线性代数英文ppt1

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线性代数英文讲义

Chapter 3 Vector Spaces1.Suppose that(6204),(3157)αβ=-=-,find vector v , let βα43=-v2.Determine whether the vector β can be a linear combination of the vectors123,,ααα，if it can be ，write down the expressions 。

（1）123(83125),(1305),(2073),(4126)βααα=--=-=-=--3. Prove that any vector 1234()b b b b β=can be a linear combination of the vectors 1234(1000),(1100)(1110),(1111)αααα====。

4.Determine whether the following vectors are linearly independent 。

（1）123(1203),(2510),(3412).ααα=-=-=。

（2）1234(3425),(2503),(5012),(3335)αααα=-=--=-=-。

（3）232323123(1),(1),(1)aa ab b bc c c ααα===。

5. If 123,,ααα are linearly independent ，prove that whether 12233123,4,5αααααα+++ are linearly independent 。

6. Suppose that 1234,,,αααα are linearly independent ，then determine whether 12233441,,,αααααααα++++ are linearly independent, show the reason.7. Prove that in n R ，if 12,,,n ααα are linearly independent ，then any vector n R ∈β can be a linear combination of12,,,n ααα .8. If 1234,,,αααα are linearly dependent ，and any three of then are linearly independent, prove that there must have nonzero numbers 1234,,,k k k k such that112233440k k k k αααα+++=。

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 =。

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1.3

Math1229A/BUnit3: Lines and Planes(text reference:Section1.3)c V.Olds201030Unit33Lines and PlanesLines inℜ2You are already familiar with equations of lines.In previous courses you will have written equations of lines in slope-point form,in slope-intercept form,and probably also in standard form for a line inℜ2.Recall that:y−y1=m(x−x1)is the slope-point form equation of the line through point(x1,y1)with slope m y=mx+b is the slope-intercept form equation of the line with slope m and y-intercept b ax+by=c is the standard form equation which either of the others can be rearranged to In this course,we don’t use the slope-point or slope-intercept forms of equations of lines.Instead, we use various other,vector-based,forms of equations.But we do still use the standard form.You already know that given any2distinct points,whether inℜ2or inℜ3,there is exactly one line which passes through both points.Suppose we have2points,P and Q.Letℓbe the line that passes through these two points.Both point P and point Q lie on lineℓ,and so do all the points between them.In fact,the directed line segment−−→P Q lies on lineℓ.When this directed line segment is translated to the origin,the resulting vector most likely doesn’t lie on lineℓ(unless the originhappens to lie on lineℓ),but if not,it does lie on a line which is parallel to lineℓ.It lies on the line parallel toℓwhich passes through the origin.So this vector does give us some information about the line.(Similar to the information given by knowing the slope of a line inℜ2,although it’s not quite the same information.)If a vector v lies on a particular line,or on a line parallel to that line,we say that v is parallel to, or is collinear with that line.And we call v a direction vector for the line.Not that the line actually has a direction associated with it.It doesn’t.It extends in both directions,but has no particular “forwards along the line”or“backwards along the line”associated with it.So don’t read too much meaning into the term direction vector.If v is a direction vector for lineℓ,then so is− v.And so is every other scalar multiple of v,except for0 v.Because of course0 v= 0which has no direction information.But every other scalar multiple of v starts at the origin,and goes either the same or the opposite direction as v and therefore also lies on the line parallel toℓwhich passes through the origin.So any such vector would be considered a direction vector for lineℓ.Deﬁnition:Any non-zero vector which is parallel to a lineℓis called a direction vectorfor lineℓ.For instance,consider the line x+y=2.The points(1,1),(2,0),(0,2),(−1,3),(3,−1),(−2,4), (4,−2),etc.,all lie on this line.So do(1/2,3/2)and(3/2,1/2)and inﬁnitely many other points. Pick any2of these points,andﬁnd the vector which is the translation to the origin of the directed line segment between them,and you have a direction vector for the line.And we know that for any points P and Q,letting p=−−→OP denote the vector from the origin to point P and q=−−→OQ denote the vector from the origin to point Q,the vector v= q− p is the translation of directed line segment −−→P Q to the origin.So for instance for points P(1,1)and Q(0,2),we have p=(1,1)and q=(0,2), and we see that v= q− p=(0,2)−(1,1)=(−1,1)is a direction vector for the line x+y=2.And other choices of P and Q give other direction vectors which are scalar multiples of this one.(Go ahead,pick some other points,and see what vectors you get.)Unit331 Point-Parallel Form32Unit3 Parametric EquationsUnit333 Example3.3.Write a point-parallel form equation for the line with parametric equationsx=1+5ty=2Solution:We use the x equation toﬁnd theﬁrst components for our point-parallel equation,and the y equation toﬁnd the second components.We need to recognize that in each equation,the number on the right hand side that isn’t multiplied by t is the coordinate of the known point,P,and that the number that is multiplied by t is the component of the direction vector, v.So from theﬁrst equation,i.e. the x-equation,we see that p1=1and v1=5.And from the second equation,since there’s no t multiplying it,the2must be p2.So where’s the t?It’s invisible,which means it must have a0 multiplier.That is,v2=0.So we have the point P(1,2)and the direction vector v=(5,0),which when we put it in the form x(t)= p+t v gives the point-parallel form equationx(t)=(1,2)+t(5,0)Two-Point Form-34Unit 3But nothing that we did here really required being in between P and Q .We could do something similar for any point on the line containing P and Q .The only diﬀerence is that t would no longer necessarily be between 0and 1.That is,for any point X on the line containing the points P and Q ,we could travel from the origin to point P ,and then travel some scalar multiple of the vector u to end up at the point X .For instance,consider the diagram shown below.As before,we have x = p +t u ,but now t is bigger than 1.Or if we needed to go the other direction along the line,from P ,then t would be negative.6X X X X X X X X X X X X X X X X X X X X ¡¡¡¡¡¡!P p Q q ¨¨¨¨¨¨¨¨¨B X x And so for any point X on the line,we havex = p +t u = p +t ( q − p )= p +t q −t p =(1−t ) p +t qfor some value t .That is,we can express the line containing two points P and Q as the line containing all points X (x,y )such that x =(1−t ) p +t q for some value t .And so we have another form of equation for the line.We call this the two-point form,and as before,we write x (t )instead of just x .Deﬁnition:The two-point form of equation for the line through points P and Q is:x (t )=(1−t ) p +t qExample 3.4.Write equations in two-point form for each of the lines in Example 3.1.Solution:(a)The line here is the line through the points P (3,1)and Q (0,6),so we have p =(3,1)andq =(0,6)and we get the two-point form x (t )=(1−t )(3,1)+t (0,6)(b)This time,we have the line through P (1,2)with direction vector v =(2,−1).We don’t know two points on the line,so we need to ﬁnd a second point.We saw in Example 3.1(b)that a point-parallel form equation for this line is x (t )=(1,2)+t (2,−1).We can choose any value of t other than 0to get another point on the line.(Notice:We don’t want to use t =0,because that will just give us the point we already know.But any other value of t will do.)For instance,using t =1we have x (1)=(1,2)+1(2,−1)=(1,2)+(2,−1)=(3,1),so we see that Q (3,1)is another point on the same line.Now that we know two points on the line,we can ﬁnd a two-point form equation.Notice,though,that since we have already been using t as the parameter for the point-parallel form equation,we should use a diﬀerent name for the parameter in the two-point form equation.(Especially since we gave t a speciﬁc value.We wouldn’t want to get confused and think the parameter in the two-point form equation was supposed to have that value too.)Notice also that it doesn’t matter in the least what letter we use to represent the parameter (which is just a scalar multiplier).So we can use s instead.We get:x (s )=(1−s )(1,2)+s (3,1)Unit335 (c)This time,we have the line through the origin with direction vector(0,1).We know that the point(0,0)is on the line,and clearly the point(0,1)is also on the line(because the vector(0,1)is on the line,since the line does pass through the origin).So a two-point form equation for this line isx(t)=(1−t)(0,0)+t(0,1)Point-Normal Form36Unit3 Example3.6.(a)Find an equation in point-normal form for the line x(t)=(0,1)+t(2,−1).(b)Write an equation in point-parallel form for the line from Example3.5.(c)Write a point-normal form equation for the line with parametric equationsx=3+t and y=2t−4Solution:(a)We have x(t)=(0,1)+t(2,−1),which we recognize as a point-parallel form equation for the line through point(0,1)parallel to the vector(2,−1).Since the vector(2,−1)is parallel to the line,then the vector(1,2),obtained by switching the components and changing one of the signs, is perpendicular to the line.That is, n=(1,2)is a normal for this line.So a point-normal form equation for the line is(1,2)•( x−(0,1))=0(b)In Example3.5we found the point-normal form equation(−1,1)•( x−(1,2))=0for a particular line.Since(−1,1)is a normal for this line,and the vector(1,1)is orthogonal to(−1,1),then the vector(1,1)is parallel to the line,i.e.is a direction vector for the line.And of course(1,2)is a point on the line.So a point-parallel form equation for this line isx(t)=(1,2)+t(1,1)(c)From the parametric equations of the line we can identify both a point on the line and a direction vector for the line.Remember,the multiplier on t is the component of the direction vector,while the number without a t is the coordinate of the known point.Keeping this in mind allows us to correctly identify both the known point and the direction vector from the parametric equations, even when they look a bit diﬀerent than we expect.Here,the parametric equations are given asx=3+ty=2t−4We’re more accustomed to seeing the form we have in the x equation.The form in the y equation, with the t term coming before the non-t term,is diﬀerent.This is just done to avoid having a “leading negative”.Equations look less tidy when theﬁrst thing on one side of the equation is a negative sign,so mathematicians often avoid writing things that way.That is,the given parametric equations are just a tidier form ofx=3+ty=−4+2tIn this form we see that the corresponding point-parallel form equation is x(t)=(3,−4)+t(1,2). So(1,2)is a direction vector for the line and therefore(2,−1)is a normal for the line.Thus we can write a point-normal form equation as(2,−1)•( x−(3,−4))=0Unit337 Standard Form38Unit3 Example3.8.Write a point-normal form equation for the line x−2y=5.Solution:We use the coeﬃcients of x and y as the components of a normal vector for the line.Of course, x−2y=1x+(−2)y,so the coeﬃcients are1and−2.That is,we get n=(1,−2)as a normal vector for the line.Now,we just need toﬁnd any point on the line.We plug in any convenient x-value and solve for y.Or we plug in any convenient y-value and solve for x.For instance,when y=0we have x−2(0)=5,so x−0=5.That is,we see that when y=0we must have x=5.So P(5,0)is a point on the line.Now we can write the point-normal form equation:(1,−2)•( x−(5,0))=0Example3.9.Write a standard form equation of the line x(t)=(3,2)+t(2,7).Solution:From the given point-parallel form equation,we see that P(3,2)is a point on the line and v=(2,7) is a direction vector for the line,i.e.is parallel to the line.And so n=(7,−2)is a normal vector to the line,so the standard form equation has7x−2y=c for some value c.And we canﬁnd c usingc=(7,−2)•(3,2)=7(3)+(−2)(2)=21−4=17Therefore the standard form equation is7x−2y=17.Lines inℜ3Of course,we can have lines in3-space,as well as in the plane.And there’s a lot that’s the same inℜ3as it was inℜ2,so we use the same terminology and notation.For instance,when we move from2dimensions to3,it’s still true that given any2points,there is exactly one line that passes through both those points.And the vector equivalent to the directed line segment between those points is parallel to that line,so we still call it a direction vector for the line.That is,we deﬁne the term direction vector the same way inℜ3as we did inℜ2.Deﬁnition:If v∈ℜ3is parallel to some lineℓinℜ3,we say that v is a directionvector forℓ.As inℜ2,we can use a direction vector for a line(i.e.a vector parallel to the line)and any one point on the line to write a point-parallel equation for the line.And from that we can write parametric equations.Or we could write a2-point form equation,instead.The only diﬀerence is that now the points have3coordinates and the vectors have3components. Of course,for parametric equations this means that we have a third equation,corresponding to the z components of the vectors.These observations are summarized in the following deﬁnitions.Unit339 Deﬁnition:Let P(p1,p2,p3)and Q(q1,q2,q3)be any points inℜ3and let v=(v1,v2,v3)be any vector inℜ3.Then:1.Ifℓis the line which passes through P parallel to v(so that v is a direction vectorforℓ),thenx(t)=(p1,p2,p3)+t(v1,v2,v3)is an equation for lineℓin point-parallel form.2.If lineℓpasses through point P and v is a direction vector forℓ,then parametricequations of lineℓare:x=p1+tv1y=p2+tv2z=p3+tv33.If points P and Q are both on lineℓthen a two-point form equation forℓisx(t)=(1−t)(p1,p2,p3)+t(q1,q2,q3)Example3.10.Letℓbe the line which passes through the points P(1,2,3)and Q(1,−1,1).Write equations of lineℓin two-point form and in point-parallel form.Solution:In two-point form,we get the equation forℓ:x(t)=(1−t)(1,2,3)+t(1,−1,1)For a point-parallel form equation of lineℓweﬁrst need toﬁnd a direction vector forℓ.The directed line segment−−→P Q is equivalent tov= q− p=(1,−1,1)−(1,2,3)=(0,−3,−2)which is parallel to(and hence is a direction vector for)ℓ.Using this direction vector and the point P which we know is on the line,we getx(t)=(1,2,3)+t(0,−3,−2)(Of course,we could have used point Q instead of point P to write the point-parallel form equation. Likewise,we could have used p− q=(0,3,2)as the direction vector.And in the two-point form equation,we could have switched the roles of P and Q.)Example3.11.Write parametric equations for the line through the point(0,1,−1)which is parallel to v=(2,1,0).Solution:An equation of the line in point-parallel form is x(t)=(0,1,−1)+t(2,1,0).This tells us that a point(x,y,z)is on this line if there is some value of t for which(x,y,z)=(0,1,−1)+t(2,1,0).So it must be true that,for the same value of t,we havex=0+2ty=1+1tz=−1+0t40Unit3 That is,we can write parametric equations of the line asx=2ty=1+tz=−1Example3.12.ℓ1is the line x(t)=(1−t)(2,1,−1)+t(0,1,2).ℓ2is the line with parametric equa-tions x=2t−2,y=1,z=5−3t.Areℓ1andℓ2the same line?Solution:Hmm.That’s diﬀerent.Let’s see.Forℓ1we recognize that what we’ve been given is a two-point form equation.(We can tell because of the(1−t)multiplier.)From it we can see that P(2,1,−1) and Q(0,1,2)are two points on lineℓ1.This also tells us that the vectorv=−−→P Q= q− p=(0,1,2)−(2,1,−1)=(−2,0,3)is parallel to lineℓ1.Forℓ2we’re given parametric equations.It may be helpful to write these equations all the same way,with“constant+multiple of t”on the right hand side.We have:x=2t−2x=−2+2ty=1⇒y=1+0tz=5−3t y=5+(−3)tFrom the rearranged set of equations,using our knowledge of the form of parametric equations,we see that the point onℓ2used to write these parametric equations is R(−2,1,5).Also,the direction vector used for these equations is u=(2,0,−3).Since u=(2,0,−3)=−(−2,0,3)=− v,we see that these vectors are scalar multiples of one an-other,so they are collinear.That is,the direction vector u used to write the equation ofℓ2is parallel to the vector which we know is parallel toℓ1.Therefore u is also parallel toℓ1,and thus linesℓ1 andℓ2are parallel to one another.It’s possible that they could be the same line.How can we tell whether they are?Sinceℓ1andℓ2are parallel,then either they have no points in common or else they are the same line and have all points in common.So all we need to do is determine whether any point which is known to be on one line is also on the other.If it is,then they are actually the same line.But if it isn’t,then they must be diﬀerent,but parallel,lines.We know that the point P(0,1,2)is on lineℓ1.Is it also on lineℓ2?If it is,then(x,y,z)=(0,1,2) must satisfy the parametric equations forℓ2,using the same value of t for each component(equation). Since the second coordinate of P is1,the equation y=1is satisﬁed.For theﬁrst coordinate,we see that we need to have x=2t−2satisﬁed for x=0.This gives0=2t−2⇒0+2=2t⇒2t=2⇒t=1Now,if we substitute t=1into the third of the parametric equations,we getz=5−3(1)=5−3=2Since z=2is the third coordinate of point P,we see that the point(x,y,z)=(0,1,2)does satisfy the parametric equations ofℓ2.That is,we have(0,1,2)=(−2,1,5)+1(2,0,−3)Unit341 so(x,y,z)=(0,1,2)satisﬁes x=−2+2t,y=1+0t and z=5−3t with t=1.Because the point (0,1,2)does satisfy the equations forℓ2,it is a point on lineℓ2.So now we know thatℓ1andℓ2are parallel lines,with a point in common,which means that they must have all points in common and be the same line.That is,sinceℓ1andℓ2are parallel and intersect at point P(0,1,2),they must intersect at all other points as well,and actually be the same line.Planes inℜ3We know that inℜ2,something of the form ax+by=c is the standard form of an equation of a line.What about the3-dimensional equivalent,ax+by+cz=d.Is that an equation of a line? Well,let’s see.Let’s think about a speciﬁc,uncomplicated,example.Consider the equation x+y+z=0. This equation is satisﬁed by the point P(2,−2,0)(because2+(−2)+0=0)and also by the point Q(3,−3,0).And we know that there’s a unique line that passes through those2points.Let’s call that lineℓing v= q− p=(3,−3,0)−(2,−2,0)=(1,−1,0)as a vector which is parallel toℓ1, we can write an equation ofℓ1asx(t)=(2,−2,0)+t(1,−1,0)Notice that for any point(x,y,z)on lineℓ1we havex=2+ty=−2−tz=0and so x+y+z=(2+t)+(−2−t)+0=2−2+t−t=0.That is,every point onℓ1satisﬁes the equation x+y+z=0.But those aren’t the only points which satisfy that equation.For instance,the point R(1,0,−1) also satisﬁes this equation.And this point is not onℓ1.The easiest way to tell is because from the parametric equations ofℓ1we can see that every point onℓ1has z=0,but the third coordinate of point R isn’t0,so it is not a point onℓ1.Hmm.Every point on lineℓ1satisﬁes x+y+z=0,but it’s not true that every point that satisﬁes x+y+z=0is on lineℓ1.So x+y+z=0cannot be an equation of lineℓ1.Then what is it?Well,actually,it’s the equation of a plane.ℜ2(i.e.2-space)is just a single plane.Butℜ3, which is to say3-space,contains inﬁnitely many planes.(For instance,think of the walls,ceilings andﬂoors of the building you’re in.And also every other building you ever have been or ever could be in.And all the ramps you’ve ever seen.And what those ramps would look like if they were knocked oﬀkilter.And...Each of those things lies in some particular plane inℜ3,and there are many other planes besides those.)So x+y+z=0is an equation of a plane.Let’s call it the planeΠ.(Planes are often named Π,which is just the Greek letter P,just like lines are namedℓ.ℓfor line,Πfor plane.Same idea.) Any plane contains inﬁnitely many lines.One of the lines that lies in the particular planeΠwe’ve been talking about is the lineℓ1.But there are many others.For instance,we saw that P(2,−2,0) and R(1,0,−1)both lie on this plane,so the line on which those points lie is another line in plane Π.We can call that oneℓ2.And the vector r− p=(1,0,−1)−(2,−2,0)=(−1,2,−1)is parallel to ℓ2so we can expressℓ2as x(t)=(1,0,−1)+t(−1,2,−1).42Unit3 Notice that x+y+z=(1,1,1)•(x,y,z).Let’s think about the vector(1,1,1)whose components are the coeﬃcients in the equation of the planeΠ.We know that(1,−1,0)is parallel to lineℓ1, which lies in planeΠ.Notice that(1,1,1)•(1,−1,0)=1(1)+1(−1)+1(0)=1−1+0=0, so the vector(1,1,1)is a normal for(i.e.is perpendicular to)lineℓ1.Likewise,we know that (−1,2,−1)is parallel to lineℓ2,which also lies in planeΠ.Notice that(1,1,1)•(−1,2,−1)= 1(−1)+1(2)+1(−1)=−1+2−1=0,so the vector(1,1,1)is also a normal for(perpendicular to) lineℓ2.However(1,−1,0)is not a scalar multiple of(−1,2,−1),so those vectors aren’t orthogonal (i.e.parallel)and therefore linesℓ1andℓ2aren’t parallel to one another.How can the same vector be perpendicular to both?Well,by being perpendicular to the whole plane in which both lines lie. This vector(1,1,1)is actually perpendicular to,i.e.a normal for,the planeΠ.Deﬁnition:A vector which is perpendicular to a particular plane inℜ3is said to benormal to the plane,and is called a normal for that plane,or a normal vector forthe plane.Point-Normal Form of an Equation of a PlaneUnit343 How do weﬁnd a normal for the plane?Well,we know from the equation ofℓ1that the vector u=(1,−1,0)is parallel toℓ1,and likewise from the equation ofℓ2that the vector v=(−1,2,−1) is parallel toℓ2.Of course any vector n which is a normal forΠ(i.e.is perpendicular to this plane) must be perpendicular to any line that lies withinΠ.So if n is a normal forΠ,then n is perpen-dicular to bothℓ1andℓ2and therefore must be orthogonal to both u and v.(That is,any vector which is perpendicular toℓ1is also perpendicular to(orthogonal to)every vector that is parallel to ℓ1.And similarly forℓ2.)So how do weﬁnd a vector which is perpendicular to both u and v?Well that’s easy.We know that the vector u× v is perpendicular to both u and v.So we can usen= u× v=(1,−1,0)×(−1,2,−1)=((−1)(−1),(0)(−1),(1)(2))−((2)(0),(−1)(1),(−1)(−1))=(1,0,2)−(0,−1,1)=(1,1,1)(Recall that we discussed previously that the vector(1,1,1)was a normal for the plane containing these linesℓ1andℓ2.)Now we know both a normal vector forΠand a point in planeΠso we can write the point-normal form equation.We get(1,1,1)•( x−(2,−2,0))=0Standard Form Equation of a Plane44Unit3 Example3.15.Write an equation in standard form for the plane with normal vector n=(1,2,3) which contains the point P(0,−1,2).Solution:Since n=(1,2,3)is a normal vector for the plane,then the standard form equation must have the form1x+2y+3z=d for some scalar d.But of course we would write that as x+2y+3z=d. How can weﬁnd the value of d?Well,we know that the point P(0,−1,2)lies on the plane,so (x,y,z)=(0,−1,2)must satisfy this equation.That is,we plug in x=0,y=−2and z=2toﬁnd the value of d.We get:x+2y+3z=d⇒0+2(−1)+3(2)=d⇒−2+6=d⇒d=6−2=4So a standard form equation of the plane is x+2y+3z=4.Notice:We could have used n• p=d,from rearranging the point-normal equation for the plane. What we did here is just another explanation of the exact same arithmetic.(Look back at the ex-amples in which we found point-normal equations of lines.We could have described the arithmetic we did there as“let x=p1and y=p2”instead of“ﬁnd x• p”.)The Plane Determined by Three PointsUnit345 Example3.16.Find both a point-normal form equation and a standard form equation of the plane determined by the points P(−1,0,1),Q(1,2,3)and R(2,−1,5).Solution:The line passing through points P and Q lies in this plane,and so any vector parallel to that line is also parallel to the plane.And if we let u= q− p,then u is such a vector.Similarly,the vector v= r− p is parallel to the line which passes through both P and R,and since that line also lies in the plane, v is another vector which is parallel to the plane we need to describe.Also,we haveu= q− p=(1,2,3)−(−1,0,1)=(1−(−1),2−0,3−1)=(2,2,2)v= r− p=(2,−1,5)−(−1,0,1)=(2−(−1),−1−0,5−1)=(3,−1,4)and we can see that since u and v are not scalar multiples of one another then they are not collinear. We use these two non-collinear vectors which are both parallel to the plane toﬁnd a normal for the plane:n= u× v=(8−(−2),6−8,−2−6)=(10,−2,−8)Now we use this normal vector and any one of the three points to write a point-normal equation of the plane.For instance,using point P,the form n•( x− p)=0gives:(10,−2,−8)•( x−(−1,0,1))=0Finally,we can also rearrange this equation to standard form.Letting x=(x,y,z),we get:(10,−2,−8)•((x,y,z)−(−1,0,1))=0⇒(10,−2,−8)•(x,y,z)−(10,−2,−8)•(−1,0,1)=0⇒10x−2y−8z=(10,−2,−8)•(−1,0,1)⇒10x−2y−8z=−10+0−8⇒10x−2y−8z=−18(Note:We might prefer to divide through the equation by2.That is,this plane would often be expressed as5x−y−4z=−9.)Determining the Distance between a Point and a Plane|| n||46Unit3That is,we simply need toﬁnd the dot product of any normal vector to the plane with the vector equivalent to the directed line segment between the point P and any known point on the plane,discard the negative sign(if there is one),and divide by the magnitude of the normal vector used.(Notice:We have not explained why this gives −−→P′P ,so you should not be trying to understand that from the above.If you’re interested,look at the explanation given in the text.All we’ve done here is to assert that it can be shown that this is true.)Theorem3.3.Consider any planeΠ.Let n be any normal vector for planeΠand let Q be any point on planeΠ.Consider any other point P which is not on the planeΠ.Then the distance between point P and planeΠis given by:distance=| n•( q− p)|√612+22+12=and so the distance from P to the plane isdistance=| n•( q− p)|√√|(1,1,−1)•((5,0,0)−(0,0,0))| √|| n||=Unit347 Finding the Distance Between a Point and a Line|| n||Example3.19.Find the distance between the point P(1,2)and the lineℓdescribed by2x+y=1. Solution:Lineℓhas normal n=(2,1).We need toﬁnd some point Q on lineℓ.Letting x=0we get 2(0)+y=1,so y=1.That is,the point on lineℓwhich has x-coordinate0has y-coordinate1,so the point Q(0,1)is a point on lineℓ.(Notice that for(x,y)=(1,2)we have2x+y=2(1)+2=4=1, so P(1,2)is not on lineℓ.)The distance between P andℓis| n•( q− p)|||(2,1)||=|(2,1)•(−1,−1)|22+12=|−2−1|4+1=35Finding the Intersection of Two Lines48Unit3 Example3.20.Find the point of intersection of the lineℓ1: x(t)=(1,0)+t(2,1)with the lineℓ2: x(s)=(1,1)+s(−1,0).Solution:Forℓ1we have parametric equations x=1+2ty=tand forℓ2we havex=1−sy=1.If some point P(x,y)is on both these lines,then it must be true that there are some values of t and s which give the same values of x and y.So we must have1+2t=1−s and t=1.Since t=1, then1+2t=3,so1−s=3and we see that s=1−3=−2.Notice that we’ve found values of the parameters,s and t,but we have not yet found the point on the line which corresponds to these values.That is,we know the value of t that gives the point on lineℓ1at which the two lines intersect,and likewise we know the value of s that gives that same point on lineℓ2.But we were asked toﬁnd the actual point at which the two lines intersect.We’re notﬁnished until we’ve done that.And we have more information than we need toﬁnd the point, since we know two ways to get it.So we can use the value of t we found,in the equation forℓ1,to get the point P.And then we can use the value of s we found,in the equation ofℓ2,to check our work.We get:t=1⇒(x,y)=(1,0)+t(2,1)=(1,0)+1(2,1)=(3,1)as the point onℓ1which we were looking for.We check that the point onℓ2is the same point: s=−2⇒(x,y)=(1,1)+s(−1,0)=(1,1)+(−2)(−1,0)=(1,1)+(2,0)=(3,1) Since we didﬁnd the same point on each line,this is the point we were looking for.We see thatℓ1 andℓ2intersect at the point P(3,1).Note:As we observed above,we found values of both parameters,but really we only need one. As we have seen,the other allows us to check our work.We’re just checking that we didn’t make an arithmetic error.If we got a diﬀerent point onℓ2than the one onℓ1that would tell us that somewhere in our calculations we made an arithmetic mistake.Either inﬁnding the points,or(more likely)inﬁnding the values of the parameters.We would need to re-do our calculations until we ﬁnd the mistake,and thenﬁnish the problem(including the check)again.Example3.21.Find the point of intersection of the lineℓ1: x(t)=(1,1,2)+t(2,1,−1)with the line ℓ2: x(s)=(0,1,2)+s(1,−1,1).Solution:Forℓ1we have x=1+2ty=1+tz=2−tand forℓ2we havex=sy=1−sz=2+s.The point of intersection ofℓ1andℓ2is a point P(x,y,z)which satisﬁes both sets of equations at the same time,so we must have:1+2t=s(1)1+t=1−s(2)2−t=2+s(3) Equation(1)says that s=1+2t,so that1−s=1−(1+2t)=0−2t=−2t.Therefore equation (2)gives1+t=−2t,so1=−3t and thus t=−13into s=1+2t。

线性代数英文课件：ch1_1 Definition

Linear Algebra
Math. Dept., Wuhan University of Technology
Textbook：工程数学-线性代数,第五版，同济大学数学系编，高等教育出版社，2011 References:
➢线性代数（第7版），S.J.Leon，机械工业出版社，2007 （Linear Algebra With Applications） ➢经济数学：线性代数，吴传生等编，高等教育出
Math. Dept., Wuhan University of Technology
Linear Algebra History
➢Leibniz introduced the definition of determinant in 17th century。 (日本数学家关孝和Seki Kowa将其概念称为“行列式”)
Example 1.
2 -1
Evaluate
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34
23
Example 2. Evaluate
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-1 4
Example 3.
23
Evaluate
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15
Math. Dept., Wuhan University of Technology
Sec.1 Determinants of Order 2 and 3
24 8 4 16 4
2 10 Example 5. Evaluate D3 1 1 4
3 2 5
Math. Dept., Wuhan University of Technology
3.Permutations &Number of Inversions
? How to generalize the definitions of 2×2 and 3×3 determinants to n×n determinants? Analyzing formula (1),we can get : (1) It’s the algebraic sum of six (which is exactly the

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

线性代数第一章初等变化和秩英文版

Chapter 1 Elementary Operation and Rank
1. Gauss Elimination of Linear Systems of Equations
2. Elementary Operation and Elementary Matrices
3. Equivalence of Matrices 4. Applications of Elementary
Solution Multiply (1) by -2 and add it to (2), multiply (1) by -1 and add it to (3) we have:
x1 2x2 5x3 19
(1)
4x2 13x3 60
(4)
x2 7x3 30
(5)
Interchange (4) and (5) we obtain:
a11
A
(
A,
)
a21
a12
a22
a1n a2n
b1 b2
am1 am2 amn bm
is called the augment matrix of LS (1.1).
An LS can uniquely determine an augment
matrix A ; on the other hand an m (n 1)
matrix A can uniquely determine an m n
LS. For example 3 4
matrix
1 A 2
2 8
5 3
19 22
1 3 2 11
uniquely determines the following 3 3 LS

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

线性代数

2
3
4
LP Decoding
LP Decoding for Non-Uniform Sources
LP Decoding for the Polya Contagion Channel
Exploiting Source Redundancy at the Decoder
Non-uniformity at the source can be exploited at the decoder. Assuming a systematic (n, k ) code C , it is possible to linearize the MAP decoding metric so as to exploit non-uniformity in an LP decoder: c ˆ = argmax P (c )P (y |c )
10
−1
10 PCE 10
−2
−3
10
−4
Standard LP Decoder Systematic LP Decoder Non−Systematic LP Decoder 10
−5
0.02
0.04
0.06
0.08
ρ
0.1
0.12
0.14
0.16
Figure: Source p1 = 0.9. Top two curves: regular systematic (200, 100) LDPC code. Bottom curve: regular (300, 100) LDPC code with the ﬁrst 100 (systematic) bits punctured.
A. Cohen, F. Alajaji, N. Kashyap, G. Takahara

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)

(完整word版)《线性代数》英文专业词汇

对角矩阵
44
an adjoint matrix
伴随矩阵
45
singular matrix
奇异矩阵
46
nonsingular matrix
非奇异矩阵
47
elementary transformations
初等变换
48
vectors
向量
ቤተ መጻሕፍቲ ባይዱ49
components
分量
50
linearly combination
坐标变换
特征向量
86
characteristic polynomial
特征多项式
87
characteristic equation
特征方程
88
scalar product
内积
89
normed vector
单位向量
90
orthogonal
正交的
91
orthogonalization
正交化
92
the Gram-Schmidt process
17
identity
恒等式
18
n-order determinants
n阶行列式
19
evaluation of determinant
行列式的求值
20
Laplace’s expansion theorem
拉普拉斯展开定理
21
cofactor
余子式
22
Algebra cofactor
代数余子式
23
the Vandermonde determinant
最大线性无关组
66
Euclidean space

线性代数ppt

ቤተ መጻሕፍቲ ባይዱ
EX5. Choose h and k such that the system has (a)no solution, (b) a unique solution, and (c) many solutions.
x1 3x2 1 2 x1 hx2 k
EX6. Mark each statement True or False. Justify each answer. a. The reduced echelon form of a matrix is unique.
b. If every column of an augmented matrix contains a pivot, then the corresponding system is consistent. c. The pivot positions in a matrix depend on whether row interchanges are used in the row reduction process. d. A general solution of a system is an explicit description of all solutions of the system. e. Whenever a system has free variables, the solution set contains many solutions. EX7. Suppose a system of linear equations has a 4 5 augmented matrix whose fifth column is not a pivot column. Is the system consistent? Why (or why not)?

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.。

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4
-3 8
5
1 4 mat rix a row mat rix
8 3 2 3 1 mat rix a column mat rix
Location
2 3 - 4 A 7 5 - 1 a13 -4, a21 7
aij is element of row i, column j

Not good！

2R1

R1+R2
Good！

1 … 0 0
1 0 … 0 1 0 0
…
1 0 0 0 1 0 0 0 1
Figure 1.3 Many solution 4x – 2y = 6 6x – 3y = 9 Both equations have the same graph. Any point on the graph is a solution. Many solutions.
Ch1_3
Definition
※ Systems of three linear equations in three variables:
Unique solution
Ch1_5
No solutions
Many solutions
Ch1_6
How to solve a system of linear equations? Gauss-Jordan elimination.
Ch1_2
Solutions for system of linear equations
Figure 1.1 Unique solution x+y= 5 2x - y = 4 Lines intersect at (3, 2) Unique solution: x = 3, y = 2.
Figure 1.2 No solution –2x + y = 3 –4x + 2 y = 2 Lines are parallel. No point of intersection. No solutions.
1 2 1 1 2 3 1 3 1 - 1 - 2 - 6 augment edmat rix
Ch1_11
Elementary Row Operations of Matrices
Elementary Transformation 1. Interchange two equations. 2. Multiply both sides of an equation by a nonzero constant. 3. Add a multiple of one equation to another equation. Elementary Row Operation Interchange two rows of a matrix.
English name of common numeral system ： natural number ,integer ,rational number, real number , complex number positive , negative
Ch1_4
A linear equation in three variables corresponds to a plane in three-dimensional space.
1 0 0 0 1 0 0 0 1 - 1 1 2
The solution is x1 -1, x2 1, x3 2.
The solution is x1 -1, x2 1, x3 2.
Ch1_14
Basic column operation symbol description ：
1 1 2 1 1 - 1 - 1 0 0 - 2 - 3 - 8
Ch1_13
Eq1+(–1)Eq2 Eq3+(2)Eq2
x1 x2 x3 2 x2 - x3 -1 - 2 x2 - 3x3 -8 x1 2 x3 3 x2 - x3 -1 - 5 x3 -10 x1 2 x3 3 x2 - x3 -1 x3 2 x1 -1 x2 1 x3 2
Ch1_7
Definition
• A matrix is a rectangular array of numbers. • The numbers in the array are called the elements of the matrix.
Matrices
2 3 - 4 A 7 5 1 7 1 B 0 5 - 8 3 6 3 5 C 0 2 5 8 9 12
R1+(–3)R2 For example: R1(first column) plus (-3) R2， So R1 is changed，R2 stays the same
4 12 1 -2 5 18 (1) 2 - 1 3 - 3 - 8 - 1 4 12 1 -2 5 18 (2) 2 - 1 3 - 3 - 8 - 1
x1 x2 x3 2
Analogous Matrix Method Augmented matrix: 1 2 1 1 1 3 2 3 1 - 1 - 2 - 6
x2 - x3 -1 R2+(–2)R1 - 2 x2 - 3x3 -8 R3+(–1)R1
Example 1
Solving the following system of linear equation. x1 x2 x3 2 2 x1 3 x2 x3 3 x1 - x2 - 2 x3 -6 row equivalent Solution Equation Method Initial system: x1 x2 x3 2 Eq2+(–2)Eq1 2 x1 3x2 x3 3 Eq3+(–1)Eq1 x1 - x2 - 2 x3 -6
Ch1_8
Row and Column
2 3 - 4 A 7 5 1
2 7 column1 3 5 column 2 - 4 - 1 column3
2
3 - 4 row1
7
5 - 1 row 2
Submatrix
1 7 4 A 2 3 0 5 1 - 2 matrixA

Ch1_16
Example 2
Solving the following system of linear equation. x1 - 2 x2 4 x3 12 2 x1 - x2 5 x3 18 - x1 3 x2 - 3 x3 -8 Solution 4 12 4 12 1 -2 1 - 2 R2 (-2)R1 5 18 3 - 3 - 6 2 -1 0 R3 R1 3 - 3 - 8 1 1 4 - 1 0
R1ห้องสมุดไป่ตู้2R2
5 - 4 14 48 2 -1 5 18 3 - 3 - 8 - 1 4 12 1 -2 4 - 5 13 42 3 - 3 - 8 - 1
Ch1_15
R2+2R1

2R1+R2

1 1 2 1 1 - 1 - 1 0 0 - 2 - 3 - 8
R1+(–1)R2 R3+(2)R2
(–1/5)R3 R1+(–2)R3 R2+R3
(–1/5)Eq3 Eq1+(–2)Eq3 Eq2+Eq3
2 3 1 0 0 1 - 1 - 1 0 0 - 5 - 10 2 3 1 0 0 1 - 1 - 1 1 2 0 0
Ch1_17
8 1 0 2 0 1 - 1 - 2 R1 (-2)R3 1 R2 R3 R3 1 3 2 0 0
1.
2. Multiply the elements of a row by a nonzero constant.
3. Add a multiple of the elements of one row to the corresponding elements of another row.
Ch1_12
Linear Algebra
Chapter 1 Linear Equations and Vectors
1.1 Matrices and Systems of Linear Equations
Definition
• An equation in the variables x and y that can be written in the form ax+by=c, where a, b, and c are real constants (a and b not both zero), is called a linear equation. • The graph of this equation is a straight line in the x-y plane. • A pair of values of x and y that satisfy the equation is called a solution . system of linear equations
matrix of coefficient and augmented matrix
x1 x2 x3 2 2 x1 3x2 x3 3 x1 - x2 - 2 x3 -6

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