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Introduction to Vector Spaces in Linear Algebra

This chapter introduces the concept of vector spaces in linear algebra, using examples in R2 and R3. It covers vector addition, scalar multiplication, properties of vectors, dot product, norm, angle, and distance between vectors.

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Introduction to Vector Spaces in Linear Algebra

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  1. Linear Algebra Chapter 4 Vector Spaces

  2. Example 1 • R2 is the collection of all sets of two ordered real numbers. • For example, (0, 0) , (1, 2) and (-2, -3) are elements of R2. • R3 is the collection of all sets of three ordered real numbers. • For example, (0,0, 0)and(-1,3, 4) are elements ofR3. 4.1 The vector Space Rn Definition 1. Let be a sequence of n real numbers. The set of all such sequences is called n-space (or n-dimensional. space) and is denoted Rn. u1 is the first component of . u2 is the second component and so on.

  3. Definition 3. Let be elements of Rn and let c be a scalar. Addition and scalar multiplication are performed as follows: Addition: Scalar multiplication : Definition 2. Let be two elements of Rn. We say that u and v are equal if u1 = v1, …, un = vn. Thus two elements of Rn are equal if their corresponding components are equal.

  4. ► The set Rn with operations of componentwise addition and scalar multiplication is an example of a vector space, and its elements are called vectors. We shall henceforth interpret Rn to bea vector space. (We say that Rn is closed underaddition and scalar multiplication). ►In general, if u and v are vectors in the same vector space, then u+ v is the diagonal of the parallelogram defined by u and v. Figure 4.1

  5. Example 3 In R2 , consider the two elements (4, 1) and (2, 3). Find their sum and give a geometrical interpretation of this sum. we get (4, 1) + (2, 3) = (6, 4). The vector (6, 4), the sum, is the diagonal of the parallelogram. Figure 4.2 Example 2 Let u = ( –1, 4, 3) and v = ( –2, –3, 1) be elements of R3. Find u + v and 3u. Solution:u + v = (–1, 4, 3) + (–2, –3, 1) = (-3 ,1 ,4) 3u = 3 (–1, 4, 3) = (-3 ,12 ,9)

  6. Example 4 Consider the scalar multiple of the vector (3, 2) by 2, we get 2(3, 2) = (6, 4) Observe in Figure 4.3 that (6, 4) is a vector in the same direction as (3, 2), and 2 times it in length. Figure 4.3

  7. Negative Vector The vector (–1)u is writing –u and is called thenegative of u. It is a vector having the same length (or magnitude) as u, but lies in the opposite direction to u. Subtraction u Subtraction is performed on element of Rn by subtracting corresponding components. -u Zero Vector The vector (0, 0, …, 0), having n zero components, is called the zero vector of Rn and is denoted 0.

  8. Figure 4.4 Commutativity of vector additionu + v = v + u Theorem 4.1 • Let u, v, and w be vectors in Rn and let c and d be scalars. • u + v = v + u • u + (v + w) = (u + v) + w • u + 0 = 0 + u = u • u + (–u) = 0 • c(u + v) = cu + cv • (c + d)u = cu + du • c(du) = (cd)u • 1u = u

  9. Solution Example 5 Let u = (2, 5, –3), v = ( –4, 1, 9), w = (4, 0, 2) in the vector space R3. Determine the vector 2u – 3v + w.

  10. Row vector: Column vector: We defined addition and scalar multiplication of column vectors in Rn in a componentwise manner: and Column Vectors

  11. Homework • Exercise set 1.3 page 32:3, 5, 7, 9.

  12. Example 1 Find the dot product of u = (1, –2, 4) and v = (3, 0, 2) Solution 4.2 Dot Product, Norm, Angle, and Distance Definition Let be two vectors in Rn. The dot product of u and v is denoted u.vand is defined by . The dot product assigns a real number to each pair of vectors.

  13. Proof 1. 4. Properties of the Dot Product • Let u, v, and w be vectors in Rn and let c be a scalar. Then • u.v = v.u • (u + v).w = u.w + v.w • cu.v = c(u.v) = u.cv • u.u 0, and u.u = 0 if and only if u = 0

  14. 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 Figure 4.5 length ofu Norm of a Vector in Rn

  15. 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. Example 2 Find the norm of each of the vectors u = (1, 3, 5) of R3 and v = (3, 0, 1, 4) of R4. Solution

  16. (a) Thus (1, 0) is a unit vector. It can be similarly shown that (0, 1) is a unit vector in R2. (b) 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). Example 3 • Show that the vector (1, 0) is a unit vector. • Find the norm of the vector (2, –1, 3). Normalize this vector. Solution

  17. Angle between Vectors ( in R2) The law of cosines gives: ← Figure 4.6

  18. Example 4 Determine the angle between the vectors u = (1, 0, 0) and v = (1, 0, 1) in R3. Solution Thus the angle between u and v is 45. Angle between Vectors (in Rn) Definition Let u and v be two nonzero vectors in Rn. The cosine of the angle between these vectors is

  19. Theorem 4.2 Two nonzero vectors u and v are orthogonal if and onlyif u.v = 0. Proof Orthogonal Vectors Definition Two nonzero vectors are orthogonal if the angle between them is a right angle .

  20. Example 5 • Show that the following pairs of vectors are orthogonal. • (1, 0) and (0, 1). • (2, –3, 1) and (1, 2, 4). Solution • (1, 0).(0, 1) = (1  0) + (0  1) = 0. The vectors are orthogonal. • (2, –3, 1).(1, 2, 4) = (2  1) + (–3  2) + (1  4) = 2 – 6 + 4 = 0. • The vectors are orthogonal.

  21. Note • (1, 0), (0,1) are orthogonalunit vectors in R2. • (1, 0, 0), (0, 1, 0), (0, 0, 1) are orthogonal unit vectors in R3. • (1, 0, …, 0), (0, 1, 0, …, 0), …, (0, …, 0, 1) are orthogonal unit vectors in Rn.

  22. Solution Let the vector (a, b) be orthogonal to (3, –1). We get Figure 4.7 Example 6 Determine a vector in R2 that is orthogonal to (3, –1). Show that there are many such vectors and that they all lie on a line. Thus any vector of the form (a, 3a) is orthogonal to the vector (3, –1). Any vector of this form can be written a(1, 3) The set of all such vectors lie on the line defined by the vector (1, 3).

  23. Figure 4.8(a) Figure 4.8(b) Theorem 4.3 • Let u and v be vectors in Rn. • Triangle Inequality: • ||u + v||  ||u|| + ||v||. • Pythagorean theorem : If u.v = 0 then ||u + v||2 = ||u||2 + ||v||2.

  24. x x-y y Example 7. Determine the distance between the points x = (1,–2 , 3, 0) and y = (4, 0, –3, 5) in R4. Distance between Points Let be two points in Rn. The distancebetween x and y is denoted d(x, y) and is defined by Note: We can also write this distance as follows. Solution

  25. Exercise 36 Let u and v be vectors in Rn. Prove that ||u|| = ||v|| if and onlyifu + v and u-v are orthogonal. Homework • Exercise set 1.5 pages 47 to 48:3, 7, 8, 9, 11, 13, 16, 17, 26.

  26. 4.3 General Vector Spaces Our aim in this section will be to focus on the algebraic properties of 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. • Let u, v, and w be arbitrary elements of V, and c and d are scalars. • Closure Axioms • 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.)

  27. Example 1 (1) V={ …, -3, -1, 1, 3, 5, 7, …} V is not closed under addition because 1+3=4  V. (2) Z={ …, -2, -1, 0, 1, 2, 3, 4, …} Z is closed under addition because for any a, b Z, a + bZ. Z is not closed under scalar multiplication because½ is a scalar, for any odd a Z, (½)a Z.

  28. 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

  29. 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. Axiom 3,4 and 7~10: trivial

  30. Let Axiom 1: u + v is a 2 2 matrix. ThusM22 is closed under addition. ► Question: Prove Axiom 2 and Axiom 7 . Vector Spaces of Matrices (Mmn) Prove that M22 is a vector space. Proof Axiom 3 and 4: From our previous discussions we know that 2 2 matrices are commutative and associative under addition (Theorem 2.2).

  31. Axiom 6: Axiom 5: The 2 2 zero matrix is , since In general: The set of m n matrices, Mmn, is a vector space.

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

  33. 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 (continued) 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.) Is the set F ={ f | f(x)=ax2+bx+c for some a,b,cR} a vector space?

  34. Theorem 4.4 (useful properties) 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.

  35. Homework • Exercise set 4.1, pages 206-207:5, 7, 11, 13, 15.

  36. 4.4 Subspaces Figure 4.9

  37. Note: ► In general, a subset of a vector space may or may not satisfy the closure axioms. ► However, any subset that is closed under both of these operations satisfies all the other vector space properties. Definition Let V be a vector space and U be a nonempty subset of V. U is said to be a subspace of V if it is closed under addition and under scalar multiplication.

  38. Example 1 Let U be the subset of R3 consisting of all vectors of the form (a, 0, 0) (with zeros as second and third components and aR ), i.e., U = {(a, 0, 0) R3 }. Show that U is a subspace of R3. Solution Let (a, 0, 0), (b, 0, 0) U, and let kR. We get (a, 0, 0) + (b, 0, 0) = (a + b, 0, 0)  U k(a, 0, 0) = (ka, 0, 0)  U The sum and scalar product are in U. Thus U is a subspace of R3. # Geometrically, U is the set of vectors that lie on the x-axis.

  39. Example 2 Let V be the set of vectors of of R3 of the form (a, a2, b), namely V = {(a, a2, b) R3 }. Show that V is not a subspace of R3. Solution Let (a, a2, b), (c, c2, d) V. (a, a2, b) + (c, c2, d) = (a+ c, a2 + c2, b + d)  (a + c, (a+ c)2, b + d) , since a2 + c2 (a + c)2. Thus (a, a2, b) + (c, c2, d) V. V is not closed under addition. V is not a subspace.

  40. Example 3 Prove that the set W of 2  2 diagonal matrices is a subspace of the vector space M22 of 2  2 matrices. Solution (+) Let W. We get  u + v W.  W is closed under addition. • () Let cR. We get • cu W.  W is closed under scalar multiplication. • W is a subspace of M22.

  41. The vector space of polynomials (Pn) Example 5.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. Example 4. Solution Let f and gPn, where ►(+) (f + g)(x) is a polynomial of degree  n. Thus f + g Pn.Then Pn is closed under addition.

  42. ►() Let cR, (cf )(x) is a polynomial of degree  n. So cf Pn. Then Pn is closed under scalar multiplication. In conclusion : By (+) and (), Pn is a subspace of the vector space F of functions. Therefore Pn is itself a vector space.

  43. Theorem 4.5 (Very important condition) Let U be a subspace of a vector space V. Ucontains the zero vector of V. Note. Let 0 be the zero vector of V. If 0U  U is not a subspace of V. If 0U  (+)() hold  U is a subspace of V.(+)() failed U is not a subspace of V. Caution. This condition is necessary but not sufficient. (See, for instance, Example 2 above and Example 5 below)

  44. Example 5 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 . This system is inconsistent it has no solution. Thus (0, 0, 0) W. (The necessary condition does not hold) W is not a subspace of R3.

  45. Let F={ f | f: RR } the vector space of functions on R. Which of the following are subspaces of F ?(a) W1={ f | f: RR, f(0)=0 }. (b) W2={ f | f: RR, f(0)=3 }. (c) W3={ f | f: RR, for some cR, f(x)=c for every x}. Homework • Exercise set 4.1, pages 207-208:19, 21, 23, 25, 27, 29, 31, 33. • Exercise set 1.3, page 32: 11, 12, 13, 15. Exercise

  46. 4.5 Linear Combinations of Vectors W={(a, a, b) | a,bR}  R3 (a, a, b) = a (1,1,0) + b (0,0,1) W is generated by (1,1,0) and (0,0,1). 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. We say that v, a vector of V, is a linear combination of v1, v2, …, vm , if there exist scalars c1, c2, …, cm such that v can be written v = c1v1 + c2v2 + … + cmvm .

  47. Example 1 The vector (5, 4, 2) is a linear combination of the vectors (1, 2, 0), (3, 1, 4), and (1, 0, 3), since it can be written (5, 4, 2) = (1, 2, 0) + 2(3, 1, 4) – 2(1, 0, 3)

  48. Suppose Example 2 Determine whether or not the vector (-1, 1, 5) is a linear combination of the vectors (1, 2, 3), (0, 1, 4), and (2, 3, 6). Solution Thus (-1, 1, 5) is a linear combination of (1, 2, 3), (0, 1, 4), and (2, 3, 6), where

  49. 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 3 Express the vector (4, 5, 5) as a linear combination of the vectors (1, 2, 3), (-1, 1, 4), and (3, 3, 2). Solution

  50. Solution Suppose  Example 4 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 the vectors (1, 2, 3), (-1, -1, -2), and (1, 4, 5).

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