Chapter 1Linear Equations and Vectors

大葉大學 資訊工程系黃鈴玲

Linear Algebra

Definition• An equation ( 方程式 ) in the variables ( 變數 ) x and y that

can be written in the form ax+by=c, where a, b, and c are real constants ( 實數常數 ) (a and b not both zero), is called a linear equation ( 線性方程式 ).

• 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 ( 解 ).

1.1 Matrices and Systems of Linear Equations

system of linear equations ( 線性聯立方程式 )

Ch1_2

Figure 1.2No solution ( 無解 )–2x + y = 3–4x + 2y = 2 Lines are parallel.No point of intersection. No solutions.

Solutions for system of linear equations

Figure 1.1Unique solution ( 唯一解 ) x + y = 5 2x y = 4 Lines intersect at (3, 2)Unique solution: x = 3, y = 2.

Figure 1.3Many solution ( 無限多解 )4x – 2y = 66x – 3y = 9 Both equations have the same graph. Any point on the graph is a solution. Many solutions. Ch1_3

Ch1_4

Definition A linear equation in n variables x1, x2, x3, …, xn has the

form

a1 x1 + a2 x2 + a3 x3 + … + an xn = b

where the coefficients ( 係數 ) a1, a2, a3, …, an and b are constants.

常見數系的英文名稱： natural number ( 自然數 ), integer ( 整數 ), rational number ( 有理數 ), real number ( 實數 ), complex number ( 複數 ) positive ( 正 ), negative ( 負 )

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A linear equation in three variables corresponds to a plane in three-dimensional ( 三維 ) space.

Unique solution

※ Systems of three linear equations in three variables:

Ch1_6

No solutions

Many solutions

Ch1_7

How to solve a system of linear equations?

Gauss-Jordan elimination. ( 高斯 - 喬登消去法 )

1.2 節會介紹

Ch1_8

Definition• A matrix ( 矩陣 ) is a rectangular array of numbers. • The numbers in the array are called the elements ( 元素 ) of

the matrix.

Matrices

1298

520

653

C

38

50

17

B 157

432A

注意矩陣左右兩邊是中括號不是直線，直線表示的是行列式。

Ch1_9

Submatrix (子矩陣 )

A matrix

215

032

471

A

Row ( 列 ) and Column ( 行 )

3column 2column 1column 2 row 1 row

1

4

5

3

7

2 157 432

157

432

A

A of ssubmatrice

25

41

1

3

7

15

32

71

RQP

Ch1_10

Identity Matrices (單位矩陣 ) diagonal ( 對角線 ) 上都是 1 ，其餘都是 0 ， I 的下標表示 size

100010001

1001

32 II

Location

7 ,4 157

4322113

aaAaij 表示在 row i, column j 的元素值也寫成 location (1,3) = 4

Size and Type

matrixcolumn a matrix row a matrix square a

matrix 13 matrix 41 matrix 33 32 :Size

2

3

8

5834

853

109

752

542

301

Ch1_11

matrix of coefficient and augmented matrix

62

3 32

2

321

321

321

xxx

xxx

xxx

Relations between system of linear equations and matrices

係數矩陣 擴大矩陣

隨堂作業： 5(f)

tcoefficien ofmatrix

211

132

111

matrix augmented

6211

3132

2111

Ch1_12

給定聯立方程式後，不會改變解的一些轉換

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.

將左邊的轉換對應到矩陣上Elementary Row Operation ( 基本列運算 )

1. Interchange two rows of a matrix. ( 兩列交換 )

2. Multiply the elements of a row by a nonzero constant. ( 某列的元素同乘一非零常數 )

3. Add a multiple of the elements of one row to the corresponding elements of another row. ( 將一個列的倍數加進另一列裡 )

Elementary Row Operations of Matrices

Ch1_13

Example 1Solving the following system of linear equation.

623322

321

321

321

xxxxxxxxx

621131322111

623322

321

321

321

xxxxxxxxx

SolutionEquation MethodInitial system:

Analogous Matrix MethodAugmented matrix:

1

2

32

321

xx

xxx

Eq2+(–2)Eq1

Eq3+(–1)Eq1

832011102111

R2+(–2)R1R3+(–1)R1

符號表示 row equivalent

832 32 xx

Ch1_14

1051

32

3

32

31

xxxxx

0150011103201

21

32

3

32

31

xxxxx

210011103201

21

1

3

2

1

xxx

2100

1010

1001

Eq1+(–1)Eq2

Eq3+(2)Eq2

(–1/5)Eq3

Eq1+(–2)Eq3

Eq2+Eq3

The solution is .2 ,1 ,1 321 xxx The solution is

.2 ,1 ,1 321 xxx

8321

2

32

32

321

xxxxxxx

832011102111

R1+(–1)R2 R3+(2)R2

(–1/5)R3

R1+(–2)R3 R2+R3

隨堂作業： 7(d)

Ch1_15

基本列運算符號說明：

R1+(–3)R2 表示將 R1( 第一列 ) 加上 (3) R2 ，

所以是 R1 這列會改變， R2 不變

For example:

83311851212421

(1) R1+2R2

83311851212421

(2) R2+2R1

8331

18512

481445

8331

421354

12421

Ch1_16

基本列運算符號說明：避免將要修改的列乘以某個倍數，這樣容易出錯

不好！ 2R1+R2

2R1

R1+R2

較好！

基本列運算的步驟： ( 盡量使矩陣一步步變成以下形式 )

…

0

0

1…

00

10

01

100

010

001

…

Ch1_17

Example 2Solving the following system of linear equation.

83318521242

321

321

321

xxxxxxxxx

Solution (請先自行練習 )

83311851212421

41106330

12421

R23

1

41102110

12421

620021108201

3100

2110

8201

R1R3

2)R1(R2

R2)1(R3

(2)R2R1

R32

1

3100

1010

2001

R3R2

2)R3(R1

.

3

1

2

solution

3

2

1

x

x

x

Ch1_18

Example 3Solve the system

7 2

328 63

441284

21

321

321

xx

xxx

xxx

7012

32863

441284

7012

32863

11321

15630

1100

11321

1100

5210

11321

1100

5210

1101

.

1100

3010

2001

.1 ,3 ,2 issolution The 321 xxx

R23

1

R12R3

3)R1(R2

R3R2

1100

15630

11321

R14

1

2)R2(R1

2R3R2

1)R3(R1

Solution (請先自行練習 )

隨堂作業： 10(d)(f)

Ch1_19

Summary

7012

32863

441284

]:[ BA

A BUse row operations to [A: B] :

.

1100

3010

2001

7012

32863

441284]:[]:[ XIBA ni.e.,

Def. [In : X] is called the reduced echelon form ( 簡化梯式 ) of [A : B].Note. 1. If A is the matrix of coefficients of a system of n equations

in n variables that has a unique solution, then A is row equivalent to In (A In).

2. If A In, then the system has unique solution.

7 2

328 63

441284

21

321

321

xx

xxx

xxx

Ch1_20

Example 4 Many SystemsSolving the following three systems of linear equation, all of which have the same matrix of coefficients.

3321

2321

1321

42

for 42

3

bxxx

bxxx

bxxx

in turn 433

,210

,11118

3

2

1

bbb

Solution

42114213111412308311

.

2

1

2

,

1

3

0

,

2

1

1

3

2

1

3

2

1

3

2

1

x

x

x

x

x

x

x

x

x

123110315210308311

212100

315210

013101

212100

131010

201001

R2+(–2)R1 R3+R1

1)R2(R3

R2R1

R32R2

R3)1(R1

隨堂作業： 13(b)The solutions to the three systems are

Ch1_21

Homework

Exercise 1.1 ：1, 2, 4, 5, 6, 7, 10, 13

Ch1_22

1.2 Gauss-Jordan EliminationDefinitionA matrix is in reduced echelon form ( 簡化梯式 ) if

1. Any rows consisting entirely of zeros are grouped at the bottom of the matrix. ( 全為零的列都放在矩陣的下層 )

2. The first nonzero element of each other row is 1. This element is called a leading 1. ( 每列第一個非零元素是 1 ，稱做 leading 1)

3. The leading 1 of each row after the first is positioned to the right of the leading 1 of the previous row. ( 每列的 leading 1 出現在前一列 leading 1 的右邊，也就是所有的 leading 1 會呈現由左上到右下的排列 )

4. All other elements in a column that contains a leading 1 are zero. ( 包含 leading 1 的行裡所有其他元素都是 0)

Ch1_23

Examples for reduced echelon form

10000

04300

03021

9100

3010

7001

3100

0000

4021

000

210

801

() ()() ()

利用一連串的 elementary row operations ，可讓任何矩陣變成 reduced echelon form

The reduced echelon form of a matrix is unique.

隨堂作業： 2(b)(d)(h)

Ch1_24

Gauss-Jordan Elimination

System of linear equations augmented matrix reduced echelon form solution

Ch1_25

41200

22200

43111

4)R1(R3

610001110052011

R2)2(R3R2R1

Example 1Use the method of Gauss-Jordan elimination to find reduced echelon form of the following matrix.

1211244129333

22200

Solution

1211244

22200

129333

R2R1

pivot ( 樞軸，未來的 leading 1)

12112442220043111

R13

1

pivot

6100050100

170011

R3R22)R3(R1

The matrix is the reduced echelon form of the given matrix.

41200

11100

43111

R221

Ch1_26

Example 2Solve, if possible, the system of equations

7537429333

321

321

321

xxxxxxxxx

Solution (請先自行練習 )

715374123111

715374129333

R131

242012103111

3)R1(R3

2)R1(R2

000012104301

R2R3

R2R11243

1243

32

31

32

31

xxxx

xxxx

The general solution ( 通解 ) to the system is

.parameter) a (callednumber real is where,

12

43

3

2

1

rrx

rx

rx

隨堂作業： 5(c)

Ch1_27

Example 3This example illustrates that the general solution can involve a numberof parameters. Solve the system of equations

642

107242

432

4321

4321

4321

xxxx

xxxx

xxxx

Solution

21000

21000

43121

64121

107242

43121

R1R3 2)R1(R1

00000

21000

20121

R2R3

3)R2(R1

., somefor ,

2

22

2

22

4

3

2

1

4

321 Rsr

x

sx

rx

srx

x

xxx

變數個數 >方程式個數 many sol.

Ch1_28

Example 4This example illustrates a system that has no solution. Let us try to solve the system

382

13

35

321

32

321

xxx

xx

xxx

Solution(自行練習 )

1000

1310

4201

2R)1(3R

2R)1(1R

The system has no solution.

0310

1310

3511

3821

1310

3511

1R)1(3R

0x1+0x2+0x3=1

隨堂作業： 5(d)

Ch1_29

Homogeneous System of linear Equations

Definition A system of linear equations is said to be homogeneous ( 齊次 ) if all the constant terms ( 等號右邊的常數項 ) are zeros.

Example:

0632

052

321

321

xxx

xxx

Observe that is a solution. 0 ,0 ,0 321 xxx

Theorem 1.1

A system of homogeneous linear equations in n variables always has the solution x1 = 0, x2 = 0. …, xn = 0. This solution is called the trivial solution.

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Homogeneous System of linear Equations

Theorem 1.2

A system of homogeneous linear equations that has more variables than equations has many solutions.

Note. 除 trivial solution 外，可能還有其他解。

隨堂作業： 8(e)

0410

0301

0632

0521

0632

052

321

321

xxx

xxx

The system has other nontrivial solutions.

rxrxrx 321 ,4 ,3

Example:

Ch1_31

Homework

Exercise 1.2:2, 5, 8, 14

1.3 The Vector Space Rn

Figure 1.5

Rectangular Coordinate System (直角座標系 )

• the origin ( 原點 ) ： (0, 0)

• the position vector ：• the initial point of : O

• the terminal point of : A(5, 3)

• ordered pair ( 序對 ): (a, b)

OA

There are two ways of interpreting (5,3)- it defines the location of a point in a plane- it defines the position vector OA

OA

OA

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Example 1

Sketch the position vectors .)4 ,3( and )2 ,5( 1), (4,

OCOBOA

Figure 1.6Ch1_33

Figure 1.7

R2 → R3

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Definition Let be a sequence of n real numbers. The set of all such sequences is called n-space and is denoted Rn.

u1 is the first component ( 分量 ) of , u2 is the second component and so on.

) ..., , ,( 21 nuuu

) ..., , ,( 21 nuuu

Example

R4 is the sets of sequences of four real numbers. For example,(1, 2, 3, 4) and (1, 3, 5.2, 0) are in R4.

R5 is the set of sequences of five real numbers. For example, (1, 2, 0, 3, 9) is in this set.

Ch1_35

Definition Let be two elements of Rn.

We say that u and v are equal if u1 = v1, …, un = vn. Thus two element of Rn are equal if their corresponding components are equal.

) ..., , ,( and ) ..., , ,( 2121 nn vvvuuu vu

Definition Let be elements of Rn and let c be a scalar. Addition and scalar multiplication are performed as follows

Addition:

Scalar multiplication :

) ..., , ,( and ) ..., , ,( 2121 nn vvvuuu vu

) ..., ,() ..., ,(

1

11

n

nn

cucucvuvu

uvu

Ch1_36

Addition and Scalar Multiplication

Note.

(1) u, v Rn u+v Rn (Rn is closed under addition)( 加法封閉性 )

(2) u Rn, c R cu Rn (Rn is closed under scalar multiplication)( 純量乘法封閉性 )

Example 2Let u = ( –1, 4, 3, 7) and v = ( –2, –3, 1, 0) be vector in R4.

Find u + v and 3u.

Solution

21) 9, 12, ,3(7) 3, 4, ,1(33

7) 4, 1, ,3(0) 1, ,3 ,2(7) 3, 4, ,1(

u

vu

Example 3

Figure 1.8

Consider the vector (4, 1) and (2, 3), we get

(4, 1) + (2, 3) = (6, 4).

Ch1_37

Figure 1.9

In general, if u and v are vectors in the same vector space, then u + v is the diagonal ( 對角線 ) of the parallelogram ( 平行四邊形 ) defined by u and v.

Ch1_38

Example 4

Figure 1.10

Consider the scalar multiple of the vector (3, 2) by 2, we get

2(3, 2) = (6, 4)

Observe in Figure 4.6 that (6, 4) is a vector in the same direction as (3, 2), and 2 times it in length.

Ch1_39

Figure 1.11

c > 1 0 < c < 1 –1 < c < 0 c < –1

In general, the direction of cu will be the same as the direction of uif c > 0, and the opposite direction to u if c < 0.The length of cu is |c| times the length of u.

Ch1_40

Special Vectors The vector (0, 0, …, 0), having n zero components, is called the zero vector of Rn and is denoted 0.

Negative Vector

The vector (–1)u is written –u and is called the negative of u. It is a vector having the same magnitude ( 量 ) as u, but lies in the opposite direction to u.

Subtraction

Subtraction is performed on elements of Rn by subtracting corresponding components. For example, in R3,(5, 3, 6) – (2, 1, 3) = (3, 2, 9)

u

u

Ch1_41

Theorem 1.3Let u, v, and w be vectors in Rn and let c and d be scalars.

(a) u + v = v + u

(b) u + (v + w) = (u + v) + w

(c) u + 0 = 0 + u = u

(d) u + (–u) = 0

(e) c(u + v) = cu + cv

(f) (c + d)u = cu + du

(g) c(du) = (cd)u

(h) 1u = u

Figure 1.12 Commutativity of vector addition u + v = v + u

Ch1_42

Linear Combinations of Vectors

Solution

)31 7, 20,(

)2276 ,0310 4,12(4

2) 0, (4,27) 3, ,12()6 10, ,4(

2) 0, (4,9) 1, ,4(3)3 5, ,2(2

w3v2u

隨堂作業： 7(e)

Let u = (2, 5, –3), v = ( –4, 1, 9), w = (4, 0, 2). Determine the linear combination 2u – 3v + w.

Example 5

We call au +bv + cw a linear combination ( 線性組合 ) of the vectors u, v, and w.

Ch1_43

Column Vectors

nnnn vu

vu

v

v

u

u

1111

We defined addition and scalar multiplication of column vectors in Rn in a componentwise manner:

and

nn cu

cu

u

uc

11

向量加法可視為矩陣運算

Row vector:

Column vector:

) ..., , ,( 21 nuuuu

nu

u

1

向量的另一種表示法

Ch1_44

Subspaces of Rn

Ch1_45

Definition A subset S of Rn is a subspace if it is closed under addition and under scalar multiplication.

We now introduce subsets of the vector space Rn that have all thealgebraic properties of Rn. These subsets are called subspaces.

Recall:

(1) u, v S u+v S (S is closed under addition)( 加法封閉性 )

(2) u S, c R cu S (S is closed under scalar multiplication) ( 純量乘法封閉性 )

Ch1_46

Example 6Consider the subset W of R2 of vectors of the form (a, 2a). Show that W is a subspace of R2.

Proof

Let u = (a, 2a), v = (b, 2b) W, and k R. u + v = (a, 2a) + (b, 2b) = (a+ b, 2a + 2b) = (a + b, 2(a + b)) Wand ku = k(a, 2a) = (ka, 2ka) WThus u + v W and ku W. W is closed under addition and scalar multiplication. W is a subspace of R2.

Observe: W is the set of vectors that can be written a(1,2). Figure 1.13

隨堂作業： 10(d)

Ch1_47

Example 7Consider the homogeneous system of linear equations. It can be shown that there are manysolutions x1=2r, x2=5r, x3=r.We can write these solutions as vectors in R3 as (2r, 5r, r).Show that the set of solutions W is a subspace of R3.Proof

Let u = (2r, 5r, r), v = (2s, 5s, s) W, and k R. u + v = (2(r+s), 5(r+s), r+s) Wand ku = (2kr, 5kr, kr) WThus u + v W and ku W. W is a subspace of R3.

Observe: W is the set of vectors that can be written r(2,5,1). Figure 1.14

02

05

03

321

32

321

xxx

xx

xxx

隨堂作業： 13

Homework

Exercise 1.3:7, 9, 10, 12, 13

Ch1_48

1.4 Basis and Dimension

Consider the vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) in R3.

These vectors have two very important properties:

(i)They are said to span R3. That is, we can write an arbitrary vector (x, y, z) as a linear combination ( 線性組合 ) of the three vectors: For any (x,y,z) R3 (x,y,z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1)

(ii)They are said to be linearly independent ( 線性獨立 ).If p(1, 0, 0) + q(0, 1, 0) +r(0, 0, 1) = (0, 0, 0) p = 0, q = 0, r = 0 is the unique solution.

Ch1_49

向量空間的某些子集合 (subset) 本身也是向量空間，稱為子空間(subspace)Basis: a set of vectors which are used to describe a vector space.

Standard Basis of Rn

A set of vectors that satisfies the two preceding properties is called a basis.

The set {(1, 0, …, 0), (0, 1, …, 0), …, (0, …, 1)} of n vectors is the standard basis for Rn. The dimension ( 維度 ) of Rn is n.

Ch1_50

There are many bases for R3 – sets that span R3 and are linearly independent.For example, the set {(1, 2, 0), (0, 1, 1), (1, 1, 2)} is a basis for R3. The set {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is the most important basis for R3. It is called the standard basis of R3.R2 : two-dimensional space ( 二維空間 )R3 : three-dimensional space

觀察 : dimension 數等於 basis 中的向量個數 ( 故成為 dimension 定義 )

隨堂作業：1

Span, Linear Independence, and Basis

The vectors v1, v2, and v3 are said to span a space if every vector v in the space can be expressed as a linear combination of them, v = av1 + bv2 + cv3.

Ch1_51

The vectors v1, v2, and v3 are said to be linearly independent if the identity pv1 + qv2 + rvm = 0 is only true for p = 0, q = 0, r = 0.

A basis for a space is a set that spans the space and is linearly independent. The number of vectors in a basis is called the dimension of the space.

Consider the subset W of R3 consisting of vectors of the form (a, b, a+b). The vector (2, 5, 7)W, whereas (2, 5, 9)W. Show that W is a subspace of R3.

Example 1

Ch1_52

Proof.

Let u=(a, b, a+b) and v=(c, d, c+d) be vectors in W and k be a scalar.

(1) u+v = (a, b, a+b) + (c, d, c+d) = (a+c, b+d, (a+c)+(b+d))

u+v W

(2) ku = k(a, b, a+b) = (ka, kb, ka+kb)

ku W

W is closed under addition and scalar multiplication. W is a subspace of R3.

Ch1_53

Separate the variables in the above arbitrary vector u.

u = (a, b, a+b) = (a, 0, a) + (0, b, b) = a(1, 0, 1) + b(0, 1, 1)

The vectors (1, 0, 1) and (0, 1, 1) thus span W.

Furthermore, p(1, 0, 1) + q(0, 1, 1) = (0, 0, 0) leads to p=0 and q=0.

The two vectors (1, 0, 1) and (0, 1, 1) are thus linearly independent.

The set {(1, 0, 1), (0, 1, 1)} is therefore a basis for W.

The dimension of W, dim(W)= 2.

Example 1 (continue)

有加法乘法封閉性

隨堂作業： 4(a)

Consider the subset V of R3 of vectors of the form (a, 2a, 3a). Show that V is a subspace of R3 and find a basis.

Example 2

Ch1_54

Sol.

Let u=(a, 2a, 3a) and v=(b, 2b, 3b) be vectors in V and k be a scalar.

(1) u+v = (a+b, 2(a+b), 3(a+b)) u+v V

(2) ku = k(a, 2a, 3a) = (ka, 2ka, 3ka) ku V

V is closed under addition and scalar multiplication. V is a subspace of R3.

(subspace)

(basis)

u = (a, 2a, 3a) = a(1, 2, 3)

{(1, 2, 3)} is a basis for V.

dim(V) = 1.隨堂作業：4(c)

Consider the following system of homogeneous linear equations.

Example 3

Ch1_55

It can be shown that there are many solutions x1 = 3r 2s, x2 = 4r, x3 = r, x4 = s.Write these solution as vectors in R4, (3r 2s, 4r, r, s).It can be shown that this set of vectors is a subspace W of R4.Find a basis for W and give its dimension.

(1) (3r 2s, 4r, r, s) = r(3, 4, 1, 0) + s(2, 0, 0, 1)

{(3, 4, 1, 0), (2, 0, 0, 1) } is a basis for W.

dim(W) = 2. 隨堂作業：11

0422

023

02

4321

431

4321

xxxx

xxx

xxxx

(2) If p(3, 4, 1, 0) + q(2, 0, 0, 1) = (0, 0, 0, 0), then p=0, q=0.

Sol.

Ch1_56

Exercise 7

State with a brief explanation whether the following statements are true or false.

(a) The set {(1, 0, 0), (0, 1, 0)} is the basis for a two-dimensional subspace of R3.(b) The set {(1, 0, 0)} is the basis for a one-dimensional subspace of R3.(c) The vector (a, 2a, b) is an arbitrary vector in the plane spanned by the vectors (1, 2, 0) and (0, 0, 1).(d) The vector (a, b, 2ab) is an arbitrary vector in the plane spanned by the vectors (1, 0, 2) and (0, 1, 1).

(e) The set {(1, 0, 0), (0, 1, 0), (1, 1, 0)} is a basis for a subspace of R3.(f) R2 is a subspace of R3.

Homework

Exercise 1.4:1, 4, 11, 12

Ch1_57

1.5 Dot Product, Norm, Angle, and Distance

Definition Let be two vectors in Rn. The dot product ( 內積 ) of u and v is denoted u‧v and is defined by .

The dot product assigns a real number to each pair of vectors.

) ..., , ,( and ) ..., , ,( 2121 nn vvvuuu vu

nnvuvu 11vu

Example

Find the dot product of u = (1, –2, 4) and v = (3, 0, 2)

Solution

11803)24()02()31( vuCh1_58

In this section we develop a geometry for the vector space Rn.

The dot product is a tool that is used to build the geometry of Rn.

Properties of the Dot ProductLet u, v, and w be vectors in Rn and let c be a scalar. Then

1. u v = v u‧ ‧

2. (u + v) w = u w + v w ‧ ‧ ‧

3. cu v = ‧ c(u v) = u‧ ‧cv

4. u u‧ 0, and u u‧ = 0 if and only if u = 0

Proof1.

uv

vuvu

numbers real ofproperty ecommutativ by the

get We). ..., , ,( and ) ..., , ,(Let

11

11

2121

nn

nn

nn

uvuvvuvu

vvvuuu

2.

. ifonly and if 0 Thus.0 , ,0 ifonly and if ,0

.0 thus,0

122

1

221

22111

0uuu

uu

uu

nn

n

nnn

uuuu

uu

uuuuuu

隨堂作業： 2(b)

Ch1_59

Norm (length) of a Vector in Rn

Definition The norm (length or magnitude) of a vector u = (u1, …, un) in Rn is denoted ||u|| and defined by

Note: The norm of a vector can also be written in terms of the dot product

221 nuu u

uuu Ch1_60

Figure 1.17

Definition A unit vector is a vector whose norm is 1. If v is a nonzero vector, then the vector

is a unit vector in the direction of v.

This procedure of constructing a unit vector in the same direction as a given vector is called normalizing the vector.

vv

u1

Find the norm of the vectors u = (1, 3, 5) of R3 and v = (3, 0, 1, 4) of R4.

Solution352591)5()3()1( 222 u

2616109)4()1()0()3( 2222 v

Example

Ch1_61

Example 1

Solution

Find the norm of the vector (2, –1, 3). Normalize this vector.

The norm of (2, –1, 3) is

The normalized vector is

The vector may also be written

This vector is a unit vector in the direction of (2, –1, 3).

.143)1(23) ,1 ,2( 222 .14

)3 ,1 ,2(141

.143

,141

,142

隨堂作業： 6(b), 10(d)Ch1_62

Angle between Vectors (R2)

Figure 1.18

The law of cosines gives

We get that

Ch1_63

Let u=(a, b) and v=(c, d). Find the angle between u and v.

0

Definition Let u and v be two nonzero vectors in Rn. The cosine of the angle between these vectors is

0 cosvuvu

Example 2Determine the angle between the vectors u = (1, 0, 0) and v = (1, 0, 1) in R3.

Solution 11) 0, (1,0) 0, 1,( vu

2101 1001 222222 vu

Thus the angle between u and v is /4 (or 45).,2

1cos

vuvu

隨堂作業： 13(c)

Angle between Vectors (R n)

Ch1_64

Definition Two nonzero vectors are orthogonal ( 正交 ) if the angle between

them is a right angle ( 直角 ).

Two nonzero vectors u and v are orthogonal if and only if u‧v = 0.

Theorem 1.4

Proof

0 0cos orthogonal are , vuvu

Ch1_65

ExampleThe vectors (2, –3, 1) and (1, 2, 4) are orthogonal since

(2, –3, 1) (1, 2, 4) = (2 ‧ 1) + (–3 0) + (1 4) = 2 – 6 + 4 = 0.

可寫成 uv

Properties of Standard basis of Rn

(1, 0), (0,1) are orthogonal unit vectors in R2.

(1, 0, 0), (0, 1, 0), (0, 0, 1) are orthogonal unit vectors in R3.

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We call a set of unit pairwise orthogonal vectors an orthonormalset.

The standard basis for Rn, {(1, 0, …, 0), (0, 1, 0, …, 0), …, (0, …, 0, 1)} is an orthonormal set.

隨堂作業： 16(d)

Ch1_67

Example 3(a) Let w be a vector in Rn. Let W be the set of vectors that are orthogonal to w. Show that W is a subspace of Rn.

(b) Find a basis for the subset W of vectors in R3 that are orthogonal to w=(1, 3, 1). Give the dimension and a geometrical description of W.

Solution

(a) Let u, vW. Since uw and vw, we have uw=0 and vw=0. (u+v)w = uw + vw = 0 u+v w u+v W

If c is a scalar, c(uw) = cuw = 0 cu w cu W W is a subspace of Rn.

(b) Let (a, b, c)W and (a, b, c)w, then (a, b, c)(1, 3, 1)=0

a+3b+c=0 W is the set {(a, b, a3b) | a, b R}

Since (a, b, a3b) = a(1, 0, 1) + b(0, 1, 3). It is clear that{(1, 0, 1), (0, 1, 3)} is a basis for W dim(W)=2

Ch1_68

Figure 1.19

Example 3 (continue)W is the plane in R3 defined by(1, 0, 1) and (0, 1, 3).

隨堂作業： 20

Distance between Points

Def. Let x=(x1, x2, …, xn) and y=(y1, y2, …, yn) be two points in Rn. The distance between x and y is denoted d(x, y) and is defined by

Note: We can also write this distance as follows.

2211 )()() ,( nn yxyxd yx

yxyx ) ,(d

Example 4

Determine the distance between the points x = (1,–2 , 3, 0) and y = (4, 0, –3, 5) in R4.

Solution

74)50()33()02()41(),( 2222 yxd

x

yxy

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The distance between x=(x1, x2) and y =(y1, y2) is2

222

11 )()( yxyx

Generalize this expression to Rn.

Example 5

Prove that the distance in Rn has the following symmetric property: d(x, y)=d(y, x) for any x, y Rn.

Solution

) ..., , ,( and ) ..., , ,( 2121 nn yyyxxx yxLet

2211 )()() ,( nn yxyxd yx

) ,()()( 2211 xydxyxy nn

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Theorem 1.5The Cauchy-Schwartz Inequality. If u and v are vectors in Rn then

Here denoted the absolute value ( 絕對值 ) of the number uv.

vuvu vu

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Theorem 1.6Let u and v be vectors in Rn.

(a) Triangle Inequality (三角不等式 ):

||u + v|| ||u|| + ||v||.

(a) Pythagorean theorem (畢氏定理 ): If u‧v = 0 then ||u + v||2 = ||u||2 + ||v||2.

Ch1_72Figure 1.21

Homework

Exercise 1.5:2, 6, 10, 13, 16, 20, 33

(1.6 節跳過 )

Ch1_73