This experiment aims to help students understand the key ideas in linear algebra, namely linear independence, basis, dimension, and coordinates. These concepts are essential while working with vector spaces. Students will learn how to determine whether vectors are linearly independent, how a basis can be formed and how to express any vector in terms of a chosen basis providing coordinates. Here V denotes a vector space over R or C.

1. Linearly independence:

1.1. Linearly dependent set:

Let ϕ ≠S⊆V. If x∈S be such that x is a linear combination of some other elements of S, then S is said to be linearly dependent. In other words, if S is linearly dependent, then for some x∈S, there exists y₁, y₂, …,y_n∈S, which are different from x such that x=α₁y₁+α₂y₂+ …+α_ny_n, for some α₁, α₂, ..., α_n∈F. Notice that in this case, (-1)x+α₁y₁+α₂y₂+ …+ α_ny_n=0, i.e. there exists a linear combination of elements of S which equals zero, but not all coefficients are zero. Any set containing zero vector is linearly dependent.

1.2. Linearly independent set:

Let ϕ≠S⊆V. Then S is said to be a linearly independent set if it is not linearly dependent. The non-empty set ϕ is defined to be linearly independent.
To show that S={a, b}⊆V is linearly independent, one needs to show that b≠αa and a≠βb, for any scalars α and β. Similarly, to show that S={a, b, c}⊆V is linearly independent one needs to show that none of a, b and c is a linear combination of the other two elements.

General method to show linear independence is provided in the proposition given below.

1.3. Proposition:

Let ϕ≠S⊆V. Then S is linearly independent if and only if [α₁x₁+ α₂x₂+ …+α_nx_n=0 ⇒ α_i=0, for all i=1, 2, …, n; where α₁, α₂, ..., α_n∈F, x₁, x₂, …, x_n∈S].

Proof: Sufficient part:

Let α_i=0, for all i=1, 2, …, n; whenever α₁x₁+ α₂x₂+ …+α_nx_n=0, where α₁, α₂, ..., α_n∈F, x₁, x₂, …, x_n∈S. To the contrary, let S be linearly dependent. By definition of a linearly dependent set, there exists x∈S, such that x=α₁y₁+ α₂y₂+ …+ α_ny_n, for some α₁, α₂, ..., α_n∈F, y₁, y₂, …, y_n∈S. Thus (-1)x+α₁y₁+α₂y₂+ …+ α_ny_n=0. By hypothesis, -1=0. This is a contradiction.

Necessary part:

Let S be linearly independent. To the contrary, let α₁x₁+ α₂x₂+ …+ α_nx_n=0 and α_i≠0, for some i=1, 2, …, n. Clearly -α_ix_i=α₁x₁+ α₂x₂+ …+ α_i-1x_i-1+α_i+1x_i+1+…+ α_nx_n. Hence x _i=-α_i^-1(α₁x₁+α₂x₂+ …+ α_i-1x_i-1+α_i+1x_i+1+…+ α_nx_n). Note that α_i^-1 exists because α_i≠0. Hence S is linearly dependent, a contradiction.

1.4. Examples-I:

Consider R² be the vector space over R, where S⊆R².

(i) S={(0, 1), (1, 2), (2, 7)} is linearly dependent.
Justification: Clearly, (2, 7) is a linear combination of (0, 1), (1, 2) as given below: (2, 7)=2(1, 2)+3(0, 1). Thus, S is linearly dependent. Notice that 2(1, 2)-3(0, 1)+1(2, 7)=0, i.e. a linear combination of elements of S is zero, but all the coefficients are not zero.
Remark. a(0, 1)+b(1, 2)+c(2, 7)=0 ⇒ (b+2c,a+2b+7c)=0 implies that b+2c=0 and a+2b+7c=0 which does not imply that a=b=c=0. Hence it does not determine whether S is linearly dependent or independent. It only gives a clue.

(ii) S={(1, 2),(1, 0)} is linearly independent.
Justification: a(1, 2)+b(1, 0)=(0, 0) ⇒ (a, 2b)+(b, 0)=(0, 0) ⇒ (a+b, 2b)=(0, 0). Thus a=0, b=0. Hence, both the coefficients are zero therefore, S is linearly independent.

1.5. Examples-II:

(i) Consider the vector space R³ over R. Then S={(1, 0, 0), (0, 1, 0), (0, 0, 1)} is linearly independent. Justification: Let α(1, 0, 0)+β(0, 1, 0)+γ(0, 0, 1)=0; for α, β, γ∈R. By solving this we get α=0, β=0, γ=0 which implies by definition, that S is linearly independent.
(ii.) Consider the vector space P₂(x) over R. Then S={1, x, x²+1} is linearly independent. Justification: Let α(1)+β(x)+γ(x2+1)=0; for α, β, γ∈R. By solving this we get α=0, β=0, γ=0 which implies by definition, that S is linearly independent.

1.6. Properties of linearly independent andb linearly dependent sets:

(i) Any set containing the zero vector is linearly dependent. In particular, {0} is linearly dependent.
(ii) Singleton set containing a non-zero vector is linearly independent.
(iii) Subset of a linearly independent is linearly independent.
(iv) Superset of a linearly dependent set is linearly dependent.

2. Basis:

A non-empty subset S of V is said to be a basis if S is a linearly independent set and spans V.

2.1. Examples:

1. Let S be the linearly independent set as given in Example 5 (i). It can be seen that S spans R³. Hence S is a basis for R³.
2. Let S be the linearly independent set as given in Example 5 (ii). It can be seen that S spans P₂(x). Hence S is a basis for P₂(x).
   

   3. Dimension:

   Let V have a basis consisting of finitely many elements. Then the number of elements in the basis of V is called the dimension of the vector space V and is denoted by Dim. The dimension of {0} is defined to be zero as it is defined to be generated by ϕ.
   

3.1. Examples:

1. In Example 5 (i) the Dim of S is 3.
   
2. In Example 5 (ii) the Dim of S is 3.
   

   3.2. Properties of basis and dimension:

3. Let V have a finite basis. Then every basis for V contains the same number of vectors.
4. If a basis of V has n elements, then any subset of V having n-1 elements does not span V.
5. If a basis has n elements, then any subset of V having n+1 elements is linearly dependent.
6. Let B be a subset of V. Then the following are equivalent.
     a. B is basis.
     b. B is a minimal generating set, that is no proper subset of B can generate V.
     c. B is a maximal linearly independent set.
   

   4. Co-ordinates:

   Let V be a vector space and x∈V and let B={ e₁, e₂} be a basis. Then x=αe₁+βe₂, for some α, β∈F. These scalars α and β are called the co-ordinates of x w.r.t. the basis {e₁, e₂}.
   

4.1. Examples:

Let R² be the vector space over R.

1. Consider a basis B={(1, 1), (1, 0)} of the vector space R over R. Then (2, 3)∈R² can be written as (2, 3)=α(1, 1)+β(1, 0). This implies that α=3 and β=-1 Thus co-ordinates of (2, 3) w.r.t. the basis B are 3, -1.
2. If B={e₁, e₂} is a basis of the vector space R² over R, then
   (i) The co-ordinates of e₁ w.r.t. the basis B are 1, 0 since e₁=1.e₁+0.e₂.
   (ii) The co-ordinates of e₂ w.r.t. the basis B are 0, 1 since e₂=0.e₁+1.e₂.