3 Vector spaces.pdf
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5.2 Vector spaces We recollect that for a non-empty set G, a binary operation on G is mapping from G G into G. Definition: A non-empty set G together with a binary operation * is called a group if the algebraic system (G, *) satisfies the following four axioms: (i) Associative axiom: (a*b)*c = a*...
5.2 Vector spaces We recollect that for a non-empty set G, a binary operation on G is mapping from G G into G. Definition: A non-empty set G together with a binary operation * is called a group if the algebraic system (G, *) satisfies the following four axioms: (i) Associative axiom: (a*b)*c = a*(b*c) for all elements a, b, c G. (ii) Identity axiom: There exists an element e G such that a*e = e*a = a for all elements a G. (iii) Inverse axiom: To each element a G there corresponds an element b G such that a*b = e = b*a. Moreover G is said to be abelian if a * b = b* a for all a, b G (similarly, G is commutative if the multiplication a b = b a for all a, b G). Examples: (i) The set of all integers with respect to addition is an abelian group. (ii) The set of all 2 2 matrices with real or complex entries is a group with respect to matrix addition. Definition: A non-empty set F is said to be a field if there exists two binary operations + and “.” on F such that (i) (F, +) is an abelian group (ii) (F \ {0},.) is a multiplicative group and (iii) for any a, b, c F, we have a (b + c) = ab + ac, (a + b)c = ac + bc. Examples: (i) Set of real numbers with usual addition and multiplication is a field. (ii) The set of all complex numbers is also a field with respect to the addition and multiplication of complex numbers. Definition: A non-empty set V is said to be a vector space over a field F (generally, we take the field as the set of real numbers or the set of complex numbers) if it satisfying the following: (i) (V, +) is an abelian group, (ii) V is closed under scalar multiplication (that is, for every F, v V we have v V) and also this scalar multiplication satisfies the following conditions: (a) (v + w) = v + w, (b) ( + ) v = v + v, (c) (v) = ()v and (d) 1.v = v for all , F and v, w V (here 1 is the identity of F with respect to multiplication). Here + is addition either in the field or in the vector space, as appropriate; and 0 is the additive identity in either. Juxtaposition indicates either scalar multiplication or the multiplication operation in the field. Note: We use F for field, also the elements of F are called scalars and the elements of V are called vectors. Definition: (i) An n-component vector a is an ordered n-tuple of numbers written as a a1 a2 row (a1,...,an) or as a column . where ai, i = 1, 2, …, n are assumed to be real . a n numbers and are called the components of the vector. (ii) A unit vector denoted by ei and defined with unity as value of its ith component and all other components zero. (iii) Null vector is a vector all of whose components are zero. (iii) A sum vector is a vector having unity as a value for each component; it will be written as 1. (iv) Let a and b be two n-component vectors. Then a and b are equal if and only if ai = bi for each i. (v) The scalar product of two n-component vectors a, b is defined to be the scalar a1b1 + n a2b2 + … + anbn = a b i 1 i i Definition: The norm of an n-component vector a = (a1, a2, …, an) is denoted and defined by ‖ a ‖ =√ = a12 a 22 ... a n2. Euclidean Space: An n-dimensional Euclidean space is defined as the collection of all vectors (points) a = (a1, a2, …, an). For these vectors, addition and multiplication by a scalar are defined as follows. En = {(a1, a2, …, an) each ai is a real or complex number, 1 ≤ i ≤ n}. Define addition and multiplication by scalar in En as, for any (a1, a2, …, an), (b1, b2, …, bn) En, (a1, a2, …, an) + (b1, b2, …, bn) = (a1 + b1, a2 + b2, …, an + bn) En, and for any scalar , (a1, a2, …, an) = (a1, a2, …, an) En. Therefore En is a vector space (over R). a1 a2 18.2.22 Example: Let M2 2 = / ai is a real number. Then M22 is a vector space a3 a4 over the set of real numbers, with respect to the addition of matrices and the multiplications of a matrix by a scalar. Definition: Let V be a vector space over F and W V. Then W is called a subspace of V if W is a vector space over F under the same operation. Definition: Suppose V is a vector space over F, vi V and i F for 1 i n. Then 1v1 + 2v2 +... + nvn is called the linear combination (over F) of {v1, v2,..., vn}. Definition: Let V be a vector space and S V. We write L(S) = {1v1 + 2v2 +... + nvn / n N, vi S and i F for 1 i n}, which is equal to the set of all linear combinations of finite number of elements of S. This L(S) is called the linear span of S. Example: Let a = (2, 3, 4, 7), b = (0, 0, 0, 1), c = (1, 0, 1, 0). Then the vector d = (5, 3, 7, 9) is a linear combination of the vectors a, b, c as follows. (5, 3, 7, 9) = (2, 3, 4, 7) + 2(0, 0, 0, 1) + 3(1, 0, 1, 0). 18.2.19 Example: Let the field K be the set R of real numbers, and let the vector space V be the Euclidean space R3. Consider the vectors e1 := (1,0,0), e2 := (0,1,0) and e3 = (0,0,1). Then any vector in R3 is a linear combination of e1, e2 and e3. Note: Let v S. Then1.v is a linear combination and hence v L(S). Therefore, S L(S). Problem: Let V be a vector space over F and W V. Then the following two conditions are equivalent. (i) W is a subspace of V (ii), F and w1, w2 W w1 + w2 W. a a Example: The set S22= / a is a real number is a subspace of M22. a a Example: The set W = {(1, 1, 0) is a real number} is a subspace of R3. Problem: If S is any subset of a vector space V, then show that L(S) is a subspace of V. Solution: Let v, w L(S) and , F. Since v, w L(S), we have that v = 1v1 +... + nvn and w = 1w1 +...+mwm for some vi S, i F for 1 i n and wj S, j F for 1 j m. Now v + w = (1v1 +... + nvn) + (1w1 +... + mwm) = 1v1 + 2v2 +... + nvn + 1w1 + 2w2 +... + mwm, is a linear combination of elements from S. Hence v + w L(S). This shows that L(S) is a subspace of V. Properties of linear span: If S and T are subsets of a vector space V then (i) S T L(S) L(T), (ii) L(S T) = L(S) + L(T), (iii) L(L(S)) = L(S). Definitions: (i) The vector space V is said to be finite-dimensional (over F) if there is a finite subset S in V such that L(S) = V. (ii) The vectors vi V for 1 i n, are linearly dependent over F if there exists elements ai F, 1 i n, not all of them equal to zero, such that a1v1 + a2v2 +... + anvn = 0. If the vectors vi, 1 i n are not linearly dependent over F then they are said to be linearly independent over F. Examples: Consider R, the set of real numbers and take V = R2, then (i) V is vector space over the field R. (ii) Consider S = {(1, 0), (1, 1), (0, 1)} R2. Then the linear span of S is L(S) = {1(1, 0) + 2(1, 1) + 3(0, 1) / i R, 1 i 3} = {(1 + 2, 2 + 3) / i R, 1 i 3} R2. If (x, y) R2 then write 1 = x, 2 = 0, 3 = y then (x, y) = (1 + 2, 2 + 3) L(S). Hence L(S) = R2 = V. This shows that R2 is finite dimensional. (iii) Write v1 = (1, 0), v2 = (2, 2), v3 = (0, 1), v4 = (3, 3). Then 1v1 + 2 v2 + 3v3 + 4v4 = 0 where 1 = 2, 2 = -1, 3 = 2, 4 = 0. Thus there exists scalars 1, 2, 3, 4 not all of them equal to zero such that 1v1 + 2 v2 + 3v3 + 4v4 = 0. Hence {vi / 1 i 4} is a linearly dependent set. (iv) Suppose v1 = (1, 0), v2 = (0, 1). Suppose 1v1 + 2v2 = 0 for some 1, 2 R. Then 1(1, 0) + 2(0, 1) = 0 = (0, 0) (1, 2) = (0, 0) 1 = 0 = 2. Hence v1, v2 are linearly independent. Lemma: Let V be a vector space over F. If v1, v2,..., vn V are linearly independent, then every element in their linear span has a unique representation in the form 1v1 + 2v2 +... + nvn with i F, 1 i n. Proof: Let S = {vi / 1 i n}. Consider L(S). Let v L(S). Then v = 1v1 +... + nvn for some i F, vi S, 1 i n (by the definition of L(S)). Uniqueness: Suppose v = 1v1 +... +nvn = 1v1 +... + nvn for some i, i F, 1i n (1v1 +... + nvn) – (1v1 +... + nvn) = 0 (1 - 1)v1 +... + (n - n)vn = 0 1 - 1 = 0 2 - 2 = 0,..., n - n = 0 (since vi, 1 i n are linearly independent) 1 = 1, 2 = 2,..., n = n. Hence every element in the linear span can be expressed as in a unique way as a linear combination of vi, 1 i n. Theorem: If vi V, 1 i n, then either they are linearly independent or some vk is a linear combination of the preceding ones v1, v2,... , vk-1. Proof: If vi, 1 i n are linearly independent then there is nothing to prove. Suppose that vi, 1 i n are not linearly independent (that is, linearly dependent). Then there exist scalars i, 1 i n not all zero, such that 1v1 + 2v2 +... + nvn = 0. Since all the i are not zero, there exists largest k such that k 0. Then k+1 = 0,... , n = 0 and so 1v1 + 2v2 +... + kvk = 0 kvk =-1v1 - 2v2 -... - k-1vk-1 vk = k-1(-1v1 - 2v2 -...- k-1vk-1) = -k-11v1 -... - k-1k-1vk-1 and k-1i F for each 1 i k-1. Hence vk is a linear combination of vi, 1 i k-1. Definition: A subset S of a vector space V is called a basis of V if (i) S consists of linearly independent elements (that is, any finite number of elements in S is a linearly independent), and (ii) S spans V (that is, V = L(S)). Note: A subset B of a vector space is a basis if and only if any of the following equivalent conditions are met: B is a minimal generating set of V, i.e., it is a generating set but no proper subset of B is. B is a maximal set of linearly independent vectors, i.e., it is a linearly independent set but no other linearly independent set contains it as a proper subset. Every vector in V can be expressed as a linear combination of vectors in B in a unique way. Examples: (i) Consider R2, the vector space of all coordinates (a, b) where both a and b are real numbers. Then a very natural and simple basis is simply the vectors e1 = (1,0) and e2 = (0,1): suppose that v = (a, b) is a vector in R2, then v = a (1,0) + b (0,1). (ii) The set of vectors {(1, 1), (−1, 2)} form a basis of R2. Verification: Part (i): First we prove that {(1, 1), (−1, 2)} is a linearly independent set. Suppose that there are numbers a, b such that: a(1, 1) + b(-1, 2) = (0, 0). (a - b, a + 2b) = (0, 0) a – b = 0 and a + 2b = 0. Subtracting we get 3b = 0 and so b = 0. From the first equation we get that a = 0. Part (ii): In this part we show that {(1, 1), (−1, 2)} generates R 2. Let (a, b) R2. Now we show that there exists two numbers x and y such that x(1, 1) + y(-1, 2) = (a, b). Then we have to solve the equations: x – y = a and x + 2y = b Subtracting the first equation from the second, we get 3y = b – a y = (b - a)/3 and x = (b + 2a)/3. Therefore {(1, 1), (−1, 2)} generates (or spans) R2. Observation: (i) Since (-1, 2) is clearly not a multiple of (1, 1) and since (1, 1) is not the zero vector, these two vectors are linearly independent. Since the dimension of R2 is 2, the two vectors already form a basis of R2 without needing any extension. (ii) The standard basis (also called natural basis or canonical basis) of the n- dimensional Euclidean space Rn is the basis obtained by taking the n basis vectors, (e1, e2, …, en) where ei is the vector with a 1 in the ith coordinate and 0 elsewhere. In many ways, it is the "obvious" basis. Example: The set of all n unit vectors e1, e2, …, en form a basis for Rn. To show that the set {e1, e2, …, en} is linearly independent: Suppose 1e1 + 2e2 + … + nen = 0. This means: 1(1, 0, …, 0) + 2(0, 1, 0, …, 0) + … + n(0, 0, …, 1) = (0, 0, …0). (1, 0, …, 0) + (0, 2, …, 0) + … + (n , …, 1) = (0, 0, …, 0). (1, 2, …,n ) = (0, 0, …, 0). This means that 1 = 0, 2 = 0, …, n = 0. Therefore {e1, e2, …, en} is linearly independent. Next to show that {e1, e2, …, en} spans Rn. Let x = (x1, x2,..., xn) Rn. Then x can be represented as a linear combination of {e1, e2, …, en} in the following way. x = x1e1 + x2e2 + … + xnen Therefore the set {e1, e2, …, en} spans Rn and hence a basis. 1 0 0 1 0 0 0 0 Example: Prove that the set B = , , , is a basis for M22. 0 0 0 0 1 0 0 1 Solution: The set B is linearly Independent: 1 0 0 1 0 0 0 0 0 0 Suppose 1 + 2 +3 + 4 = . This implies that 1 = 0, 0 0 0 0 1 0 0 1 0 0 2 = 0, 3 = 0, 4 = 0. We show that the set B spans M22: a b Let M22. Then c d a b 1 0 0 1 0 0 0 0 = a + b + c + d . Therefore the set B spans M22, and c d 0 0 0 0 1 0 0 1 hence a basis. Example: Test whether the set B = {(1, 1, 0), (3, 0, 1), (5, 2, 2)} forms a basis for R 3. If so represent (1, 2, 3) in terms of basis vectors. Solution: Consider the matrix determinant of the coefficients. 1 3 5 D = 1 0 2 = 1(-2) – 3(2) + 5(1) = -3 0. 0 1 2 Let (a, b, c) R3. Now (a, b, c) = 1(1, 1, 0) + 2(3, 0, 1) + 3(5, 2, 2) This means 1 + 32 + 53 = a 1 + 22 =b 2 + 23 = c. a 3 5 b 0 2 c 1 2 2a b 6c Solving, by Cramer’s rule 1 = = D 3 1 a 5 1 b 2 0 c 2 2b 2a 3c 2 = = and D 3 1 3 a 1 0 b 0 1 c a b 3c 3 = = D 3 when (a, b, c) = (1, 2, 3), we have 1 = -14/3, 2 = -11/3, 3 = 10/3. Therefore (1, 2, 3) = -14/3(1, 1, 0)-11/3(3, 0, 1) + 10/3(5, 2, 2). Problem : Any vector in Rn can be expressed as a linear combination of a set of vectors in only one way. Proof: Let b Rn and {a1, a2, …, ar} a set of basis vectors. Suppose that b is expressed in two ways as follows: b = 1a1 + 2a2 + … + nar and b = 1a1 + 2a2 + … +nar where ’s and ’s are scalars. Now (1 - 1)a1 + (2 -2)a2 + … + (r - r) ar = 0. Since {a1, a2, …, ar} is linearly independent, we have that 1 - 1 = 2 -2 = … r - r = 0. This means 1 = 1, 2 = 2, …, r = r = 0. Remark: If V is a finite dimensional vector space over F, then any two bases of V have that same number of elements. Problem: Define minimal spanning set of vectors. Prove that a minimal spanning set of vectors forms a basis. Solution: Mimimal spanning set: A subset S of a vector space V is said to be a minimal spanning set if (i) S is a spanning set for V, and (ii) S \ {v} do not span V for any v S. Let S = {v1, v2,..., vn} be a minimal spanning set. This means L(S) = V. In order to prove S is a basis, it suffices to prove S is linearly independent. In a contrary way, suppose that S is not linearly independent. Then there exists vj (for some j, 1 j n) is a linear combination of its preceding ones. That is., vj = 1v1 + 2v2 +...+ j-1 vj-1 for some i F, 1 i (j –1). Clearly L({v1, v2,..., vj-1, vj+1,..., vn}) L({vi / 1 i n}) = L(S). On the other hand, take x L(S). Then x = 1v1 + 2v2 +... + nvn for some i F, 1 i n x = 1v1 +... + j – 1vj-1 + j(1v1 + 2v2 +...+ j-1vj-1) + j+1vj+1 +... + nvn x = (1 + j1)v1 +... + (j-1 + jj-1)vj-1 + j+1vj+1 +... + nvn L({v1, v2,..., vj-1, vj+1,..., vn}). Therefore, L({v1, v2,..., vj-1, vj+1,..., vn}) = L(S) = V, which is a contradiction to the fact that n is minimum with S spans V. Therefore S is linearly independent. Problem: Define a maximal linearly independent set. Prove that a maximal linearly independent set is a basis. Solution: Maximal linearly independent set : A subset S of a vector space V is said to be a maximal linearly independent set if (i) S is a linearly independent set, and (ii) S {v} is linearly dependent for any v V \ S. Let S = {v1, v2,..., vn} be a maximal linearly independent set. In order to prove S is a basis, it suffices to prove S spans V. Take v V. Suppose 1v1 + 2v2 +...+ nvn + v = 0. If = 0, then since v1, v2,..., vn are linearly independent, we get i = 0 for all 1 i n. This means v, v1, v2,..., vn are (which are n + 1, in number) linearly independent, a contradiction to the maximality of n. Therefore, 0. Now v = (-1v1) + (-2v2) +...+ (-nvn). This implies v = (-1-1)v1 + (-2-1)v2 +...+ (n-1)vn. Therefore S spans V. Hence S is a basis. Problem 2: Whether the vectors {a1= (4, 2, 1), a2 = (2, -6, -5), a 3 = (1, -2, 3)} are linearly independent? Solution: The set {a1, a2, a3} is linearly independent if 𝜆 𝑎 + 𝜆 𝑎 + 𝜆 𝑎 = 0 implies each of the 𝜆 = 0 for i = 1, 2, 3. That is, 𝜆 (4, 2, 1) + 𝜆 (2, −6, −5) + 𝜆 (1, −2, 3) = 0 This yields the following system of homogeneous equations 4𝜆 + 2𝜆 + 𝜆 = 0 2𝜆 − 6𝜆 − 2𝜆 = 0 𝜆 − 5𝜆 + 3𝜆 = 0 The above system will have trivial solution 𝜆 = 𝜆 = 𝜆 = 0 if and only if |A| 4 2 1 ≠0, where A is the coefficient matrix 2 −6 −2 1 −5 3 4 2 1 We observe that |𝐴| = 2 −6 −2 = −132 ≠ 0. Hence 𝜆 = 𝜆 = 𝜆 = 0. 1 −5 3 Thus the given vectors are linearly independent. Note: 1. A set of vectors are linearly independent if |𝐴| ≠ 0 where A is the matrix formed from the given vectors. 2. Maximum number of linearly independent vectors in En is n. Hence any set of n+1 vectors are always linearly dependent. Problem Verify whether or not, the following set of vectors is linearly independent? {(4, 2, -1), (3, -6, -5)} Ans. The vectors are linearly independent. Problem Verify whether or not, the following set of vectors is linearly independent? {(1, 1, 1), (0, 2, 3), (1, -2, 3)} Ans: The set of vectors is linearly independent. Problem 5 Verify whether or not, the following set of vectors is linearly independent? {(1, 2, 3), (1, 1, 1) , (1, 0, 1)} Ans: linearly independent. Problem 6 Verify whether or not, the following set of vectors is linearly independent? {(1, 3, -2), (2, -1, 4) , (1, -11, 14)} Ans: The set is not linearly independent Problem: Show that if a set of vectors are linearly independent then every subset is also linearly independent. Solution: Suppose that set of vectors {𝑎 , 𝑎 ,…, 𝑎 } are linearly independent. Then 𝜆 𝑎 + 𝜆 𝑎 + ⋯ + 𝜆 𝑎 = 0 implies that all 𝜆 are zero. Consider the subset {𝑎 , 𝑎 , 𝑎 }. If these vectors are linearly dependent then 𝜆 𝑎 + 𝜆 𝑎 + 𝜆 𝑎 = 0 implies there exists say𝜆 ≠ 0. But this contradicts that {𝑎 , 𝑎 ,…, 𝑎 } are linearly independent. The proof is complete. Exercises: 1. Test whether or not the following set of vectors is linearly independent? i) {(2, 2, 3), (-1, -2, 1), (0, 1, 0)} ii) {(4, 2, 1), (2, -6, -5)} iii) {(4, 2, 1), (2, -6, -5) , (1, -2, 3) , (1,-1,2)} 2. Show that if a set of vectors are linearly dependent then every superset is also linearly dependent. 3. Which of the following set of vectors forms a basis for E3 ? Express (3,1,2) as linear combination of basis vectors. i) {(2, 2, 3), (-1, -2, 1) , (0, 1, 0)} ii) {(4, 2, 1), (2, -6, -5)} iii) {(4, 2, 1), (2, -6, -5) , (1, -2, 3) , (1,-1,2)} Exercises 1. Prove that the set A 2, 1,0 , 3,5,1 , 1,1, 2 forms a basis for 3 and express 2, 4,5 in terms of elements of A. 2. Check whether the following set of vectors form a basis for 3. i B 2,1, 4 , 1, 1, 2 , 3,1, 2 ii C 1, 2,1 , 2,1, 0 , 1, 1, 2 3. Let S = {vi / 1 i n} is a subset of vector space V. If vj linear combination of its preceding ones, then prove that L({v1, v2,..., vj-1, vj+1,..., vn}) = L(S). 4. Prove that for a vector space V over a field K, if vi V, 1 i n, then either they are linearly independent or some vk is a linear combination of the preceding ones v1, v2,... , vk-1.
