Question

Consider two subsets of ℝ² given as, S1 = {[1, -2], [3, 5]} and S2 = {[1, 1], [0, 0]}. Then,
1. S₁ is not a basis for ℝ² but S₂ is a basis for ℝ².
2. neither S₁ nor S₂ are bases for ℝ².
3. both S₁ and S₂ are bases ℝ².
4. S₁ is a basis for ℝ² but S₂ is not a basis for ℝ².

Answer 1

Correct Answer - Option 4 : S₁ is a basis for ℝ² but S₂ is not a basis for ℝ².

Concept:

Every basis for the vector space Rⁿ consists of n vectors.

Let S be a subset of a vector space Rⁿ, the basis is a linearly independent spanning set for Rⁿ.

A set of vectors {v₁, v₂,…, v_p} in a vector space V is said to be linearly independent if the vector equation c₁v₁ + c₂v₂ +…+ c_pv_p = 0 has only one trivial solution c₁ = 0, c₂ = 0,…, c_p = 0;

The set is said to be linearly dependent if there exists weights c₁, c₂,…, c_p not all 0, such that c₁v₁ + c₂v₂ +…+ c_pv_p = 0

Calculation:

Given S1 = {[1, -2], [3, 5]} and S2 = {[1, 1], [0, 0]} 

Let S₁ = (v₁, v₂) and the vector equation be av₁ + bv₂ = 0

S1 = {[1, -2], [3, 5]} ⇒ v1 = [1, -2], v2 = [3, 5];

⇒ a + 3b = 0, -2a + 5b =0 ⇒ a = 0, b = 0

Since, both the constants are zero, S1 will be linearly independent. 

Let S₂ = (v1, v2) and the vector equation be av1 + bv2 = 0

S2 = {[1, 1], [0, 0]} ⇒ v1 = [1, 1], v2 = [0, 0];

⇒ a = 0, a = 0 ⇒ a = 0 and b can be any real number

Since many trivial solutions, S2 will be linearly dependent.

∴ S1 is a basis for ℝ2 but S2 is not a basis for ℝ2.