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Elementary Linear Algebra Anton & Rorres, 10 th Edition Lecture Set 05 Chapter 4: General Vector Spaces 1006003 คณ ตศาสตร ว ศวกรรม 3 สาขาว ชาว ศวกรรมคอมพ วเตอร ป การศ กษา 1/2554 1006003 คณตศาสตรวศวกรรม 3 สาขาวชาวศวกรรมคอมพวเตอร ปการศกษา 1/2554 ผศ.ดร.อร ญญา ผศ.ดร.สมศ กด วล ยร ชต

Chapter Content Real Vector Spaces Subspaces Linear Independence Coordinates and Basis Dimension Row Space, Column Space, and Null Space Rank, Nullity, and the Fundamental Matrix Spaces Matrix Transformations from R n to R m 2011/7/6 Elementary Linear Algebra 2

4-1 Vector Space LtV Let V be an arbitrary nonempty set of objects on which h two operations are defined: Addition Multiplication by scalars If the following axioms are satisfied by all objects u, v, w in V and all scalars k and l, then we call V a vector space and we call the objects in V vectors. (see Next Slide) 2011/7/6 Elementary Linear Algebra 3

4-1 Vector Space (continue) 1. If u and v are objects in V, then u + v is in V. 2. u + v = v + u 3. u + (v + w) = (u + v) + w 4. There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u for all u in V. 5. For each u in V, there is an object -uu in V, called a negative e of u, such that u + (-u) = (-u) + u = 0. 6. If k is any scalar and u is any object in V, thenku is in V. 7. k (u + v) = ku + kv 8. (k + l) ) u = ku + lu 9. k (lu) = (kl) (u) 10. 1u = u Slide 7 Slide16 2011/7/6 Elementary Linear Algebra 4

4-1 Remarks Depending on the application, scalars may be real numbers or complex numbers. Vector spaces in which the scalars are complex numbers are called complex vector spaces, and those in which hthe scalars must be real are called real vector spaces. Any kind of object can be a vector, and the operations of addition and scalar multiplication may not have any relationship or similarity to the standard vector operations on R n. The only requirement is that the ten vector space axioms be satisfied. 2011/7/6 Elementary Linear Algebra 5

To Show that a Set with Two Operations is a Vector Space Step1. Identify the set V of objects that will become vectors. Step 2. Identify the addition and scalar multiplication operations on V. Step 3. Verify Axioms 1(closure ( l under addition) i and 6 (closure under scalar multiplication) ; that is, adding two vectors in V produces a vector in V, and multiplying a vector in V by a scalar also produces a vector in V. Step 4. Confirm that Axioms 2,3,4,5,7,8,9 and 10 hold.

4-1 Example 1 (R n Is a Vector Space) The set V = R n with the standard operations of addition and scalar multiplication is a vector space. Axioms 1 and 6 follow from the definitions of the standard operations on R n ; the remaining axioms follow from Theorem 3.1.1. The three most important special cases of R n are R (the real numbers), R 2 (the vectors in the plane), and R 3 (the vectors in 3-space). slide4 2011/7/6 Elementary Linear Algebra 7

4-1 Example 2 (22 Matrices) Show that the set V of all 22 matrices with real entries is a vector space if vector addition is defined to be matrix addition and vector scalar multiplication is defined to be matrix scalar multiplication. u 11 u u 12 u 21 u22 Let and v 11 v v 12 v 21 v22 To prove Axiom 1, we must show that u + v is an object in V; that is, we must show that u + v is a 22 matrix. u u11 u12 v11 v12 u11 v11 u12 v v u21 u22 v21 v22 u21 v21 u22 v 12 22 2011/7/6 Elementary Linear Algebra 8

4-1 Example 2 (continue) Similarly, Axiom 6 hold because for any real number k we have u u ku ku u 21 u 22 ku21 ku 22 11 12 11 12 ku k so that ku is a 22 matrix and consequently is an object in V. Axioms 2 follows from Theorem 1.4.1a since u v u 11 u 12 v 11 v 12 v 11 v 12 u 11 u u21 u22 v21 v22 v21 v22 u21 u 12 v u Similarly, Axiom 3 follows from part (b) of that theorem; and Axioms 7, 8, and 9 follow from part (h), (j), and (l), respectively. 22 2011/7/6 Elementary Linear Algebra 9

Section 1.4 Algebraic Properties of Matrices Axiom3 Axiom7 Axiom8 Axiom9

4-1 Example 2 (continue) To prove Axiom 4, let 0 Then Similarly, il l u + 0 = u. 0 0 0 0 0 0 u11 u12 u11 u12 0 u u 0 0 u21 u22 u21 u22 u11 u12 To prove Axiom 5, let u u 21 u 22 Then u ( u) Similarly, (-u) + u = 0. For Axiom 10, 1u = u. u11 u12 u11 u12 0 0 u21 u22 u21 u22 0 0 0 2011/7/6 Elementary Linear Algebra 11

4-1 Example 3 (Vector Space of mn Matrices) The arguments in Example 2 can be adapted to show that the set V of all mn matrices with real entries, together th with the operations matrix addition and scalar multiplication, is a vector space. The mn zero matrix is the zero vector 0 If u is the mn matrix ti U, then matrix ti U is the negative u of the vector u. We shall denote this vector space by the symbol M mn 2011/7/6 Elementary Linear Algebra 12

4-1 Example 4 (Vector Space of Real-Valued Functions) Let V be the set of real-valued functions defined on the entire real line (-,). If f = f (x) and g = g (x) are two such functions and k is any real number The sum function : (f + g)(x) = f (x) + g (x) The scalar multiple : (k f)(x)=k f(x). In other words, the value of the function f + g at x is obtained by adding together the values of f and g at x (Figure 5.1.1 a). 2011/7/6 Elementary Linear Algebra 13

4-1 Example 4 (continue) The value of k f at x is k times the value of f at x (Figure 5.1.1 b). This vector space is denoted by F(-,). If f and g are vectors in this space, then to say that f = g is equivalent to saying that f(x) = g(x) for all x in the interval (-,). The vector 0 in F(-,) is the constant function that identically zero for all value of x. The negative of a vector f is the function -f f = -f(x). Geometrically, the graph of -ff is the reflection of the graph of f across the x-axis (Figure 5.1.1 c). 2011/7/6 Elementary Linear Algebra 14

4-1 Example 5 (Not a Vector Space) Let V = R 2 and define addition and scalar multiplication operations as follows: If u = (u 1, u 2 ) and v = (v 1, v 2 ), then define u + v = (u 1 + v 1, u 2 + v 2 ) and if k is any real number, then define k u = (ku 1, 0) There are values of u for which Axiom 10 fails to hold. For example, if u = (u 1, u 2 ) is such that u 2 0,then 1u =1(u 1, u 2 )=(1u u 1, 0) = (u 1, 0) u Thus, V is not a vector space with the stated operations. 2011/7/6 Elementary Linear Algebra 15

4-1 An Unusual Vector Space Let V be the set of positive real numbers, and define the operation on V to be u + v = uv [vector addition is numerical multiplication] ku = u k [ scalar multiplication is numerical exponentiation] Prove: Axiom 4 : zero vector 0 Axiom 5 : negative of a vector u Leave the other as your HW slide4 2011/7/6 Elementary Linear Algebra 16

4-1 Example 6 Every Plane Through the Origin Is a Vector Space Let V be any plane through the origin in R 3. Since R 3 itself is a vector space, Axioms 2, 3, 7, 8, 9, and 10 hold for all points in R 3 and consequently for all points in the plane V. We need only show that Axioms 1, 4, 5, and 6 are satisfied. 1. If u and v are objects in V, then u + v is in V. 4. There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = uforalluinv. u 5. For each u in V, there is an object -u in V, called a negative of u, such that u + (-u) = (-u) + u = 0. 6. If k is any scalar and u is any object in V, then ku is in V. (Slide 22) 2011/7/6 Elementary Linear Algebra 17

4-1 Example 7 (The Zero Vector Space) Let V consist of a single object, which we denote by 0, and ddefine 0 + 0 = 0 and k 0 = 0 for all scalars k. We called this the zero vector space. 2011/7/6 Elementary Linear Algebra 18

Theorem 4.1.1 Let V be a vector space, u be a vector in V, and k a scalar; then: 0 u = 0 k 0 = 0 (-1) u = -u If k u = 0, then k = 0 or u = 0. 2011/7/6 Elementary Linear Algebra 19

4-2 Subspaces A subset W of a vector space V is called a subspace of V if W is itself a vector space under the addition and scalar multiplication defined on V. Theorem 4.2.1 If W is a set of one or more vectors from a vector space V, then W is a subspace of V if and only if the following conditions hold: a) If u and v are vectors in W, then u + v is in W. b) If k is any scalar and u is any vector in W, then ku is in W. Remark W is a subspace of V if and only if W is a closed under addition (condition (a)) and closed under scalar multiplication (condition (b)). 2011/7/6 Elementary Linear Algebra 21

4-2 Example 1 Let W be any yplane through the origin and let u and v be any vectors in W. u + v must lie in W since it is the diagonal of the parallelogram determined by u and v, and k u must line in W for any scalar k since k u lies on a line through u. Thus, W is closed under addition and scalar multiplication, so it is a subspace of R 3. 2011/7/6 Elementary Linear Algebra 22

4-2 Example 2 A line through the origin of R 3 is a subspace of R 3. Let W be a line through the origin of R 3. 2011/7/6 Elementary Linear Algebra 23

4-2 Example 3 (Not a Subspace) Let W be the set of all points (x, y) in R 2 such that x 0 and y 0. These are the points in the first quadrant. The set W is not a subspace of R 2 since it is not closed under scalar multiplication. For example, v = (1, 1) lines in W, but its negative e (-1)v = -v = (-1, -1) does not. 2011/7/6 Elementary Linear Algebra 24

4-2 Subspace Remarks Think about set and empty set! Every nonzero vector space V has at least two subspace: V itself is a subspace, and the set {0} consisting of just the zero vector in V is a subspace called the zero subspace. Examples of subspaces of R 2 and R 3 : Subspaces of R 2 : {0} Lines through the origin R 2 Subspaces of R 3 : {0} Lines through the origin Planes through origin R 3 They are actually the only subspaces of R 2 and R 3 2011/7/6 Elementary Linear Algebra 25

4-2 Example 4 (Subspaces of M nn) ) The sum of two symmetric matrices is symmetric, and a scalar multiple of a symmetric matrix is symmetric. => the set of nn n symmetric matrices is a subspace of the vector space M nn of nn matrices. Subspaces of M nn the set of nn upper triangular matrices the set of nn lower triangular matrices the set of nn diagonal matrices 2011/7/6 Elementary Linear Algebra 26

4-2 Linear Combination If w is a vector in a vector space V, then w is said to be a linear combination of the vectors v 1, v 2,, v r if it can be expressed in the form w = k 1 v 1 + k 2 v 2 + + k r v r where k 1, k 2,, k r are scalars. These scalars are called the coefficients of the linear combination Vectors in R 3 are linear combinations of i, j, and k Every vector v = (a, b, c) in R 3 is expressible as a linear combination of the standard basis vectors i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) since v = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = a i + b j + c k 2011/7/6 Elementary Linear Algebra 27

4-2 Example : Linear Combinations Consider the vectors u = (1, 2, -1) and v = (6, 4, 2) in R 3. Show that w = (9, 2, 7) is a linear combination of u and v and that w = (4, -1, 8) is not a linear combination of u and v. Solution. 2011/7/6 Elementary Linear Algebra 28

Linear Combination and Spanning If S = {v 1, v 2,, v r} is a set of vectors in a vector space V, then the subspace W of V containing of all linear combination of these vectors in S is called the space spanned by v 1, v 2,, v r, and we say that the vectors v 1, v 2,,v v r span W. To indicate that W is the space spanned by the vectors in the set S = {v 1, v 2,, v r }, we write W = span(s) or W = span{v 1, v 2,, v r }. Ex. R 3 = span {i, j, k} 2011/7/6 Elementary Linear Algebra 29

4-2 Example 10 If v 1 and v 2 are non-collinear vectors in R 3 with their initial points at the origin span{v 1, v 2 }, which consists of all linear combinations k 1 v 1 + k 2 v 2 is the plane determined by v 1 and v 2. p y 1 2 Similarly, if v is a nonzero vector in R 2 and R 3, then span{v}, which is the set of all scalar multiples kv, is the linear determined by v. 2011/7/6 Elementary Linear Algebra 30

4-2 Example Testing for Spanning p.186 Determine whether v 1 = (1, 1, 2), v 2 = (1, 0, 1), and v 3 = (2, 1, 3) span the vector space R 3. 2011/7/6 Elementary Linear Algebra 31

4-2 Solution Space Solution Space of Homogeneous Systems If Ax = 0 is a system of the linear equations, then each vector x that satisfies this equation is called a solution vector of the system. Theorem e 4.2.4. shows that the solution o vectors of a homogeneous linear system form a vector space, which we shall call the solution space of the system. 2011/7/6 Elementary Linear Algebra 32

Theorem 4.2.4 If Ax = 0 is a homogeneous linear system of m equations in n unknowns, then the set of solution vectors is a subspace of R n. 2011/7/6 Elementary Linear Algebra 33

4-2 Example 7 Find the solution spaces of the linear systems. 1-2 3 x 0 1-2 3 x 0 (a ) y 0 y 0 2-4 6 (b ) -3 7 8 3-6 9 z 0-2 4-6 z 0 1-2 3 x 0 0 0 0 x 0 (c) y 0-3 7-8 (d) 0 0 0 y 0 4 1 2 z 0 0 0 0 z 0 Each of these systems has three unknowns, so the solutions form subspaces of R 3. Geometrically, each solution space must be a line through the origin, a plane through the origin, the origin only, or all of R 3. 2011/7/6 Elementary Linear Algebra 34

4-2 Example 7 (continue) Solution. () (a) x = 2s - 3t, y = s, z = t x = 2y - 3z or x 2y + 3z = 0 This is the equation of the plane through the origin with n = (1, -2, 3) as a normal vector. (b) x = -5t, y = -t, z =t which are parametric equations for the line through the origin parallel to the vector v = (-5, -1, 1). (c) The solution is x = 0, y=0 0, z=0 0, so the solution space is the origin only, that is {0}. (d) The solution are x = r, y = s, z = t, where r, s, and t have arbitrary values, so the solution space is all of R 3. 2011/7/6 Elementary Linear Algebra 35

4.3 Linearly Dependent & Independent If S = {v 1, v 2,, v r } is a nonempty set of vector, then the vector equation k 1 v 1 + k 2 v 2 + + k r v r = 0 has at least one solution, namely k 1 = 0, k 2 = 0,, k r = 0. If this the only solution, then S is called a linearly independent set. If there are other solutions, then S is called a linearly dependent set. Example 1 If v 1 = (2, -1, 103) 0, 3), v 2 = (125 (1, 2, 5, -1), and v 3 = (7, -1, 158) 5, 8). Then the set of vectors S = {v 1, v 2, v 3 } is linearly dependent, since 3v 1 + v 2 v 3 = 0. 2011/7/6 Elementary Linear Algebra 37

4.3 Example 3 Let i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) in R 3. Consider the equation k 1 i + k 2 j + k 3 k = 0 k 1 (1, 0, 0) + k 2 (0, 1, 0) + k 3 (0, 0, 1) = (0, 0, 0) (k 1, k 2, k 3 )=(000) (0, 0, 0) The set S = {i, j, k} is linearly independent. Similarly the vectors e 1 = (1, 0, 0,,0), e 2 = (0, 1, 0,, 0),, e n = (0, 0, 0,, 1) form a linearly independent set in R n. 2011/7/6 Elementary Linear Algebra 38

4.3 Example 4 p.191 Determine whether the vectors v 1 = (1, -2, 3), v 2 = (5, 6, -1), v 3 = (3, 2, 1) form a linearly dependent set or a linearly independent set. 2011/7/6 Elementary Linear Algebra 39

Theorem 4.3.1 p.193 A set with two or more vectors is: Linearly dependent if and only if at least one of the vectors in S is expressible as a linear combination of the other vectors in S. Linearly independent if and only if no vector in S is expressible as a linear combination of the other vectors in S. 2011/7/6 Elementary Linear Algebra 40

4.3 Example 6 If v 1 = (2, -1, 0, 3), v 2 = (1, 2, 5, -1), and v 3 = (7, -1, 5, 8). the set of vectors S = {v 1, v 2, v 3 } is linearly dependent In this example each vector is expressible as a linear combination of the other two since it follows from the equation 3v 1 +v 2 -v 3 =0 that v 1 =-1/3v 2 +1/3v 3, v 2=-3 v 1+v 3 3, and v 3 =3v 1 +v 2 2011/7/6 Elementary Linear Algebra 41

4.3 Example 7 Let i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) in R 3. The set S = {i, j, k} is linearly independent. Suppose that k is expressible as k=k 1 i+k 2 j Then, in terms of components, (001)=k (0, 0, 1 (1, 0, 0)+k 2 (0, 1, 0) or (0, 0, 1)=(k 1, k 2, 0) 2011/7/6 Elementary Linear Algebra 42

Theorem 4.3.2 Sets that contain the zero vector and sets with one or two vectors A finite set of vectors that contains the zero vector is linearly dependent. A set with exactly two vectors is linearly independently if and only if neither vector is a scalar multiple of the other. 2011/7/6 Elementary Linear Algebra 43

Geometric Interpretation of Linear Independence In R 2 and R 3, a set of two vectors is linearly independent if and only if the vectors do not lie on the same line when they are placed with their initial points at the origin. In R 3, a set of three vectors is linearly independent if and only if the vectors do not lie in the same plane when they are placed with their initial points at the origin. 2011/7/6 Elementary Linear Algebra 44

The Wronskian

4-4 Nonrectangular Coordinate Systems The coordinate system establishes a one-to-one correspondence between points in the plane and ordered pairs of real numbers. Although perpendicular coordinate axes are the most common, any two nonparallel lines can be used to define a coordinate system in the plane. 2011/7/6 Elementary Linear Algebra 47

4-4 Nonrectangular Coordinate Systems A coordinate system can be constructed by general vectors: v 1 and v 2 are vectors of length 1 that points in the positive direction of the axis: OP a v b v 1 2 Similarly, the coordinates (a, b, c) of the point P can be obtained by expressing OP as a linear combination of the vectors 2011/7/6 Elementary Linear Algebra 48

4-4 Basis If V is any vector space and S = {v 1, v 2,,v n } is a set of vectors in V, then S is called a basis for V if the following two conditions hold: S is linearly independent. S spans V. Theorem 4.4.1 (Uniqueness of Basis Representation) If S = {v 1, v 2,,vv n } is a basis for a vector space V, then every vector v in V can be expressed in the form v = c 1 v 1 + c 2 v 2 + + c n v n in exactly one way. 2011/7/6 Elementary Linear Algebra 49

4-4 Coordinates Relative to a Basis If S = {v 1, v 2,, v n } is a basis for a vector space V, and v = c 1 v 1 + c 2 v 2 + + c n v n is the expression for a vector v in terms of the basis S, then the scalars c 1, c 2,,c c n, are called the coordinates of v relative to the basis S. The vector (c 1, c 2,, c n ) in R n constructed from these coordinates is called the coordinate vector of v relative to S; it is denoted by (v) S = (c 1, c 2,, c n ) 2011/7/6 Elementary Linear Algebra 50

4-4 Example 1 (Standard Basis for R 3 ) Suppose that i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) S ={i {i, j, k} is a linearly independent set in R 3. S also spans R 3 since any vector v = (a, b, c) in R 3 can be written as v = (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = ai + bj + ck Thus, S is a basis for R 3 ; it is called the standard basis for R 3. Looking at the coefficients of i, j, and k, (v) S = (a, b, c) Comparing this result to v = (a, b, c), v =(v) S 2011/7/6 Elementary Linear Algebra 51

4-4 Example 2 (Standard Basis for R n ) If e 1 = (1, 0, 0,, 0), e 2 = (0, 1, 0,, 0),, e n = (0, 0, 0,, 1), S = {e 1, e 2,, e n } is a linearly independent set in R n S also spans R n since any vector v = (v 1, v 2,, v n ) in R n can be written as v = v 1 e 1 + v 2 e 2 + + v n e n Thus, S is a basis for R n ;itiscalledthestandard is the standard basis for R n. The coordinates of v = (v 1, v 2,,v v n ) relative to the standard basis are v 1, v 2,, v n, thus (v) S = (v 1, v 2,, v n ) => v = (v) s A vector v and its coordinate vector relative to the standard basis for R n are the same. 2011/7/6 Elementary Linear Algebra 52

4-4 Example 3 Let v 1 =(121)v (1, 2, 1), v 2 =(290)andv (2, 9, 0), v 3 =(334) (3, 3, 4). Show that the set S = {v 1, v 2, v 3 } is a basis for R 3. 2011/7/6 Elementary Linear Algebra 53

4-4 Example 4 (Representing a Vector Using Two Bases) Let S = {v 1, v 2, v 3 } be the basis for R 3 in the preceding example. Find the coordinate vector of v = (5, -1, 9) with respect to S. Find the vector v in R 3 whose coordinate vector with respect to the basis S is (v) s = (-1, 3, 2). v = c 1 v 1 + c 2 v 2 + + c n v n (v) S = (c 1, c 2,, c n ) 2011/7/6 Elementary Linear Algebra 54

4-4 Example 6 (Standard Basis for M mn) ) Let 1 0 0 1 0 0 0 0 M1, M2, M3, M4 0 0 0 0 1 0 0 1 The set S = {M 1, M 2, M 3, M 4 } is a basis for the vector space M 22 of 2 2 matrices. To see that S spans M 22, note that an arbitrary vector (matrix) can be written as a c b d 1 a 0 0 0 b 0 0 1 0 c 0 1 0 0 d 0 0 0 1 am 1 bm 2 cm 3 a c dm To see that S is linearly independent, assume am 1 + bm 2 + cm 3 + dm 4 = 0. It follows that a b 0 0. Thus, a = b = c = d = 0, so S is lin. indep. c d 0 0 4 b d 2011/7/6 Elementary Linear Algebra 55