑一般项即可．推论1 行列式的某一行（或列）中所有元素都乘同一数k ，等于用k 乘此行列式．推论2 某行（或列）全为零的行列式的值为零．推论3 两行（或列）成比例的行列式的值为零．性质4 若行列式中某一行（或列）的元素都是两项之和，则该行列式可按此行（或列）分拆成两个行列式之和，即nnn in i nnnn in i n nnn in in i i n a a c c a a a a b b a a a a c b c b a a1111111111111111+=++．证明与性质2的证明类似，考虑一般项即可．性质5 将行列式某一（如第j ）行（或列）每个元素的常数l 倍加到另一（如第i ）行（或列）相应的元素上去，其值不去，即nn ijnnn jn in j i na D a a la a la a a a ⨯==++(111111)．证由性质4，左边的行列式可分拆成两个行列式之和，一个为D ，而另一个为0111111=nnn jn j jn j na a a a la la a a（因其第i 行与第j 行成比例）；从而得证．二、行列式的计算—化为三角形行列式定理1 任何一个行列式均可利用性质2和5化为上（或下）三角形行列式，从而计算其值．证（1）若a ij ＝0（i ，j ＝1，2，…，n ），则00==⨯n n D ；（2）若0≠∃ij a ，则可用性质2（先第1行与第i 行互换，再第1列与第j 列互换）将a ij 调到左上角；（3）若011≠a ，则可用性质5将第1列（或行）的其余n －1个元素化为零（“打洞”）；（4）对右下角的n －1阶行列式重复（1）～（3）的步骤，如此下去（归纳），即可将D 化为上（或下）三角形行列式．以下以（r i ，r j ）表示互换i ，j 行；r i ＋hr j 表示将第j 行的h 倍加到i 行．例1（1）4130211021102011)r 2r (),r r (0112012121102011)r ,r (0112012120112110141321-------+-----------;4)2()2()1(12000420021102011r r 2200420021102011)r 3r (),r r (342423=-⋅-⋅-⋅-=-------------++（2）)r ,r (72160112064802131)r 5r (),r r (3315112043512131)c ,c (335111024315211332141221------+--------------)r 4r (108003200112021315)r ,r (1510001080011202131)r 8r (),r 4r (7216064801120213134432423-----⋅-----+-----402221520003200112021315=⋅⋅⋅⋅=---⋅．（3）48222162002000020111164,3,2i ),r r (31111311113111116)r r r (r 31111311113111131i 4321=⋅⋅⋅⋅=⋅=-⋅+++ ; （4）xaa aa a x a a a a a x a a a a a x a a n x n i r r xaa aaa x a a a a a x a a a a a x a a a a a x i11111])1([,,2),(1⋅-+=+11)(])1([000000000000011111])1([,,2),(--⋅-+=----⋅-+=-n i a x a n x ax ax a x ax a n x n i ar r;（5）cb a b a ac b a b a a c b a b a a dc b a i r rd c b a c b a b a a d c b a c b a b a a d c b a c b a b a a d c b a i i +++++++++=-+++++++++++++++++++3630232001,2,3),(361036323423214341030020002,3),(a aa aa r rb a a b a a a a i r r i i =*-++*=-+．例2 证明奇数（n ）阶反对称行列式（a ji ＝－a ij ）的值为零，即000021212112=---n nnna a a a a a ．证 0)1(=⇒-=⋅-==D D D D D n T ．例3 解方程 (a 1≠0)113211232113221132111321=-+-+-+-+-------xa a a a a a a xa a a a a a a x a a a a a a a xa a a a a a a a n n n n n n n n n n n n解将左边行列式的第1行的相反数分别加到第2～n 行，得左边x)-x)(-(x)-x)(-(00000000000001-n 2-n 21112211321a a a a a xa xa x a x a a a a a a n n n n=----=---故原方程的解为)1,,2,1(-==n i a x i i ，共n －1个解．三、按行、列展开定理1．代数余子式设nn ji a D ⨯=，把D 中元素a ij 所在的第i 行和第j 行划去后，余下的n －1阶行列式叫做a ij 在D 中的余子式，记作M ij ，记A ij ＝（－1）i ＋j M ij ，叫做a ij 在D 中的代数余子式．例４（1）213132321----=D 的52113)1(1111=--⋅-=+A12312)1(2112=---=+A ,71332)1(3113-=---=+A ；（2）5021011321014321---=D的19521013201)1(3113=---=+A ,521013421)1(3223---=+A =- 63，18521201421)1(3333=--=+A ．10013201421)1(3443-=--=+A ；2．按行、列展开定理引理若n 阶行列式nn ij a D ⨯=的元素a ij 所在第i 行（或第j 列）的其他所有元素全为零，则ij ij A a D =．证（1）当i ＝j ＝1，即D 的第1行（或第1列）除a 11外所有元素全为零，则由§3例1（7）知1111A a D ⋅=；（2）一般地，设nnnjn ijnja a a a a a a D1111100=，将D 的第i 行依次与第i －1，i －2，…，2，1行对换，再将第j 列依次与第j －1，j －2，…，2，1列对换，使a ij 调到左上角，所得的新行列式D D D j i j i ⋅-=⋅-=+-+)1()1(21，而a ij 在D 1中的余子式即为a ij 在D 中的余子式M ij ，由（1）ij ij ji ij ij A a D D M a D =-=⇒=+11)1(．定理2 n 阶行列式nn ija D ⨯=的值等于其任一行（或列）的每一个元素分别与其相应的代数余子式的乘积之和，即),,2,1(111n i A a A a A aD in in i i nj ij ij=++==∑=或∑==++==ni nj nj j j j ij n j A a A a Ai a D 111),,2,1( ．证（1）nnn n i nnnn n in i i n a a a a a a a a a a a a a a a a D2111121121211121100000000=+++++++++=),,2,1(00002211211121121211211n i A a A a A a a a a a a a a a a a a a a a in in i i i i nnn n in nnnn n i n=++++++引理（2）由行列式的性质1立即得对列的等式也成立．例4 （3）对（1）中的18)7(312513)2(1131211-=-⋅+⋅-⋅=⋅+⋅-+⋅=A A A D ；对（2），24)10(018)1()63(1193-=-⋅+⋅-+-⋅+⋅=D ．定理3 设nn ija D ⨯=，则（1）D A a A a A aik kn in k i nj kj ij⋅=++=∑=δ 111；（2）∑=⋅=++=ni jr nr nj r j ir ijD A a A a A a111δ证（1）由定理1知当i ＝k 时成立．当i ≠k 时，将nn k ia a a a a a a a a a a a nn n n in i i in i i n⨯=212121112110按第k 行展开即得∑==nj kj ijA a10，即∑=≠⋅=nj kj ij k i D A a 1)(0；故得证．由行列式的性质1立即得对列的结论（2）也成立．定理2、3表明，行列式D 的任一行（或列）的每一个元素与其相应的代数余子式的乘积之和等于D 的值，而D 的任一行（或列）的每一个元素与另外一行（或列）的每一个元素的代数余子式的乘积之和等于零．例４（4）对（1）中的D 有 0)7()1(1352)1(32131211=-⋅-+⋅+⋅-=⋅-+⋅+⋅-A A A ， 0)7(211532)1(3131211=-⋅+⋅-⋅=⋅+⋅-+⋅A A A ；对（2）中的D 有0)10(1183)63(1191131143332313=-⋅+⋅+-⋅+⋅=⋅+⋅+⋅+⋅A A A A ， 0)10(2181)63(01922)1(024*******=-⋅+⋅--⋅+⋅=⋅+⋅-+⋅+⋅A A A A ， 0)10(5180)63(2194)5(024********=-⋅-⋅+-⋅+⋅=⋅-+⋅+⋅+⋅A A A A ．3. 行列式的归纳定义 11111a a D ==,21122211212112221111222112112)1()1(a a a a a a a a a a a a D -=⋅-⋅+-⋅==++，当n ≥2时，n n nn ijn A a A a A a a D 1112121111+++==⨯ ，其中j j j M A 111)1(+-=，M 1j 为a 1j 在D n 中的余子式（n －1阶行列式）．可以证明如上定义的n 阶行列式与前面的定义n 阶行列式是完全一样的．4．行列式的简化计算首先利用性质将某行（或列）化为仅有一个元素可能非零，再按该行（或列）展开，降为n －1阶行列式，如此下去，直到化为二阶或一阶，即可计算其值．例5（1）527211417)1()1(5207011321014107)2(),2(5021011321014321233431---⋅-=----++---+r r r r 241861926)1(110921126)2(),(222321-=--=-⋅-⋅=-+-+r r r r ．（2）)4)(1(22)1(202001120020001100112002000110011212--=-=-=--k k kk k k k k kk k r r k k k ．（3）0551*******3550100131111115)(),2(33511102431521133431----=----+-------c c c c 40552605502611512=---=----r r ．例6（1）dd c dcb a b a a dcd c dc b a baba D n 000012⋅=行展开按第)1(2)1(21)12()1(2)()1(000--+---=--=⋅-n n n n D bc ad D bc adD cd c dc b a bab（递推公式）)()()()(221)2(22bc ad dc ba D bc ad D bc ad D bc ad n n n -==-=-==-=-- ．（2）n ≥2，ba b ba bb ab a ba b a a a b b a ba ba D n n 0)1(010001+-⋅+⋅=列展开按第n n b a )(--=．（3）++---+---=----xa a a a xxxx xa a a a a xxxx D n n n n n n 122112211111111列展开按第n n n n n nn a a xD x a xD x xa +++=--------+)((1111)1(1211递推公式）＝).1,(21212211111a x a x xa a x D x a D a x a x a x n n n n ++=+-=+=++++==--例7（1）计算V andermonde 行列式)2(≥n ：Ｄn =),,,(21n x x x V =),(111112112222121j i x x x x x x x x x x x j i n nn n n n ≠≠---；解 D 2 =122121211),(x x x x x x V -==，将D n 中依次第i 行减去第i －1行的x n 倍。

线性代数英文课件2.2

Deﬁnition: • A collection of a ﬁnite number of linear equations involving the same variables is called a system of linear equations, often abbreviated as SLE. • A system of m linear equations in n variables is called an m × n SLE (pronounced “m by n”). • An n-vector (x1 , x2 , ..., xn ) is called a solution to a particular SLE with n variables if it satisﬁes all of the equations in the system.
Math 1229A/B
Unit 5: Systems of Linear Equations
(text reference: Section 2.2)
Байду номын сангаас
c V. Olds 2010
Unit 5
61
5
Systems of Linear Equations
You know what the standard form of an equation of a line in ℜ2 looks like. It has the form ax + by = c, for some constants a, b and c. Any equation in 2 variables in which each variable is only multiplied by a constant is the standard form equation of a line in ℜ2 . The variables are usually called x and y , but they could be called x1 and x2 , or m and n, or anything. Of course, we can have other kinds of equations with 2 variables, and you’ve probably seen some of those before. For instance, x2 + y 2 = 1 is the equation of the circle with radius 1. More complicated curves in 2-space have equations like x2 + xy + y 2 = 4 or 5x4 y − 3x2 y 2 + 2xy 5 = 0 or √ √ y x − x y + y = 3. And then there are equations like x + x − y22 = 6, and 2x+y = 4 and x + sin y = 1. But none of those are linear equations. They’re not lines. They’re curvy things. You also know that in ℜ3 , an equation like x + y + z = 1 is not a line. It’s a plane. But ... what is a plane? It’s just a whole bunch of lines side-by-side. Well, okay, we wouldn’t really think of a plane that way, but a plane does have some characteristics which make it similar to a line in some ways. It’s ﬂat. No curvy bits. And the same goes for a hyperplane in ℜm . Any plane or hyperplane, in standard form, has an equation which is just the sum of a bunch of constant multiples of variables, set equal to some constant. There are never any more complicated things done to the variables, like squaring, or taking the square root, or multiplying two variables together, or dividing by a variable. Just a constant times a variable, plus a constant times another variable, plus ... and equal to some constant. The most interesting things that happen are that sometimes a constant is negative, so that we’re actually subtracting, or sometimes a constant is 0, so that the variable isn’t even there. Equations like this are called linear equations, because the relationship between the variables is always like the relationship between the variables in an equation of a line in ℜ2 . We’re going to be working with linear equations a lot in this course. In fact, the textbook is called Elementary Linear Algebra, because the kind of algebra we’re studying all relates to these linear equations. So we need a careful deﬁnition of what these are. But it’s not any more complicated than what we’ve already said. Deﬁnition: A linear equation in the variables x1 , x2 , ... xn is an equation which can be put into the form a1 x1 + a2 x2 + ...an xn = b for some constants a1 , a2 , ..., an and b, where not all of the ai values are 0. Note: That last bit just means that the equation 0 = 0 isn’t considered to be a linear equation. There has to be at least one variable actually appearing in the equation. Also, as you already know, the variables don’t necessarily have to have the names shown in the deﬁnition. They could be x, y, z, w, r and t, or they could be subscripted y ’s or s’s instead of x’s. The names of the variables don’t matter. Any equation which says, or can be rearranged to say, that the sum of scalar multiples of variables is equal to a constant is a linear equation.

线性代数英文课件3.3

1 3 5
Solution: We row-reduce each of the given matrices to get the RREF matrix. (You’ve had lots of practice with row-reducing by now, so the details of the reduction are not shown here. Of course, you should feel free to check that the RREF matrices shown here are correct.) 1 1 −1 1 0 0 RREF 2 1 − (a) 1 − − − → 0 1 0 −2 −3 4 0 0 1 The RREF of A does not contain any zero rows (i.e. rows containing only zeroes), so there are 3 non-zero rows. Therefore r(A) = 3, and since A has 3 columns, A has full rank. 1 1 1 1 1 1 RREF (b) 2 2 2 − − − − → 0 0 0 3 3 3 0 0 0 This time, there are some zero rows in the RREF of A. In fact, the RREF of A has only one non-zero row, so r(A) = 1. (And since A has more than 1 column, A does not have full rank.) 1 (c) 2 3 1 0 1 1 RREF 3 3 − − − − → 0 1 3 5 0 0 0 0 1

线性代数英文课件：ch4_Review

x1
x3
1 3
x5 ;
x2
x3
2 3
x5 ;
x3
x3 ;
x4
5 3
x5 ;
x5
x5
1 0 1 0 1/3
0
1
1
0
2
/
3
0 0 0 1 5/3
0
0
0
0
0
x1 x2 x3 x4 x5
c1
1
1
1
0
0
c2
0
1 3 2 3
which means
(k k1 knr ) * k11 kn-r n-r 0 (2)
From question(1) we know
*,1, ,n-r are linearly independent,
(k k1 knr ) 0; k1 0, , kn-r 0. k k1 kn-r 0.
1 0 1 0 1/3
0
1
1
0
2
/
3
0 0 0 1 5/3
0 0 0 0 0
x1 x3 1 / 3;
x2
x3
2 / 3;
x3
x3 ;
x4 5 / 3.
x1 1 1 / 3
x2 x3 x4
c
1
1
0
2 0 5
/ /
3
3
,
c
R.
(6)Let matrix A=(1 ,2 ,3 ,4 ,5 )，solve Ax 0.
1, ,n-r are linearly independent.
k1 knr 0,
So *,1, ,n-r are linearly independent.

线性代数英文课件：ch5-1Eigenvalues and Eigenvectors

1. Eigenvalues and Eigenvectors Definition 1 Let A be an n×n matrix. A number
is called an eigenvalue of A if there exists a nonzero
n-dimensional vector such that
How many eigenvalues does a matrix A have?
1
A
4
2
3
,
1 AE
4
2
3
(5 )(1 )
So the eigenvalues of A is: 1 5, 2 1
A
1 0
0
3
AE 1
0
0 (1 )(3 )
3
The eigenvalues of a diagonal matrix are the elements in the diagonals (对角矩阵的特征值为对角线上的元素)
So the eigenvector is k11(k1 R, and k1 0).
When 2 3 1,
solve the LS( A E ) x 0
1 1 1
1 1 1
A
E
1
1
1
r
0
0
0
1 1 1
0 0 0
The equivalent LS is x1 x2 x3 0
2.How to find the Eigenvalues and Eigenvectors of a matrix?
Two Steps:
(1) A E 0,or E A 0,
To find out the eigenvalues of A;

线性代数英文讲义

Example The set {(1, 1, 1)T, (2, 1, -3)T, (4, -5, 1)T} is an
orthogonal set in R3.
Theorem 5.5.1 If { v1, v2, …, vn} is an orthogonal set of
nonzero vectors in an inner product space V, then v1, v2, …,vn are linearly independent.
Example Let X be the subspace of R3 spanned by e1, and
let Y be the subspace spanned by e2.
Example Let X be the subspace of R3 spanned by e1 and e2,
and let Y be the subspace spanned by e3.
Corollary 5.1.2 ( Cauchy-Schwarz Inequality)
If x and y are vectors in either R2 or R3 , then (2) xTy≤‖x‖‖y‖
with equality holding if and only if one of the vectors is 0 or one vector is a multiple of the other.
Corollary 5.5.3 Let { u1, u2, …, un} be an orthonoemal basis
α=
u， v v
and the vector projection of u onto v is given by

北京邮电大学国际学院线性代数讲义Lecture 04

0 1 0 Example: E1 1 0 0 is an elementary matrix of type I. Let A be a 3 3 0 0 1 matrix. Then
0 1 0 a11 a12 a13 a21 a22 a23 E1 A 1 0 0 a21 a22 a23 a11 a12 a13 0 0 1 a 31 a32 a33 a31 a32 a33 a11 a12 a13 0 1 0 a12 a11 a13 AE1 a21 a22 a23 1 0 0 a22 a21 a23 a 31 a32 a33 0 0 1 a32 a31 a33
11
Elementary Matrices
Definition: A matrix B is row equivalent to A if there exists a finite sequence E1 , E2 ,, Ek of elementary matrices such
6
Elementary Matrices
Type II. Type II elementary matrices are obtained by multiplying a row of I by a nonzero constant.
1 0 0 Example: E 2 0 1 0 is an elementary matrix of type II. Then 0 0 3 a12 a13 1 0 0 a11 a12 a13 a11 0 1 0 a a22 a23 a21 a22 a23 21 0 0 3 a 31 a32 a33 3a31 3a32 3a33 a11 a12 a13 1 0 0 a11 a12 3a13 a21 a22 a23 0 1 0 a21 a22 3a23 a 0 0 3 a 31 a32 3a33 31 a32 a33

高等数学【线性代数】英文版课件1

Ordinary Differential Equations Lecture Notes
1.1. How Differential Equations Arise
In this section the idea of a differential equation is introduced through the mathematical formulation of some problems. Newton’s Second Law of Motion Statement: For an object of constant mass m, the sum of applied forces acting on the object is equal to the mass of the object multiplied by the acceleration of the object. Assumption: The object is moving in one dimension under the inﬂuence of a force F. Then the mathematical statement of this law is m dv = F, dt (1.1.1)
Ordinary Differential Equations Lecture Notes
To formulate this law mathematically, let T(t) denote the temperature of the object at time t, and Tm (t) denote the temperature of the surrounding medium. Then Newton’s law of cooling can be expressed as the differential equation dT = −k(T − Tm ), dt where k is a constant. Note that the minus sign in front of the constant k is traditional: it ensures that k is always positive. Indeed, (i) If T > Tm , then the object will cool, so that dT/dt < 0. Hence, from equation (1.1.3), k must be positive. (ii) Similarly, if T < Tm , then dT/dt > 0, and once more equation (1.1.3) implies that k must be positive. (1.1.3)

线性代数英文课件4.1

Math 1229A/B
Unit 10: Determinants
(text reference: Section 4.1)
c V. Olds 2010
128
Unit 10
10
Determin are a special class of matrices. We have already seen one instance of a concept which is deﬁned only for square matrices — the inverse matrix. That is, only a square matrix may have an inverse. In this unit we will (begin to) learn about another concept which is deﬁned only for square matrices — the determinant of a matrix.
Unit 10 d, then det A = ad − cb. That is, if A is a 2 × 2 matrix, then det A = a11 a22 − a21 a12 . Example 10.1. Find the determinants of the following matrices: 1 2 2 0 (a) A = [5] (b) B = (c) C = 3 4 1 3
The number which is the determinant of a square matrix measures a certain characteristic of the matrix. In a more advanced study of matrix algebra, this characteristic is used for various purposes. In this course, the only way in which we will use this number is in its connection to the existence of the inverse of the matrix, and through that it’s application to SLE’s in which the coeﬃcient matrix is a square matrix. For these purposes, what will matter to us is whether or not this number, the determinant of the matrix, is 0. But of course, in order to determine whether or not the determinant of a particular matrix is 0, we need to know how to calculate that number. Calculating the determinant of a square matrix is somewhat complicated. The deﬁnition is recursive, meaning that the calculation is deﬁned in a straightforward way for small matrices, and then for larger matrices, the determinant is deﬁned as being a calculation involving the determinants of smaller matrices, which are certain submatrices of the matrix. We could express this recursive deﬁnition of the determinant of a square matrix of order n as applying for all n ≥ 2, speciﬁcally deﬁning only the determinant of a square matrix of order 1, i.e. a (square) matrix containing only a single number. However, the calculation for a 2 × 2 matrix is very straightforward — easier to think of as a special deﬁnition all on its own — so instead we use speciﬁc deﬁnitions for n = 1 and n = 2, and then deﬁne the determinant of a square matrix of order n > 2 in terms of determinants of submatrices of order n − 1, which are found by expressing them in terms of determinants of successively smaller submatrices until we get down to submatrices of order 2. The calculation of det A as deﬁned in this way, when A is a square matrix of order n > 2, is not really as complicated as it will look. It’s just a matter of applying a certain formula carefully, as many times as necessary until we have expressed det A in terms of the determinants of 2 × 2 matrices. Those determinants are easy to ﬁnd. So we start by deﬁning det A for square matrices of order 1 and of order 2. When A is a 1 × 1 matrix, i.e. a matrix containing only one number, ﬁnding the particular number det A which is associated with that matrix is trivial. That number is the only number around — the single number that’s in the matrix. For a square matrix of order 2, i.e. a matrix containing 4 numbers arranged in a square, we have to do a little more work. But it’s a simple calculation. In fact, we can think of the calculation as “down products minus up products”, which is something we have seen before. But this time there’s only one down product, and only one up product, so it’s actually just “down product minus up product”.

线性代数英文课件：ch2-2 Inverse of a Matrix

Example 5 For a system of linear equations:
2 x1 x2 3x3 1
2
x1
x2
x3
5
4 x1 x2 2 x3 5
It can also be denoted as:
Ax b, Where
2 1 3 x1 1
A
2 4
1 1
1 2
A1 AXBB1 A1CB1
That is, X=A-1CB-1
3
2
So
X
3 2
1
3 5
2
1
2
2
1 0
2
0 1
3 5
1
1
2
2
0 2
2 1
0 1
3 5
10
1 2
11 3
4
4 1
By premultiplying (左乘)a matrix A by a matrix B, we mean multiplying A on the left by B, that is, forming the product BA.
Solution A 2 0, and B 1 0
so A and B is invertible, and
3
2
A1 3
2
1
2
3
1
5 2
2
1
B1
3 5
1 2
Premultiplying matrix equation AXB C by matrix
A-1, and postmultiplying B-1 ,
Suppose B,C are both inverse matrices of A, that is,

线性代数英文课件4.2

144
Unit 11
Theorem 11.1. If matrix B is obtained from square matrix A by multiplying one row or column of A by some non-zero scalar c, then det B = c(det A).
Math 1229A/B
Unit 11: Properties of Determinants
(text reference: Section 4.2)
c V. Olds 2010
Unit 11
143
11
Properties of Determinants
In this section, we learn more about determinants. First, we observe some properties of determinants that allow us to calculate determinants more easily. We examine the eﬀects on the determinant when the various kinds of elementary row operations are performed, so that we can easily see how the determinants of the various row-equivalent matrices are related to one another as we perform these operations. This allows us to calculate the determinant of a matrix by row-reducing the matrix (a procedure we already know well) to obtain a matrix whose determinant is easily calculated using facts we’ve already learnt in the previous section. We also learn some useful properties which allow us to calculate the determinant of a matrix from the determinants of one or more other matrices whose determinants we may already know. And ﬁnally we examine the relationship between determinants and inverses, which allows us to relate determinants to systems of linear equations, using what we already know about the implications of the existence of the inverse of a matrix for the number of solutions to the SLE which has that matrix as its coeﬃcient matrix. Throughout all of this, of course, it is important to remember that we are only dealing with square matrices when we talk about determinants. That is, it is only for a square matrix that the characteristic “the determinant of the matrix” is deﬁned. First, let’s think about what eﬀect multiplying some row of a matrix by a non-zero scalar will have on the determinant. That is, let’s think about the relationship between det A and det B if matrix B is identical to matrix A except that one of the rows in B is the corresponding row of A multiplied by some c = 0. So suppose we have some n × n matrix A = [aij ]. Let B = [bij ] be the matrix obtained by multiplying one row, row k , by some non-zero scalar c. Then we know that bkj = cakj and bij = aij for all i = k . We can calculate det B by expanding along row k . Notice that when we form submatrices of B by deleting row k (and also some column of B ), the one row that’s diﬀerent than in matrix A is deleted, so that in the submatrix of B obtained, each entry is just the corresponding entry from matrix A and therefore the entire submatrix of B is simply the corresponding submatrix of A. That is, we have Bkj = Akj . So when we expand along row k we get:

线性代数英文讲义

Chapter 3 Vector SpacesThe operations of addition and scalar multiplication are used in many diverse contexts in mathematics. Regardless of the context, however, these operations usually obey the same set of algebraic rules. Thus a general theory of mathematical systems involving addition and scalar multiplication will have applications to many areas in mathematics.§1. Examples and DefinitionNew words and phrasesVector space 向量空间Polynomial 多项式Degree 次数Axiom 公理Additive inverse 加法逆1.1 ExamplesExamining the following sets:(1) V=2R : The set of all vectors 12x x ⎛⎫ ⎪⎝⎭ (2) V=m n R ⨯: The set of all mxn matrices(3) V=[,]a b C : The set of all continuous functions on the interval [,]a b(4) V=n P : The set of all polynomials of degree less than n.Question 1: What do they have in common?We can see that each of the sets, there are two operations: addition and multiplication, i.e. with each pair of elements x and y in a set V, we can associate a unique element x+y that is also an element in V, and with each element x and each scalar α, we can associate a unique element αin V. And the operations satisfy some algebraic rules.xMore generally, we introduce the concept of vector space. .1.2 Vector Space Axioms★Definition Let V be a set on which the operations of addition and scalar multiplication are defined. By this we mean that, with each pair of elements x and y in a set V, we can associate a unique element x+y that is also an element in V, and with each element x and each scalar α, we can associate a unique element xαin V.The set V together with the operations of addition and scalar multiplication is said to form a vector space if the following axioms are satisfied.A1. x+y=y+x for any x and y in V.A2. (x+y)+z=x+(y+z) for any x, y, z in V.A3. There exists an element 0 in V such that x+0=x for each x in V.A4. For each x in V, there exists an element –x in V such that x+(-x)=0. A5. α(x+y)= αx+αy for each scalar αand any x and y in V.A6. (α+β)x=αx+βx for any scalars αandβand any x in V.A7. (αβ)x=α(βx) for any scalars αandβand any x in V.A8. 1x=x for all x in V.From this definition, we see that the examples in 1.1 are all vector spaces. In the definition, there is an important component, the closure properties of the two operations. These properties are summarized as follows:C1. If x is in V and αis a scalar, then αx is in VC2. If x, y are in V, then x+y is in V.An example that is not a vector space:Let {}=, on this set, the addition and W a a(,1)| is a real numbermultiplication are defined in the usually way. The operation + and scalar multiplication are not defined on W. The sum of two vector is not necessarily in W, neither is the scalar multiplication. Hence, W together with the addition and multiplication is not a vector space.In the examples in 1.1, we see that the following statements are true.Theorem 3.1.1 If V is a vector space and x is any element of V, then (i) 0x=0(ii) x+y=0 implies that y=-x (i.e. the additive inverse is unique). (iii)(-1)x=-x.But is this true for any vector space?Question: Are they obvious? Do we have to prove them?But if we look at the definition of vector space, we don’t know what the elements are, how the addition and multiplication are defined. So theorem above is not very obvious.Proof(i)x=1x=(1+0)x=1x+0x=x+0x, (A6 and A8)Thus –x+x=-x+(x+0x)=(-x+x)+0x (A2)0=0+0x=0x (A1, A3, and A4)(ii)Suppose that x+y=0. then-x=-x+0=-x+(x+y)Therefore, -x=(-x+x)+y=0+y=y(iii)0=0x=(1+(-1))x=1x+(-1)x, thusx+(-1)x=0It follows from part (ii) that (-1)x=-xAssignment for section 1, chapter 3Hand in: 9, 10, 12.§2. SubspacesNew words and phrasesSubspace 子空间Trivial subspace 平凡子空间Proper subspace 真子空间Span 生成Spanning set 生成集Nullspace 零空间2.1 DefinitionGiven a vector space V , it is often possible to form another vector space by taking a subset of V and using the operations of V . For a new subset S of V to be a vector space, the set S must be closed under the operations of addition and scalar multiplication.Examples (on page 124)The set 1212|2x S x x x ⎧⎫⎛⎫⎪⎪==⎨⎬ ⎪⎪⎪⎝⎭⎩⎭together with the usual addition and scalar multiplication is itself a vector space .The set S=| and are real numbers a a a b b ⎧⎫⎛⎫⎪⎪ ⎪⎨⎬ ⎪⎪⎪ ⎪⎝⎭⎩⎭together with the usual addition and scalar multiplication is itself a vector space.★Definition If S is a nonempty subset of a vector space V , and S satisfies the following conditions:(i) αx ∈S whenever x ∈S for any scalar α(ii) x+y ∈S whenever x ∈S and y ∈Sthen S is said to be a subspace （子空间）of V .A subspace S of V together with the operations of addition and scalar multiplication satisfies all the conditions in the definition of a vector space. Hence, every subspace of a vector space is a vector space in its own right. Trivial Subspaces and Proper SubspacesThe set containing only the zero element forms a subspace, called zero subspace, and V is also a subspace of V . Those two subspaces are called trivial subspaces of V . All other subspaces are referred to as proper subspaces.Examples of Subspaces(1) the set of all differentiable functions on [a,b] is a subspace of [,]a b C(2) the set of all polynomials of degree less than n (>1) with the property p(0) form a subspace of n P .(3) the set of matrices of the form a b b c ⎛⎫⎪-⎝⎭ forms a subspace of 22R ⨯. (4) the set of all mxm symmetric matrices forms a subspace of m m R ⨯(5) the set of all mxm skew-symmetric matrices form a subspace of m m R ⨯2.2 The Nullspace of a MatrixLet A be an mxn matrix, and{}()|,0n N A X X R AX =∈=.Then N(A) form a subspace of n R . The subspace N(A) is called the nullspace of A.The proof is a straightforward verification of the definition.2.3 The Span of a Set of VectorsIn this part, we give a method for forming a subspace of V with finite number of vectors in V .Given n vectors 12n v ,v ,,v in a vector space of V , we can form a newsubset of V as the following.{}12n 1122n n Span(v ,v ,,v )v v v |' are scalars i s αααα=+++It is easy to show that this set forms a subset of V. We call this subspace the span of 12n v ,v ,,v , or the subspace of V spanned by12n v ,v ,,v .Theorem 3.2.1 If 12n v ,v ,,v are elements of a vector space of V , then{}12n 1122n n Span(v ,v ,,v )v v v |' are scalars i s αααα=+++ is a subspace of V .For example, the subspace spanned by two vectors 100⎛⎫ ⎪ ⎪ ⎪⎝⎭and010⎛⎫ ⎪ ⎪ ⎪⎝⎭is the subspace consisting of the elements 120x x ⎛⎫ ⎪ ⎪ ⎪⎝⎭.2.4 Spanning Set for a Vector Space★Definition If 12n v ,v ,,v are vectors of V andV=12n Span(v ,v ,,v ), then the set {}12n v ,v ,,v is called a spanning set（生成集）for V .In other words, the set {}12n v ,v ,,v is a spanning set for V if andonly if every element can be written as a linear combination of 12n v ,v ,,v .The spanning sets for a vector space are not unique.Examples (Determining if a set spans for 3R )(a) (){}1231,2,3,T e e e (b) ()()(){}1,1,1,1,1,0,1,0,0T T T (c) ()(){}1,0,1,0,1,0T T (d) ()()(){}1,2,4,2,1,3,4,1,1T T T -To do this, we have to show that every vector in 3R can be written as a linear combination of the given vectors.Assignment for section 2, chapter 3 Hand in: 6, 8, 13, 16, 17, 18, 20Not required: 21Chapter 3---Section 3 Linear Independence§3. Linear IndependenceNew words and phrasesLinear independence 线性无关性Linearly independent 线性无关的Linear dependence 线性相关性Linearly dependent 线性相关的3.1 MotivationIn this section, we look more closely at the structure of vector spaces. We restrict ourselves to vector spaces that can be generated from a finite set of elements, or vector spaces that are spans of finite number of vectors. V=Span(v,v,,v)12nThe set {}v,v,,v is called a generating set or spanning set(生成集).12nIt is desirable to find a minimal spanning set. By minimal, we mean a spanning set with no unnecessary element.To see how to find a minimal spanning set, it is necessary to consider how the vectors in the collection depend on each other. Consequently we introduce the concepts of linear dependence and linear independence. These simple concepts provide the keys to understanding the structure of vector spaces.Give an example in which we can reduce the number of vectors in a spanning set.Consider the following three vectors in 3R.11x 12⎛⎫ ⎪=- ⎪ ⎪⎝⎭ 22x 31-⎛⎫ ⎪= ⎪⎪⎝⎭31x 38-⎛⎫⎪= ⎪ ⎪⎝⎭ These three vectors satisfy(1) 312x =3x +2xAny linear combination of 123x ,x ,x can be reduced to a linear combination of 12x ,x . Thus S= Span(123x ,x ,x )=Span(12x ,x ). (2) 1233x +2x +(1)x 0-= (a dependency relation)Since the three coefficients are nonzero, we could solve for any vector in terms of the other two. It follows thatSpan(123x ,x ,x )=Span(12x ,x )=Span(13x ,x )=Span(23x ,x )On the other hand, no such dependency relationship exists between12x and x . In deed, if there were scalars 1c and 2c , not both 0, such that(3) 1122c x +c x 0=then we could solve for one of the two vectors in terms of the other. However, neither of the two vectors in question is a multiple of the other. Therefore, Span(1x ) and Span(2x ) are both proper subspaces of Span(12x ,x ), and the only way that (3) can hold is if 12c =c =0.Observations: (I)If 12n v ,v ,,v span a vector space V and one of these vectors can be written as a linear combination of the other n-1 vectors, then those n-1 vectors span V .(II) Given n vectors 12n v ,v ,,v , it is possible to write one of thevectors as a linear combination of the other n-1 vectors if and only if there exist scalars 12n c ,c ,,c not all zero such that1122n n v v v 0c c c +++=Proof of I: Suppose that n v can be written as a linear combination of the vectors 12n-1v ,v ,,v .Proof of II: The key point here is that there at least one nonzero coefficient.3.2 Definitions★Definition The vectors 12n v ,v ,,v in a vector space V are said to be linearly independent （线性独立的） if1122n n v v v 0c c c +++=implies that all the scalars 12n c ,c ,,c must equal zero. Example: 12n e ,e ,,e are linearly independent.Definition The vectors 12n v ,v ,,v in a vector space V are said to be linearly dependent （线性相关的）if there exist scalars 12n c ,c ,,c not all zero such that1122n n v v v 0c c c +++=.Let 12n e ,e ,,e ,x be vector in n R . Then 12n e ,e ,,e ,x are linearlydependent.If there are nontrivial choices of scalars for which the linear combination 1122n n v v v c c c +++ equals the zero vector, then 12n v ,v ,,vare linearly dependent. If the only way the linear combination1122n n v v v c c c +++ can equal the zero vector is for all scalars 12n c ,c ,,cto be 0, then 12n v ,v ,,v are linearly independent.3.3 Geometric InterpretationThe linear dependence and independence in 2R and 3R .Each vector in 2R or 3R represents a directed line segment originated at the origin.Two vector are linearly dependent in 2R or 3R if and only if two vectors are collinear. Three or more vector in 2R must be linearly dependent.Three vectors in 3R are linearly dependent if and only if three vectors are coplanar. Four or more vectors in 3R must be linearly dependent.3.4 Theorems and ExamplesIn this part, we learn some theorems that tell whether a set of vectors is linearly independent.Example: (Example 3 on page 138) Which of the following collections of vectors are linearly independent?(a) (){}1231,2,3,Te e e(b) ()()(){}1,1,1,1,1,0,1,0,0TTT(c) ()(){}1,0,1,0,1,0T T(d) ()()(){}1,2,4,2,1,3,4,1,1T TT-The problem of determining the linear dependency of a collection of vectors in m R can be reduced to a problem of solving a linear homogeneous system.If the system has only the trivial solution, then the vectors are linearly independent, otherwise, they are linearly dependent, We summarize the this method in the following theorem:Theorem n vectors 12n x ,x ,,x in m R are linearly dependent if the linear system Xc=0 has a nontrivial solution, where 12n X=(x ,x ,,x ). Proof: 1122n n c x +c x +c x 0+= ⇔ Xc=0.Theorem 3.3.1 Let 12n x ,x ,,x be n vectors in n R and let12n X=(x ,x ,,x ). The vectors 12n x ,x ,,x will be linearly dependent if andonly if X is singular. (the determinant of X is zero)Proof: Xc=0 has a nontrivial solution if and only X is singular.Theorem 3.3.2 Let 12n v ,v ,,v be vectors in a vector space V. A vector v in Span(12n v ,v ,,v ) can be written uniquely as a linear combination of12n v ,v ,,v if and only if 12n v ,v ,,v are linearly independent.(A vector v in Span(12n v ,v ,,v ) can be written as two different linear combinations of 12n v ,v ,,v if and only if 12n v ,v ,,v are linearly dependent.)（Note: If---sufficient condition ; Only if--- necessary condition ） Proof: Let v ∈ Span(12n v ,v ,,v ), then 1122n n v v v v ααα=+++Necessity: (contrapositive law for propositions)Suppose that vector v in Span(12n v ,v ,,v ) can be written as two different linear combination of 12n v ,v ,,v , then prove that 12n v ,v ,,v are linearly dependent. The difference of two different linear combinations gives a dependency relation of 12n v ,v ,,vSuppose that 12n v ,v ,,v are linearly dependent, then there exist twodifferent representations. The sum of the original relation plus the dependency relation gives a new representation.Assignment for section 3, chapter 3Hand in : 5, 11, 13, 14, 15, ; Not required: 6, 7, 8, 9, 10,§4. Basis and DimensionNew words and phrasesBasis 基Dimension 维数Minimal spanning set 最小生成集Standard Basis 标准基4.1 Definitions and TheoremsA minimal spanning set for a vector space V is a spanning set with no unnecessary elements (i.e., all the elements in the set are needed in order to span the vector space). If a spanning set is minimal, then its elements are linearly independent. This is because if they were linearly dependent, then we could eliminate a vector from the spanning set, the remaining elements still span the vector space, this would contradicts the assumption of minimality. The minimal spanning set forms the basic building blocks for the whole vector space and, consequently, we say that they form a basis for the vector space（向量空间的基）.★Definition The vectorsv,v,,v form a basis for a vector space V12nif and only if(i)v,v,,v are linearly independent12n(ii)v,v,,v span V.12nA basis of V actually is a minimal spanning set（最小张成集）for V.We know that spanning sets for a vector space are not unique. Minimal spanning sets for a vector space are also not unique. Even though, minimal spanning sets have something in common. That is, the number of elements in minimal spanning sets.We will see that all minimal spanning sets for a vector space have the same number of elements.Theorem 3.4.1 If {}12n v ,v ,,v is a spanning set for a vector space V , then any collection of m vectors in V , where m>n, is linearly dependent.Proof Let {}12m u ,u ,,u be a collection of m vectors in V . Then each u i can be written as a linear combination of 12n v ,v ,,v .i 1122n u =v +v ++v i i in a a aA linear combination 1122m u + u u m c c c ++can be written in the formnnn11j 22j j j=1j=1j=1v + v v j j m nj c a c a c a ++∑∑∑Rearranging the terms, we see that 1122m j 11u + u u ()v nmm ij i j i c c c a c ==++=∑∑Then we consider the equation 1122m m c u + c u c u 0++= to see if we canfind a nontrivial solution (12n c ,c ,,c ). The left-hand side of the equation can be written as a linear combination of 12n v ,v ,,v . We show that thereare scalars 12n c ,c ,,c , not all zero, such that 1122m m c u + c u c u 0++=.Here, we have to use a theorem: A homogeneous linear system must have a nontrivial solution if it has more unknowns than equations. Corollary 3.4.2 If {}12n v ,v ,,v and {}12m u ,u ,,u are both bases for a vector space V , then n=m. (all the bases must have the same number of vectors.)Proof Since 12n v ,v ,,v span V , if m>n, then {}12m u ,u ,,u must be linearly dependent. This contradicts the hypothesis that {}12m u ,u ,,u is linearly independent. Hence m n ≤. By the same reasoning, n m ≤. So m=n.From the corollary above, all the bases for a vector space have the same number of elements (if it is finite). This number is called the dimension of the vector space.★Definition Let V be a vector space. If V has a basis consisting of n vectors, we say that V has dimension n (the dimension of a vector space of V is the number of elements in a basis.) The subspace {0} of V is said to have dimension 0. V is said to be finite-dimensional if there is a finite set of vectors that spans V; otherwise, we say that V is infinite-dimensional.Recall that a set of n vector is a basis for a vector space if two conditions are satisfied. If we know that the dimension of the vector space is n, then we just need to verify one condition.Theorem 3.4.3 If V is a vector space of dimension n>0:I.Any set of n linearly independent vectors spans V (so this setforms a basis for the vector space).II.Any n vectors that span V are linearly independent (so this set forms a basis for the vector space).ProofProof of I: Suppose thatv,v,,v are linearly independent and v is12nany vector in V. Since V has dimension n, the collection of vectorsv,v,,v,v must be linearly dependent. Then we show that v can be 12nexpressed in terms ofv,v,,v.12nProof of II: Ifv,v,,v are linearly dependent, then one of v’s can12nbe written as a linear combination of the other n-1 vectors. It follows that those n-1 vectors still span V. Thus, we will obtain a spanning set with k<n vectors. This contradicts dimV=n (having a basis consisting of n vectors).Theorem 3.4.4 If V is a vector space of dimension n>0:(i) No set of less than n vectors can span V .(ii)Any subset of less than n linearly independent vectors can be extended to form a basis for V .(iii) Any spanning set containing more than n vectors can be pareddown (to reduce or remove by or as by cutting) to form a basis for V . Proof(i): If there are m (<n) vectors that can span V , then we can argue that dimV<n. this contradicts the assumption.(ii) We assume that 12k v ,v ,,v are linearly independent ( k<n). Then Span(12k v ,v ,,v ) is a proper subspace of V . There exists a vector1v k + that is in V but not in Span(12k v ,v ,,v ). We can show that12k v ,v ,,v ,1v k + must be linearly independent. Continue this extensionprocess until n linearly independent vectors are obtained.(iii) The set must be linearly independent. Remove (eliminate) one vector from the set, the remaining vectors still span V . If m-1>n, we can continue to eliminate vectors in this manner until we arrive at a spanning set containing n vectors.4.2 Standard BasesThe standard bases（标准基）for n R, m nR .Although the standard bases appear to be the simplest and most natural to use, they are not the most appropriate bases for many applied problems. Once the application is solved in terms of the new basis, it is a simple matter to switch back and represent the solution in terms of the standard basis.Assignment for section 4, chapter 3Hand in : 4, 7, 9，10，12，16，17,18Not required: 11，13，14, 15，§5. Change of BasisNew words and phrasesTransition matrix 过渡矩阵5.1 MotivationMany applied problems can be simplified by changing from one coordinate system to another. Changing coordinate systems in a vector space is essentially the same as changing from one basis to another. For example, in describing the motion of a particle in the plane at a particular time, it is often convenient to use a basis for 2R consisting of a unit tangent vector t and a unit normal vector n instead of the standard basis. In this section we discuss the problem of switching from one coordinate system to another. We will show that this can be accomplished by multiplying a given coordinate vector x by a nonsingular matrix S.5.2 Changing Coordinates in 2RThe standard basis for 2R is 12{e ,e }. Any vector in 2R can be written as a linear combination 12{e ,e }1122x=e +e x x .The scalars 12 and x x can be thought of as the coordinates （坐标） of x with respect to the standard basis. Actually, for any basis 12{u ,u } for 2R , a given vector x can be represented uniquely as a linear combination1122x=u +u c cThe scalars 12 and c c are the coordinates of x with respect to the basis12{u ,u }. Let us denote the ordered bases by [12e ,e ] and [12u ,u ]. 12(,)T x x iscalled the coordinate vector of x with respect to [12e ,e ],12(,)T c c the coordinate vector of x with respect to [12u ,u ].We wish to find the relationship between the coordinate vectors x and c.11122122x=e +e (e ,e )x x x x ⎛⎫= ⎪⎝⎭11122122x=u +u (u ,u )c c c c ⎛⎫= ⎪⎝⎭11121222(e ,e )(u ,u )x y x y ⎛⎫⎛⎫= ⎪ ⎪⎝⎭⎝⎭111222(u ,u )x c x c ⎛⎫⎛⎫= ⎪ ⎪⎝⎭⎝⎭Or simply, x=UcThe matrix U is called the transition matrix （过渡矩阵）from the ordered basis [12u ,u ] to [12e ,e ].The matrix U is nonsingular since 12u ,u are linearly independent. By the formula x=Uc, we see that if given a vector 1122u +u c c , its coordinate vector with respect to [12e ,e ] is given by Uc.Conversely if given a vector 12(,)T x x , then its coordinate vector with respect to [12u ,u ] is given by -1U xNow let us consider the general problem of changing from one basis[12v ,v ] to another basis [12u ,u ]. In this case, we assume that 112212x v +v (v ,v )c c c == and 112212x u +u (u ,u )d d d == ThenVc=UdIt follows that1d U Vc -=.Thus, given a vector x in 2R and its coordinate vector c with respect to the ordered basis [12v ,v ], to find the coordinate vector of x with respect to the new basis [12u ,u ], we simply multiply c by the transition matrix1S U V -=.where 12V=(v ,v ) and 12U=(u ,u ) Example (example 4 on page 156) Given two bases15v 2⎛⎫= ⎪⎝⎭, 27v 3⎛⎫= ⎪⎝⎭and 13u 2⎛⎫= ⎪⎝⎭, 21u 1⎛⎫= ⎪⎝⎭(1) Find the coordinate vectors c and d of the vector ()x=12,5Twith respect to the bases [12v ,v ] and [12u ,u ], respectively.12[e ,e ]12[v ,v ]12[u ,u ]1U -UV1U V -(2) And find the transition matrix S corresponding to the change of basis from [12v ,v ] to [12u ,u ]. (3) Check that d=Sc.Solution: The coordinate vector with respect to the basis [12v ,v ] is15712371212352551--⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫== ⎪ ⎪ ⎪⎪ ⎪-⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭The coordinate vector with respect to the basis [12u ,u ] is13112111272152359--⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫== ⎪ ⎪ ⎪⎪ ⎪--⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭The transition matrix corresponding to the change of the basis from [12v ,v ] to [12u ,u ] isS=131571157342123232345--⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫== ⎪ ⎪ ⎪⎪ ⎪---⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭Check that73419451⎛⎫⎛⎫⎛⎫= ⎪ ⎪⎪---⎝⎭⎝⎭⎝⎭.The discussion of the coordinate changes in 2R can be easily generalized to that in n R . We summarize it as follows.12n [v ,v ,,v ]1U -UV1U V -12n [e ,e ,,e ]12n [u ,u ,,u ]where 12V=(v ,v ,,v )n and 12U=(u ,u ,,u )nInterpretation: if x=12(,,,)T n x x x is a vector in n R , then the coordinate vector c of x with respect to 12[v ,v ,,v ]n is given by x=Vc, (c=-1V x ), the coordinate vector d of x with respect to 12[u ,u ,,u ]n is given by x=Ud, (d=-1U x ). The transition matrix from 12[v ,v ,,v ]n to 12[u ,u ,,u ]n is given by S=1U V -.5.3 Change of Basis for a General Vector Space★Definition (coordinate) Let V be a vector space and let E=[12n v ,v ,,v ] be an ordered basis for V . If v is any element of V , then v can be written in the form121122n n 12n n v v v v [v ,v ,,v ]c c c c c c ⎛⎫ ⎪ ⎪=+++= ⎪ ⎪⎝⎭(this is a formal multiplication since vectors here are not necessarily column vectors in n R ) where 12n c ,c ,,c are scalars. Thus we can associate with each vector v a unique vector c=12n (c ,c ,,c )T in n R . The vector c defined in this way is called theE [v]. The i c ’s are called coordinates of v relative to E . Transition MatrixLet E=[12n w ,w ,,w ], F=[12n v ,v ,,v ] be two ordered bases for V .Then11112121212122221122w v v v w v v v w v v v n n n n n n n nn ns s s s s s s s s =+++=+++=+++Formally, this change of bases can be written as111212122212n 12n 12[w ,w ,,w ][v ,v ,,v ]n n n n nn s s s s s s s s s ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭(The multiplication is formal matrix multiplication. If the vector space is the Euclidean space, then the multiplication becomes the actual multiplication.)This is called the change of basis from E=[12n w ,w ,,w ] to F =[12n v ,v ,,v ].A vector v has different coordinate vectors in different bases. Let x=E [v], i.e. 1122n v w +w ++w n x x x = and y=F [v], 1122n v v +v ++v n y y y =, then 1122n 111v ()v +()v ++()v nnnj j j j nj j j j j s x s x s x ====∑∑∑1ni ij j j y s x ==∑In matrix notation, we have y=Sx, whereS=111212122212n n n n nn s s s ss s s s s ⎛⎫ ⎪ ⎪⎪ ⎪⎝⎭This matrix is referred to as the transition matrix corresponding to the change of basis from E=[12n w ,w ,,w ] to F =[12n v ,v ,,v ] S is nonsingular, since Sx=y if and only if1122n n 1122n n w +w ++w v +v ++v x x x y y y =Sx=0 implies that 1122n n w +w ++w 0x x x =. Hence x must be zero. 1S y x -=1S - is the transition matrix corresponding to the change of base from F=[12n v ,v ,,v ] to E=[12n w ,w ,,w ]Any nonsingular matrix can be thought of as a transition matrix. If S is an nxn nonsingular matrix and [12n v ,v ,,v ] is an ordered basis for V , then define [12n w ,w ,,w ] by111212122212n 12n 12[w ,w ,,w ][v ,v ,,v ]n n n n nn s s s s s s s s s ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭Then12nw ,w ,,w arelinearly independent. Suppose that1122n n w +w ++w 0x x x =Then1122n 111()v +()v ++()v 0nnnj j j j nj j j j j s x s x s x ====∑∑∑By the linear independence of 12n v ,v ,,v , it follows that10nij jj s x==∑or , equivalentlySx=0Since S is nonsingular, x must equal zero. Therefore, 12n w ,w ,,w are linearly independent and hence they form a basis for V . The matrix S is the transition matrix corresponding to the change from the ordered basis [12n w ,w ,,w ] to [12n v ,v ,,v ]. Example Let 110u 01⎛⎫= ⎪⎝⎭ 210u 01⎛⎫= ⎪-⎝⎭ 301u 10⎛⎫= ⎪⎝⎭401u 10⎛⎫= ⎪-⎝⎭; 110v 00⎛⎫=⎪⎝⎭ 201v 00⎛⎫= ⎪⎝⎭ 301v 10⎛⎫= ⎪⎝⎭ 410v 01⎛⎫= ⎪-⎝⎭.Find the transition matrix corresponding to the change of base from E=[1234u ,u ,u ,u ] to F =[1234v ,v ,v ,v ]In many applied problems it is important to use the right type of basis for the particular application. In chapter 5 we will see that the key to solving least squared problems is to switch to a special type of basis called an orthonormal basis. In chapter 6 we will consider a number of applications involving the eigenvalues and eigenvectors associated with an nxn matrix A. The key to solving these types of problems is to switch to a basis for n R consisting of eigenvectors of A.Chapter 3---Section 5 Change of Basis Assignment for section 5, chapter 3 Hand in: 6, 7, 8, 11 ,Not required; 9, 10,§6. Row Space and Column SpaceNew words and phrasesRow space 行空间Column space 列空间Rank 秩6.1 DefinitionsWith an mxn matrix A, we can associate two subspaces.Definition If A is an mxn matrix, the subspace of 1n R ⨯ spanned by the row vectors of A is called the row space of A, the subspace of m R spanned by the column vectors of A is called the column space of A.Theorem 3.6.1 Two row equivalent matrices have the same row space. Proof 21kE E E A B =The row vectors of B must be a linear combination of the row vectors of A. Consequently, the row space of B must be a subspace of the row space of A. By the same reasoning, the row space of A is a subspace of the row space of B. So, they are the same.★Definition The rank （秩）of a matrix of A is the dimension of the row space of A.。

线性代数 英文讲义

线性代数 英文讲义

麻省理工上课讲义 线性代数[第1集]An overview of key ideas

线性代数英文课件：ch1_3 Cofactor Expansion

线性代数 英文讲义

线性代数英文课件：ch3-1 Elementary Operations

线性代数讲义

线性代数英文课件2.2

线性代数英文课件3.3

线性代数英文课件：ch4_Review

线性代数英文课件：ch5-1Eigenvalues and Eigenvectors

线性代数 英文讲义

北京邮电大学国际学院线性代数讲义Lecture 04

高等数学【线性代数】英文版课件1

线性代数英文课件4.1

线性代数英文课件：ch2-2 Inverse of a Matrix

线性代数英文课件4.2

线性代数 英文讲义

线性代数英文讲义

线性代数英文讲义

麻省理工上课讲义线性代数[第1集]An overview of key ideas

线性代数英文讲义

线性代数英文讲义

线性代数英文讲义