Chapter 4 General Vector Spaces

Linear Algebra Chapter 4General Vector Spaces 大葉大學資訊工程系黃鈴玲

4.1 General Vector Spaces and Subspaces Our aim in this section will be to focus on the algebraic properties of Rn. We draw up a set of axioms (公理) based on the properties of Rn.Any set that satisfies these axioms will have similar algebraic properties to the vector space Rn. • Definition • A vector space is a set V of elements calledvectors, having operations of addition and scalar multiplication defined on it that satisfy the following conditions. (u, v, and w are arbitrary elements of V, and c and d are scalars.) • Closure Axioms (最重要！) • The sum u + v exists and is an element of V. (V is closed under addition.) • cu is an element of V. (V is closed under scalar multiplication.)

補充範例 (1) V={ …, -3, -1, 1, 3, 5, 7, …} (所有奇數構成的集合) V is not closed under addition because 1+3=4  V. (2) Z={ …, -2, -1, 0, 1, 2, 3, 4, …} (所有整數構成的集合) Z is closed under addition because for any a, b Z, a + bZ. Z is not closed under scalar multiplication because ½ is a scalar, for any odd a Z, (½)a Z. 隨堂作業：14

Definition of Vector Space (continued) Addition Axioms 3. u + v = v + u (commutative property) 4. u + (v + w) = (u + v) + w (associative property) 5. There exists an element of V, called thezero vector, denoted 0, such that u + 0 = u. 6. For every element u of V there exists an element called thenegative of u, denoted-u, such that u + (-u) = 0. Scalar Multiplication Axioms 7.c(u + v) = cu + cv 8. (c + d)u =cu + du 9. c(du) = (cd)u 10.1u = u

補充： Prove that W is a vector space. Let , for some a, bR. A Vector Space in R3 Proof Axiom 1: u + v W. Thus W is closed under addition. W. Thus W is closed under scalar multiplication. Axiom 2: Axiom 5: Let 0 = (0, 0, 0) = 0(1,0,1), then 0 W and 0+u = u+0 = u for any u  W. Axiom 6: For any u =a(1,0,1)  W. Let -u = -a(1,0,1), then -u W and (-u)+u = 0. 隨堂作業：5 Axiom 3,4 and 7~10: trivial

Let Axiom 1: u + v is a 2 2 matrix. ThusM22 is closed under addition. Vector Spaces of Matrices Prove that M22 is a vector space. Proof Axiom 3 and 4: From our previous discussions we know that 2 2 matrices are communicative and associative under addition (Theorem 2.2).

Axiom 6: Axiom 5: The 2 2 zero matrix is , since 推廣：The set of m n matrices, Mmn, is a vector space. 隨堂作業：7

Vector Spaces of Functions (跳過) Prove that V={ f | f: RR } is a vector space. Let f, g V, c  R. For example: f: RR, f(x)=2x, g: RR, g(x)=x2+1. Axiom 1: f + g is defined by (f + g)(x) = f(x) + g(x).  f + g: RR  f + g  V. Thus V is closed under addition. Axiom 2: cf is defined by (cf)(x) = c f(x).  cf: RR  cf  V. Thus V is closed under scalar multiplication.

Axiom 6:Let the function –f defined by (-f )(x) = -f (x). Thus [f + (-f )] = 0, -f is the negative of f. Vector Spaces of Functions (跳過) Axiom 5: Let 0 be the function such that 0(x) = 0 for every xR. 0 is called the zero function. We get (f + 0)(x) = f(x) + 0(x) = f(x) + 0 = f(x) for every xR. Thus f + 0 = f. 0 is the zero vector. 隨堂作業：13 V={ f | f(x)=ax2+bx+c for some a,b,cR}

The Complex Vector Space Cn (跳過) It can be shown that Cn with these two operations is acomplex vector space.

Proof (a) 0v + 0v = (0 + 0)v = 0v (0v + 0v) + (-0v) = 0v + (-0v) 0v + [0v + (-0v)] = 0, 0v + 0 = 0, 0v = 0 (c) (-1)v + v = (-1)v + 1v = [(-1) + 1]v = 0v = 0 Theorem 4.1 Let V be a vector space, v a vector in V, 0 the zero vector of V, c a scalar, and 0 the zero scalar. Then (a) 0v = 0 (b)c0 = 0 (c) (-1)v = -v (d) If cv = 0, then either c = 0 or v = 0.

Subspaces In general, a subset of a vector space may or may not satisfy the closure axioms. However, any subset that is closed under bothof these operations satisfies all the other vector space properties. Definition Let V be a vector space and U be a nonempty subset of V. If U is a vector space under the operations of addition and scalar multiplication of V it is called a subspace of V. Note. U只要有加法與純量乘法的封閉性，其他axiom都會滿足。

Example 1 Let W be the subset of R3 consisting of all vectors of the form (a, a, b). Show that W is a subspace of R3. Solution Let (a, a, b), (c, c, d) W, and let kR. We get (a, a, b) + (c, c, d) = (a+c, a+c, b+d)  W k(a, a, b) = (ka, ka, kb)  W Thus W is a subspace of R3.

Example 1’ Let W be the set of vectors of the form (a, a2, b). Show that W is not a subspace of R3. Solution Let (a, a2, b), (c, c2, d) W. (a, a2, b) + (c, c2, d) = (a+ c, a2 + c2, b + d)  (a + c, (a+ c)2, b + d) Thus (a, a2, b) + (c, c2, d) W. W is not closed under addition. W is not a subspace. 隨堂作業：18(a), 20(b)

Example 2 Prove that the set U of 2  2 diagonal matrices is a subspace of the vector space M22 of 2  2 matrices. Solution (+) Let U. We get  u + v U.  U is closed under addition. • () Let cR. We get • cu U.  U is closed under scalar multiplication. • U is a subspace of M22. 隨堂作業：27(a)

Example 3 (跳過) Let Pn denoted the set of real polynomial functions of degree  n. Prove that Pn is a vector space if addition and scalar multiplication are defined on polynomials in a pointwise manner. Solution Let f and gPn, where (+) (f + g)(x) is a polynomial of degree  n. Thus f + g Pn.Pn is closed under addition.

(跳過) () Let cR, (cf)(x) is a polynomial of degree  n. Thus cf Pn.Pn is closed under scalar multiplication. By (+) and (), Pn is a subspace of the vector space V of functions. Therefore Pn is a vector space.

Theorem 4.2 Let U be a subspace of a vector space V. U contains the zero vector of V. Proof Let u be an arbitrary vector in U and 0 be the zero vector of V. Let 0 be the zero scalar. By Theorem 4.5(a) we know that 0u = 0. Since U is closed under scalar multiplication, this means that 0 is in U. Note. Let 0 be the zero vector of V, U is a subset of V. If 0U  U is not a subspace of V. If 0U  check (+)() to determine if U is a subspace of V.

Example 4 Let W be the set of vectors of the form (a, a, a+2). Show that W is not a subspace of R3. Solution If (a, a, a+2) = (0, 0, 0), then a = 0 and a + 2 = 0.   (0, 0, 0) W. W is not a subspace of R3. 隨堂作業：24(a,b), 27(b)

Homework • Exercise 4.1:5, 7, 14, 18, 19, 20, 24, 27

4.2 Linear Combinations W={(a, a, b) | a,bR}  R3 (a, a, b) = a (1,1,0) + b (0,0,1) W中的任何向量都可以用 (1,1,0) 及 (0,0,1)來表示e.g., (2, 2, 3) = 2 (1,1,0) + 3 (0,0,1) (-1, -1, 7) = -1 (1,1,0) + 7 (0,0,1) Definition Let v1, v2, …, vm be vectors in a vector space V. The vector v in V is a linear combination (線性組合) of v1, v2, …, vm if there exist scalars c1, c2, …, cm such that v can be written v = c1v1 + c2v2 + … + cmvm W中的任何向量都是 (1,1,0) 及 (0,0,1) 的 linear combination.

Example The vector (7, 3, 2) is a linear combination of the vector (1, 3, 0) and (2, -3, 1) since (7, 3, 2) = 3(1, 3, 0) + 2(2, -3, 1) The vector (3, 4, 2) is not a linear combination of (1, 1, 0) and (2, 3, 0) because there are no values of c1 and c2 for which (3, 4, 2) = c1(1, 1, 0) + c2(2, 3, 0) is true.

Example 1 Determine whether or not the vector (8, 0, 5) is a linear combination of the vectors (1, 2, 3), (0, 1, 4), and (2, -1, 1). Solution Suppose c1(1, 2, 3)+c2(0, 1, 4)+c3(2, -1, 1)=(8, 0, 5). Thus (8, 0, 5) is a linear combination of (1, 2, 3), (0, 1, 4), and (2, -1, 1). 隨堂作業：2(a)

決定一個向量是否是某些向量的linear combination 求解聯立方程式  唯一解表示 linear combination 的係數唯一無限多解表示 linear combination 的係數不唯一無解表示不是 linear combination

Suppose Thus (4, 5, 5) can be expressed in many ways as a linear combination of (1, 2, 3), (-1, 1, 4), and (3, 3, 2): Example 2 Determine whether the vector (4, 5, 5) is a linear combination of the vectors (1, 2, 3), (-1, 1, 4), and (3, 3, 2). Solution

Solution Suppose  Example 2’ Show that the vector (3, -4, -6) cannot be expressed as a linear combination of the vectors (1, 2, 3), (-1, -1, -2), and (1, 4, 5). This system has no solution. Thus (3, -4, -6) is not a linear combination of (1, 2, 3),(-1, -1, -2), and (1, 4, 5). 隨堂作業：2(c)

Spanning a Vector Space Definition Let v1, v2, …, vm be vectors in a vector space V. These vectors spanV if every vector in V can be expressed as a linear combination of them. {v1, v2, …, vm} is called a spanning set of V.

Example 3 Show that the vectors (1, 2, 0), (0, 1, -1), and (1, 1, 2) span R3. Solution Let (x, y, z) be an arbitrary element of R3. Suppose   The vectors (1, 2, 0), (0, 1, -1), and (1, 1, 2) span R3. 隨堂作業：4(a)

Theorem 4.3 Let v1, …, vm be vectors in a vector space V. Let U be the set consisting of all linear combinations of v1, …, vm . Then U is a subspace of V spanned the vectors v1, …, vm . U is said to be the vector space generated by v1, …, vm. It is denoted Span{v1, …, vm}. Proof (+) Let u1 = a1v1 + … + amvm and u2 = b1v1 + … + bmvmU. Then u1 + u2 = (a1v1 + … + amvm) + (b1v1 + … + bmvm) = (a1 + b1) v1 + … + (am+ bm) vm u1 + u2 is a linear combination of v1, …, vm . u1 + u2U. U is closed under vector addition.

()Let cR. Then cu1 = c(a1v1 + … + amvm) = ca1v1 + … + camvm) cu1 is a linear combination of v1, …, vm . cu1U. U is closed under scalar multiplication. Thus U is a subspace of V. By the definition of U, every vector in U can be written as a linear combination of v1, …, vm . Thus v1, …, vm span U.

We can visualize U. U is made up of all vectors in the plane defined by the vectors (-1, 5, 3) and (2, -3, 4). Example 4 Consider the vectors (-1, 5, 3) and (2, -3, 4) in R3. Let U =Span{(-1, 5, 3), (2, -3, 4)}. U will be a subspace of R3 consisting of all vectors of the form c1(-1, 5, 3) + c2(2, -3, 4). The following are examples of some of the vectors in U, which are obtained by given c1 and c2 various values.

Figure 4.1

Figure 4.2 We can generalize this result. Let v1 and v2 be vectors in the space R3. The subspace U generated by v1 and v2 is the set of all vectors of the form c1v1 + c2v2. If v1 and v2 are not colinear, then U is the plane defined by v1 and v2 .

Example 5 Let v1 and v2 span a subspace U of a vector V. Let k1 and k2 be nonzero scalars. Show that k1v1 and k2v2 also span U. Solution Choose any vector vU. Since v1 and v2 span U,  There exist a, bR such that v = av1 + bv2 We can write k1v1 and k2v2 span U. 隨堂作業：20

Determine whether the matrix is a linear combination of the matrices in the vector space M22 of 2  2 matrices. Example 6 Solution Suppose Then

This system has the unique solution c1 = 3, c2 = -2, c3 = 1. Therefore

Example 7 (跳過) Show that the function h(x) =4x2+3x-7 lies in the space Span{f, g} generated by f(x) = 2x2-5 and g(x) = x+1. Solution Suppose Then Therefore the function h(x) lies in Span{f, g}.

Homework • Exercise 4.2:2, 4, 6, 18, 20

4.3 Linear Dependence and Independence The concepts of dependence and independence of vectors are useful tools in constructing “efficient” spanning sets for vectorspaces – sets in which there are no redundant vectors. • Definition • The set of vectors { v1, …, vm } in a vector space V is said to be linearly dependent(線性相依) if there exist scalars c1, …, cm, not all zero, such that c1v1 + … + cmvm = 0 • The set of vectors { v1, …, vm } is linearly independent(線性獨立) if c1v1 + … + cmvm = 0 can only be satisfied when c1 = 0, …, cm = 0.

Example 1 Determine whether the set{(1, 2, 0), (0, 1, -1), (1, 1, 2)} is linearly dependent in R3. Solution Suppose c1 = 0c2 = 0 is the unique solution.c3 = 0 Thus the set of vectors is linearly independent. Note.不需真的求解，只需判斷是唯一解或無限多解，故當係數矩陣是方陣時，求算係數矩陣的行列式即可。此題行列式0  唯一解  線性獨立若行列式=0 無限多解  線性相依隨堂作業：1(c,e)

Example 2 (跳過) (a) Show that the set {x2+1,3x–1, –4x+1} is linearly independent in P2.(b) Show that the set {x+1, x–1, –x+5} is linearly dependent in P1. Solution (a) Suppose c1(x2 + 1) + c2(3x – 1) + c3(– 4x + 1) = 0  c1x2+(3c2 – 4c3)x + c1– c2+ c3 = 0  c1 = 0, c2 = 0, c3 = 0 is the unique solution  linearly independent (b) Suppose c1(x+1) + c2(x –1) + c3(–x+5) = 0  (c1+ c2 – c3)x + c1– c2+5c3 = 0  many solutions  linearly dependent

Proof () Let the set { v1, v2, …, vm } be linearly dependent. Therefore, there exist scalars c1, c2, …, cm, not all zero, such that c1v1 + c2v2 + … + cmvm = 0 Assume that c1 0. The above identity can be rewritten Thus, v1 is a linear combination of v2, …, vm. Theorem 4.4 A set consisting of two or more vectors in a vector space is linearly dependent if and only if it is possible to express one of the vectors as a linearly combination of the other vectors.

() Conversely, assume that v1 is a linear combination of v2, …, vm. Therefore, there exist scalars d2, …, dm, such that v1 = d2v2 + … + dmvm Rewrite this equation as 1v1 + (-d2)v2 + … + (-dm)vm = 0 Thus the set {v1, v2, …, vm} is linearly dependent, completing the proof.

Linear Dependence of {v1, v2} {v1, v2} linearly dependent; vectors lie on a line {v1, v2} linearly independent; vectors do not lie on a line Figure 4.3 Linear dependence and independence of {v1, v2} in R3.

Linear Dependence of {v1, v2, v3} {v1, v2, v3} linearly dependent; vectors lie in a plane {v1, v2, v3} linearly independent; vectors do not lie in a plane Figure 4.4 Linear dependence and independence of {v1, v2, v3} in R3.

Proof Consider the set {0, v2, …, vm}, which contains the zero vectors. Let us examine the identity Theorem 4.5 Let V be a vector space. Any set of vectors in V that contains the zero is linearly dependent. We see that the identity is true for c1 = 1, c2 = 0, …, cm = 0 (not all zero). Thus the set of vectors is linearly dependent, proving the theorem.

Proof Since the set {v1, …, vm} is linearly dependent, there exist scalars c1, …, cm, not all zero, such that Consider the set of vectors {v1, …, vm, vm+1, …, vn}, which contains the given vectors. There are scalars, not all zero, namely c1, …, cm, 0, …, 0 such that Thus the set {v1, …, vm, vm+1, …, vn} is linearly dependent. Theorem 4.6 Let the set {v1, …, vm} be linearly dependent in a vector space V. Any set of vectors in V that contains these vectors will also be linearly dependent.

Since {v1, v2} is linearly independent Thus system has the unique solution a = 0, b = 0. Returning to identity (1) we get that {v1 + v2, v1 – v2} is linearly independent. Example 3 Let the set {v1, v2} be linearly independent. Prove that {v1 + v2, v1 – v2} is also linearly independent. Solution Suppose (1) We get 隨堂作業：12

Homework • Exercise 4.3:1, 3, 8, 12

Proof (a) () Assume that v1, …, vn are linearly independent. Let vV. Since v1, …, vn span V, we can express v as a linear combination of these vectors. Suppose we can write 4.4 Properties of Bases Theorem 4.7 Let thevectors v1, …, vn span a vector space V. Each vector in V can be expressed uniquely as a linear combination of these vectors if and only if the vectors are linearly independent.   Since v1, …, vn are linearly independent, a1 – b1 = 0, …, an – bn = 0, implying that a1 = b1, …, an = bn.  unique 得證

Chapter 4 General Vector Spaces