Chapter 1�Linear Equations and Vectors

大葉大學 資訊工程系

黃鈴玲

Linear Algebra

1.1 Matrices and Systems of Linear Equations

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).

system of linear equations (線性聯立方程式)

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Solutions for system of linear equations

Figure 1.1

Unique solution (唯一解)

x + y = 5

2x − y = 4

Lines intersect at (3, 2)

Unique solution: � x = 3, y = 2.

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Definition

A linear equation in n variables x₁, x₂, x₃, …, x_n has the form

a₁ x₁ + a₂ x₂ + a₃ x₃ + … + a_n x_n = b

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

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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)

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

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※ Systems of three linear equations in three variables:

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How to solve a system of linear equations?

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Gauss-Jordan elimination. (高斯-喬登消去法)

1.2節會介紹

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.

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• Matrices

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。

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• Submatrix (子矩陣)

• Row (列) and Column (行)

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• matrix of coefficient and augmented matrix

Relations between system of linear equations and matrices

係數矩陣

擴大矩陣

隨堂作業：5(f)

給定聯立方程式後，

不會改變解的一些轉換

• Elementary Transformation
• Interchange two equations.
• Multiply both sides of an equation by a nonzero constant.
• 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)
• Interchange two rows of a matrix.� (兩列交換)
• Multiply the elements of a row by a nonzero constant.� (某列的元素同乘一非零常數)
• 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)

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Elementary Row Operations of Matrices

Example 1

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Solving the following system of linear equation.

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隨堂作業：7(d)

基本列運算符號說明：

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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:

基本列運算符號說明：

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避免將要修改的列乘以某個倍數，這樣容易出錯

不好！

較好！

基本列運算的步驟： (盡量使矩陣一步步變成以下形式)

≈…≈

≈…≈

≈…≈

Example 2

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Solving the following system of linear equation.

Solution (請先自行練習)

Example 3

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Solve the system

Solution (請先自行練習)

隨堂作業：10(d)(f)

Summary

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Use row operations to [A: B] :

Def. [I_n : 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 I_n (A ≈ I_n).

2. If A ≈ I_n, then the system has unique solution.

Example 4 Many Systems

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Solving the following three systems of linear equation, all of which have the same matrix of coefficients.

Solution

隨堂作業：13(b)

The solutions to �the three systems are

Homework

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

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1.2 Gauss-Jordan Elimination

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Definition

A 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)

• Examples for reduced echelon form

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(✓)

(✓)

(🗴)

(🗴)

• 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)

Gauss-Jordan Elimination

• System of linear equations �⇒ augmented matrix �⇒ reduced echelon form�⇒ solution

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Example 1

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

Solution

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

Example 2

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Solve, if possible, the system of equations

Solution (請先自行練習)

The general solution (通解) to the system is

隨堂作業：5(c)

Example 3

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This example illustrates that the general solution can involve a number

of parameters. Solve the system of equations

Solution

變數個數 >�方程式個數�⇒ many sol.

Example 4

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

Solution(自行練習)

The system has no solution.

隨堂作業：5(d)

Homogeneous System of linear Equations

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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.

Homogeneous System of linear Equations

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Note. 除 trivial solution 外，可能還有其他解。

隨堂作業：8(e)

Homework

• Exercise 1.2:�2, 5, 8, 14

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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)

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

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Example 1

Sketch the position vectors .

Figure 1.6

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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.

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Definition

Let be two elements of Rⁿ.

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

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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 2

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

Find u + v and 3u.

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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.

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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.

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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 u�if c > 0, and the opposite direction to u if c < 0.

The length of cu is |c| times the length of u.

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Special Vectors

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

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Theorem 1.3

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

1. u + v = v + u
2. u + (v + w) = (u + v) + w
3. u + 0 = 0 + u = u
4. u + (–u) = 0
5. c(u + v) = cu + cv
6. (c + d)u = cu + du
7. c(du) = (cd)u
8. 1u = u

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Linear Combinations of Vectors

隨堂作業：7(e)

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

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Column Vectors

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Subspaces of Rⁿ

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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 the�algebraic 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)

Example 6

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Consider 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)) ∈ W

and

ku = k(a, 2a) = (ka, 2ka) ∈ W

Thus 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)

Example 7

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Consider the homogeneous system of linear �equations. It can be shown that there are many�solutions 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) ∈ W

and ku = (2kr, 5kr, kr) ∈ W

Thus 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

隨堂作業：13

Homework

• Exercise 1.3:�7, 9, 10, 12, 13

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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:

1. 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)
2. 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.

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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.

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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.

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The vectors v₁, v₂, and v₃ are said to be linearly independent if the identity pv₁ + qv₂ + rv_m = 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

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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.

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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

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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

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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

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

Sol.

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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, 2a−b) 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

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1.5 Dot Product, Norm, Angle, and Distance

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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 Product

Let 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

隨堂作業：2(b)

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Norm (length) of a Vector in Rⁿ

Definition

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

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

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Example

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.

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

Solution

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Example 1

Solution

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

隨堂作業：6(b), 10(d)

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Angle between Vectors (R²)

Figure 1.18

The law of cosines gives

We get that

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

0 ≤ θ ≤ π

Angle between Vectors (R ⁿ)

Definition

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

​

​

隨堂作業：13(c)

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Definition

Two nonzero vectors are orthogonal (正交) if the angle between them is a right angle (直角).

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

可寫成u⊥v

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 orthonormal�set.

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)

Example 3

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(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, v∈W. Since u⊥w and v⊥w, we have u⋅w=0 and v⋅w=0.

(u+v)⋅w = u⋅w + v⋅w = 0 ⇒ u+v ⊥w ⇒ u+v ∈W

If c is a scalar, c(u⋅w) = cu⋅w = 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, −a − 3b) | a, b ∈R}

Since (a, b, −a − 3b) = 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

Example 3 (continue)

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Figure 1.19

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, …, x_n) and y=(y₁, y₂, …, y_n) 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.

​

Solution

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

Generalize this expression to Rⁿ.

Solution

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Theorem 1.5

The Cauchy-Schwartz Inequality.

If u and v are vectors in Rⁿ then

​

Here denoted the absolute value (絕對值) of the number u⋅v.

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Theorem 1.6

Let u and v be vectors in Rⁿ.

1. Triangle Inequality (三角不等式):

||u + v|| ≤ ||u|| + ||v||.

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

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Figure 1.21

Homework

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

(1.6節跳過)

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