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Short Notes on Linear Algebra 1 Vector Spaces, Independence

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Short Notes on Linear Algebra 1 Vector Spaces, Independence Short notes on Linear Algebra by Sanand D ||||||||||||||||||||||||||||||||||||||||- 1 Vector spaces, independence, dimension, basis, linear operators 1.1 Introduction Definition 1.1. Let V be a set whose elements are called vectors with an operation of vector addition (+) such that + is commutative and associative and the following properties are satisfied • f0g 2 V such that 0 + v = v 8v 2 V. • For all v 2 V, −v 2 V such that v + (−v) = 0. • For all v1; v2 2 V, v1 + v2 2 V. Let F be a field such that there is an operation of scalar multiplication (:) between elements of F and V satisfying the following properties • 1:v = v 8v 2 V, 1 2 F • (a1a2)v = a1(a2v), 8a1; a2 2 F • a(v + w) = av + aw, 8a 2 F and 8v; w 2 V • (a1 + a2)v = a1v + a2v, 8a1; a2 2 F. A set V satisfying the properties above is said to be a vector space over a field F. Example 1.2. Cn; Rn; Cn×m; Rn×m, solutions of homogeneous linear equations, solutions of homo- geneous odes, set of real/complex valued continuous/differentiable/analytic functions are examples of vector spaces. A set of polynomials with real/complex coefficients form a vector space over real/complex numbers. Definition 1.3. A subset W of V satisfying the properties above is called a subspace of V. Example 1.4. R; R2 form a subspace of R3. A set of differentiable functions form a subspace of a vector space of continuous functions. A set of complex polynomials of degree at most n form a subspace of C[x] (Geometrically, one can think of vector spaces as euclidean spaces and subspaces are planes passing through the origin.) Suppose v1; : : : ; vk 2 V. Then < v1; : : : ; vk > is the span of fv1; : : : ; vkg which is a collection of all linear combinations of fv1; : : : ; vkg. The span of any set forms a subspace. Definition 1.5. A set of non-zero vectors v1; : : : ; vk is said to be independent if α1v1+:::+αkvk = 0 implies that all αis (1 ≤ i ≤ k) are equal to zero. A set of vectors which is not independent is said to be dependent. (Geometrically if v depends on v1; : : : ; vk, then v lies in the space spanned by v1; : : : ; vk. Whereas if v is independent on v1; : : : ; vk, then v lies outside the space spanned by v1; : : : ; vk.) Definition 1.6. A maximal linearly independent set is called a basis. 1 Basis is not unique. By the definition of basis and linear independence, each v 2 V can be represented as a unique linear combination of basis vectors. Any linearly independent set can be extended to a basis by adjoining linearly independent vectors to the previous set. [ ] [ ] [ ] T T T Example 1.7. e1 = 1 0 ::: 0 ; e2 = 0 1 ::: 0 ; : : : ; en = 0 0 ::: 1 form a basis for Cn. This is called the standard basis. Definition 1.8. Dimension of a vector space is the cardinality of its basis. Example 1.9. Dimension of Cn over C is n. Dimension of C[x] over C is infinite with basis 1; x; : : : ; xn;:::. Dimension of a subspace of polynomials of degree at most n is n + 1 with basis 1; x; : : : ; xn. If V1 and V2 are two subspaces of V, then V1 +V2 and V1 \V2 are subspaces of V. By choosing a basis for V1 and extending to basis of V by taking linearly independent vectors in V2 which are not in V1, it follows that dim(V1 + V2) = dim(V1)+ dimV2− dim (V1 \ V2) Definition 1.10. If V1 + V2 = V and V1 \ V2 = f0g, then we say that V is a direct sum of V1 and V2. It is denoted by V = V1 ⊕ V2. Example 1.11. Cn = C ⊕ C ⊕ ::: ⊕ C i.e. direct sum of n copies of C. 1.2 Co-ordinates and linear maps on vector spaces Co-ordinates: Let fv1; : : : ; vng be a basis of V. Let v 2 V. Therefore, v can be expressed uniquely [as a linear combination] of vis, v = α1v1 + ::: + αnvn. Thus, w.r.t. this basis, v has co-ordinates T α1 α2 : : : αn . Matrix representation of linear operators: Let A : V ! V. A is said to be linear if for all v; w 2 V and c1; c2 2 F, A(c1v + c2w) = c1Av + c2Aw. Let fv1; : : : ; vng be a basis of V. Then for any v 2 V, Av = α1Av1 +:::+αnAvn. Thus, it is enough to define the action of A on basis vectors. 2 3 α1 6 7 [ ] 6 α2 7 6 7 Av = Av1 Av2 : : : Avn 6 : 7 : (1) 4 : 5 αn Let Avi = a1iv1 + ::: + anivn for 1 ≤ i ≤ n. Then, the matrix representation of A w.r.t. the basis above is 2 3 a11 a12 : : : a1n 6 7 6 a21 a22 : : : a2n 7 6 7 A = 6 :::::: 7 (2) 4 :::::: 5 an1 an2 : : : ann and, co-ordinates of Av w.r.t. the given basis are 2 3 2 3 a11 a12 : : : a1n α1 6 7 6 7 6 a21 a22 : : : a2n 7 6 α2 7 6 7 6 7 Av = 6 :::::: 7 6 : 7 : (3) 4 :::::: 5 4 : 5 an1 an2 : : : ann αn Definition 1.12. Kernel of an operator A : V ! W is the set of vectors v 2 V such that Av = 0. 2 Kernel of an operator A is denoted by ker(A) and it is a subspace of V. Image of A is set of all w 2 W such that w = Av for some v 2 V. Im(A) is also a subspace. When A is represented by a matrix w.r.t. some basis, then any vector in the image space of A is a linear combination of columns of A. Thus, Im(A) is spanned by columns of A. Definition 1.13. Rank of a matrix is equal to the number of its linearly independent columns. Consequently, it is equal to the dimension of the image space of A. Thus, Ax = b has a solution iff b lies in the column span of A iff the augmented matrix [A b] has the same rank as that of A. Note that A is onto , the rank of A is equal to the dimension of the codomain space and A is 1 − 1 , the dimension of ker(A) is equal to zero. It is 1 − 1 and onto , the dimension of ker(A) is equal to zero and the rank of A is equal to the dimension of the codomain space. (Note that Im(A) is a subspace of the codomain. If rank(A) = dimension of Im(A) is equal to the dimension of the codomain, then Im(A) = codomain space.) Theorem 1.14. (Rank-Nullity) Let A : V ! V where V is an n−dimensional vector space. Then rank(A)+dim(ker(A)) = n. Proof. Let v1; : : : ; vk for a basis for ker(A). Extend this set to a full basis of V say v1; : : : ; vn. Claim: fAvk+1; : : : ; Avng are linearly independent. Suppose not, therefore, αk+1Avk+1 +:::+αnAvn = 0. This implies that A(αk+1vk+1 +:::+αnvn) = 0, hence αk+1vk+1 +:::+αnvn 2 ker(A). Therefore, αk+1vk+1 +:::+αnvn = β1v1 +:::+βkvk. But this contradicts the linear independence of fv1; : : : ; vng. Therefore, fAvk+1; : : : ; Avng are linearly independent. Since rank (A) = dim(Im(A)), rank(A) = n − k and dim(ker(A)) = k. Suppose A : V ! W. Let fv1; : : : ; vng be a basis of V and fw1; : : : ; wmg be a basis for W. Then to find a matrix representation for A, it is enough to define Avi (1 ≤ i ≤ n). Avi = a1iw1 + ::: + amiwm (1 ≤ i ≤ n). Thus, A is represented by an m × n matrix A = [aij]. Rank- Nullity theorem holds for these linear maps as well where n is the dimension of the domain. Dual spaces: For a vector space V, its dual V∗ consists of all linear maps from V to F (the underlying field). If V is an n dimensional vector space whose elements are represented by column vectors w.r.t. some basis, then elements of V∗ are represented by 1 × n matrices which can be thought of as row vectors. Thus, the dual space of column vectors is row vectors and vice versa. f g V f ∗ ∗g V∗ ∗ For any basis v1; : : : ; vn of , there exists a dual basis v1; : : : ; vn of such that vi (vj) = 1 if ∗ 6 V ! W ∗ W∗ ! V∗ i = j and vi (vj) = 0 if i = j. A linear map A : induces a map A : . The action ∗ ∗ ∗ ∗ of A is defined as A w (v) := w (Av). Let A = [aij] be a matrix representation of A w.r.t. bases ∗ v1; : : : ; vn and w1; : : : ; wm of V and W respectively and [bij] be a representation of A w.r.t. dual ∗ ∗ ∗ W∗ ∗ bases. Consider the action of A on basis vectors w1; : : : ; wm of . Since A = [bij] is a matrix representation of A∗, ∗ ∗ ∗ ∗ A wi = (b1iv1 + ::: + bnivn) (4) ) ∗ ∗ ∗ ∗ A wi (vj) = (b1iv1 + ::: + bnivn)vj = bji: (5) Observe that since A∗w∗(v) = w∗(Av) from the definition of A∗, ∗ ∗ ∗ ∗ A wi (vj) = wi (Avj) = wi (a1jw1 + ::: + amjwm) = aij (6) (7) ∗ Thus, aij = bji and the matrix representation of induced map A is the transpose of A. Observe that (V∗)∗ = V i.e. the double dual of V is V itself for finite dimensional vector spaces. 3 2 Change of basis Let e1; : : : ; en be the standard basis of a vector space V. Let this be an old basis and let Aold be a matrix representation of a linear operator A : V ! V.
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