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INTRODUCTION OF VECTOR SPACES. DR. PRASANTA SINHA DEPARTMENT OF MATHEMATICS For B.Sc Part I.
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INTRODUCTION OF VECTOR SPACES DR. PRASANTA SINHA DEPARTMENT OF MATHEMATICS For B.Sc Part I
The study of vectors relative to addition and multiplication by real numbers can be generalized in two directions. Firstly it is not necessary for us to restrict our study to ordered pairs or ordered triples, instead one may take ordered triples. Secondly we do not need confine ourselves to co-ordinates being real numbers. The co-ordinates may be taken from any field or more generally from any division ring.
In Analytical Geometry and Mechanics, one come across the concept of a vector as a directed magnitude, we do not need not recall that definition here. From algebraic point of view the vectors so defined have following salient features. • In three dimensional Euclidean space, a vector v is determined uniquely by its three components (ξ, η, λ ) (relative to a definite coordinate system) and conversely given any order triple (ξ, η, λ ) of real numbers there exists a vector having ξ, η, λ as its coordinates. Similarly in the two dimensional Euclidean space, a vector has two components ad is determined uniquely by its co-ordinates. Futher, the real numbers are called scalars.
There are three fundamental operations , namely addition of vectors, the multiplications of a vector by a scalars and the scalar product of vectors. Thus if V = (α, β, γ ) and V/ = (α/, β/, γ/ ) are two vectors in R3, the three dimensional Euclidean space, c is any scalar, then we define V + V/ = (α + α/, β + β/ , γ + γ/) and cV = (cα, cβ, cγ ) and scalar product (V, V/ ) = (αα/, ββ/, γ γ/ ). First two operations on vectors play very significant part in geometry, we endeavour to give a treatment of vectors in a more general settings.
DEFINITION • Definition 1.1 A system (V,F, +, .) where V is a non void set, F is a field, + is a binary composition on V and . Is a mapping of F x V into which for each element v ϵ V, α ϵ F determines an element α.v ϵ V is called a vector space over a field F if it satisfies the following conditionsV-1) (V, +) is an abelian group. • V-2) For all x,y ϵ V, α,β ϵ F we have • (i) α.(x+y) = α.x + α.y • (ii) (α + β).x = α.x+β.x • (iii) (αβ).x = α.(β.x) • (iv) 1.x = x • Every element of V is called a vector, every member of F is called a scalar.
Example 1. For any field F, let V = {(α1, α2) : αi ϵ F }. Then V is a vector space over a field F under vector addition and multiplication of a vector by a scalar given by (α1, α2) + (β 1, β2) =( α1 + β 1, α2 + β2) α (α1, α2) =( α α1, α α2). Example 2. Let V be the additive group of polynomials in one variable over a field F. For any α ϵ F and f(x) = α0 +α1x+ α2x2 +……………. + αnxn ϵ F[x] define αf(x) = αα0 + αα1x+ αα2x2 +……………. + ααnxn Then V becomes a vector space over F. Example 3. Let V be a set of all continuous real-valued continuous function defined on the closed interval [0,1]. For any f, g ϵ V and α ϵ R, define f +g and αf by (f+g )(x) = f(x) +g(x) (αf)(x) = αf(x) for all x in [0,1] Under these operations, V becomes a vector space over R. Example 4. For any field F and any positive integer n, the set of all n-tuples (α1, α2, ………. , αn) of elementsof F, is avector space over F under addition and multiplication of a vector by a scalar given by (α1, α2, ………. , αn) + (β 1, β2,…………., β n) =( α1 + β 1, α2 + β2,…………., αn+ β n) α (α1, α2, ………. , αn) =( α α1, α α2, …………. , α αn). This vector space is usually denoted by F(n) or by Fn.
NOTATION 1.1 : VF will denote the vector space VF over the field F. Remark 1.For the sake of convenience, in future, we shall write αv in place of α.v. Remark 1.2. We give below an example to show that the condition V-2 (iv) viz. 1.x =x ∀ x ϵ V is independent of the remaining conditions V-1 and V-2 (i), ()ii , (iii). Example 1.5. Consider V = {(α, β, γ) : α, β, γ ϵ R} Define (α, β, γ) + (α/, β/, γ/) = (α + α/ , β+ β/, γ + γ/) λ (α, β, γ) = (λα, λβ, 0) ∀ α, β, γ, α/, β/, γ/ , λ ϵ R, It can be easily verified that all the condition V-1, V-2 (i), (ii), and (iii) are satisfied Now (3,2,6) ϵ V but 1(3,2,6) = (3,2,0) ≠ (3,2,6). Hence V is not a vector space over R.
Lemma1. 6. Let V be a vector space over a field F and let 0V and 0F be additive identities of V and F respectively, the the following hold. • r 0V = 0V ∀ rϵ F • 0F v = 0V ∀ v ϵ V • –v = (-1) v ∀ v ϵ V • (-α)v = α (-v) = - ( αv) • If αv = 0V then either α = 0F or v = 0V. • NOTE 1.1. In future we shall use the same symbol 0 and O for the zero vector 0V and for the zero scalar 0F, for any vector vF. it should be clear to the readers from the context for which thing 0 or O is being used at a particular occasion.
SUBSPACES • Definition 1.4 a non void sub set W of a vector space of VF is said to be a subspace of VF if W is a vector space itself over the same field . The equivalent description of a subspace is given below • S-1. For any a,b ϵ W a+b ϵ W • S-2. For any a ϵ W and r ϵ F, ra ϵ W. • It is immediate from the definition that (0)and V are both subspaces of V these are trivial subspace of V. Any subspace W of VF which is different from(0) and V is called a proper subspace of V.
Lemma 1.2. If W is a subspace of a vector space VF, then W is a subgroup of (V, +) and W is vector space over F under induced addition and multiplication of vectors by scalars. Proof. For any x, y ϵ W, we have (-1)y ϵ W by S-2; so by S-1, x + (-1)y ϵ W → x – y ϵ W. This proves that W is a subgroup of (V,+). The property V-2 in definition 1.1 is satisfied by elements of W, since W ⊆ V and the same property holds for V and also the composition in W are all induced by those in V. Hence W is a vector space over F under induced addition and the multiplication of vector by scalars.
Examples • Example1. 6. Consider R3 = {(a1, a2, a3) : aiϵ R} Let W1 = {(0, a2, a3) : aiϵ R } W2 = {(a1, a2, a3) : aiϵ R and a1 + a2, + a3 =0}. It can be easily seen that W1 and W2 both satisfy the condition which define a subspace. Consequently W1 and W2 are both subspaces of R3 • Example 1.7. Let C be the set of all complex numbers, C is a vector space over R, the field of real numbers. W = {ib : b ϵ R , i= √ (-1)} is a subspace of C.
Theorem 1.1. A non void subset W of a vector space VF is a subspaceof VF iff it satisfies the following condition: a, b ϵ F; x, y ϵ W impliesax +by ϵ W. Proof. It is immediate. Theorem 1.2. Intersection of any family of subspaces of a vector space is a subspace. Proof. It is quite obvious.
Question: Does the above result hold good for finite union of subspaces? Answer: Unfortunately the answer is in negation. in fact, it does not hold for union any two subspaces, which follows from the following example. Example 1.8. Consider R2 is a vector space over the field of real numbers R and two subspaces W1 = {(a,0) : a ϵ R } W2 = {(0,b) : b ϵ R } are two subspaces of R2. Let W = W1 U W2. Now (1,0) ϵ W1 , (0,1) ϵ W2, but (1,1) is neither in W1 W2. So, that, (1,1)= (1,0) +(0,1) ∉ W = W1 U W2. Consequently, W1W2 is not a subspace of R2.
Definition 1.5. Let X be subset of a vector space V, then a subspace W of V is said to be spanned by or generated by X if (i) X ⊆ W (ii)For any subspace W/ of V , X ⊆ W/ implies W ⊆ W/ . i.e., smallestsubspace which generates X. Notation 1.2 . <X> will denote the subspace spanned by X. Further if X consists of elements x1, x2,…. xn, then we simply write <X> = < x1, x2,…. xn>.
Theorem 1.3. For any nonvoid subset X = {x1, x2,…. xn} the subspace spanned by or generated by X is the set of all of all vectors of the form a1x1 + a2x2 +…….+ anxn, where ai ϵ F, for i=1,2,…,n. Proof. Definition 1.5. For any finite number of vectors x1, x2,…. xn in a vector space V and the scalar a1, a2,…., an in F, the vectora1x1 + a2x2 +…….+ anxn is called a linear combination of x1, x2,…. xn. Definition 1.6. For any two subspaces W1 andW2 of a vector space V, their sum W1 + W2 is the set given by {w1 + w2 : w1 ϵ W1 , w2 ϵ W2}
Theorem 1.3. For any two subspaces W1 andW2 of a vector space V , W1 + W2 is a subspace of V spanned by W1 U W2. Proof. Firstly we show that W1 + W2 is a subspace of V. Clearly 0 = 0+0 ϵ W1 + W2 So, W1 + W2 ≠ φ . Let x, y ϵ W1 + W2 , a, b ϵ F . Then x = x = w1 +w2, y = w3 +w4, w1, w3 ϵ W1 , w2, w4 ϵ W2. Thus ax + by = (a w1 + b w3) + (a w2 + b w4) ϵ W1 + W2. Hence, W1 + W2 is a subspace of V. Now we show that W1 + W2 is spanned by W1U W2. For any w1 ϵ W1 as 0 ϵ W2 we have w1 = w1 + o ϵ W1 + W2 so, we have W1 ⊆ W1 + W2, Similarly, we can show that W2 ⊆ W1 + W2, So, W1 U W2 ⊆ W1 + W2. Let W be any subspace of V containing W1 U W2. Then given w1 ϵ W1, w2 ϵ W2, we have w1, w2 ϵ W1 U W2 ⇛ w1, w2 ϵ W ⇛ w1 +w2 ϵ W, since V is closed under vector addition, it implies that W1 + W2 ⊆ W.
Example 1.9. For any field, consider the vector space F(3) . e1 = (1,0, 0), e2 = (0,1, 0), e3 = (0,0, 1), are in (1,0, 0),. Then x = a1(1,0, 0) + a2(0,1, 0) + a3(0,0, 1) = a1e1 + a2 e2+ a3e3. Hence every member of F(3) is a linear combination of e1, e2, e3. Consequently, { e1, e2, e3} spans F(3). In general if we consider F(n) and define e1 = (1,0,….., 0), e2 = (0,1, 0,….,0), ……….en = (0,0,…, 1), then F(n) is spanned by the n vectors e1, e2,……,en. Example 1.10. Consider R(3) , x1 = (1,1,0), x2 = (1,2,0) ϵ R3. Let W be the subspace spanned by { x1 , x2}. Then x1 , x2 ϵ W ⇛ x2 – x1 ϵ W ⇛ (0,1,0) ϵ W ⇛ x1 – (0.1.0) ϵ W. Thus, e1 = (1,0, 0), e2 = (0,1, 0) ϵ W ⇛ for every a, b ϵ R, a e1 + b e2 = (a,b,0) ϵ W. Let W/ = {(a,b,0): a,b ϵ R}. W/ is a subspace of R3 and as seen above W/ ⊆ W. . Now x1 = e1 + e2 ϵ W/, x2 = e1 + 2e2 ϵ W/ since e1 , e2 ϵ W/. Thus the subspace W = < x1 , x2> ⊆ W/. Hence W =W/ .
Linear Independence • Definition 1.6. Let x1, x2,…. xnbe a finite number of members (not necessarily all distinct) of avector space V. these vectors are said to be linearly dependent (L.D) if for some a1, a2,……, an ϵ F with at least one of them non zero, a1 x1 + a2 x2 +……+ anxn = 0. These vectors are said to be linearly Independent (L.I) if for all ai ϵ F (I = 1,2,….,n), a1 x1 + a2 x2 +……+ anxn = 0 ⇛ ai =0 for all i. • Example 1.8. Consider V = {(a,b) : a,b ϵ F} where F is a field. The vectors e1 = (1, 0), e2 = (0,1) in V are L.I, since for any r,s ϵ F, r e1 + se2 = 0 ⇛ (r,0) + (0,s) = (0,0) ⇛ r = 0 , s = 0. Further notice that any vector (a, b) ϵ V can be expressed as ae1 + be2 as a linear combination of e1 ande2.
Theorem 1.4. If v1, v2,……….. vn are L.I. in a vector space V, then each element of the subspace W spanned by them is expressible uniquely as a linear combination of v1, v2,……….. vn. Theorem 1.5. Let u1, u2,………, un be any n linearly independent vectors in a vector space V. Any n+1 vectors v1,v2,……….. vn, vn+1, each of which is a linear combination of u1, u2,………, un are L.D. Corollary 1.6. If v1, v2,……….. vn is a L.I. subset of a vector space V, then any subset W of V having more than n vectors each of which can be expressed as a linear combination of v1, v2,……….. vnmust be linearly dependent.
BASE • We conclude section with the concept of Base. • Definition 1.7. A subset B of a vector space V is called a basis of V if (i) B is linearly independent and (ii) B spans V. • Definition 1.8. A vector space V is finitely generated if it has a finite subset which spans V itself. • Example 1.9. For any field F, consider F(n), the vector space consisting of all n-tuples of the type (α1, α2,. ……, αn) over F. Let e1 = (1,0,0,……..,0), e2 = (0,1,0,……..,0),……..en = (0,0,0,……..,1). Then it can be easily seen that e1, e2,…….,en are L.I. and they span F(n). Thus { e1, e2,…….,en } is a basis of F(n) and F(n) is finitely generated.