Correct Answer - Option 4 : S
1 is a basis for ℝ
2 but S
2 is not a basis for ℝ
2.
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,…, vp} in a vector space V is said to be linearly independent if the vector equation c1v1 + c2v2 +…+ cpvp = 0 has only one trivial solution c1 = 0, c2 = 0,…, cp = 0;
The set is said to be linearly dependent if there exists weights c1, c2,…, cp not all 0, such that c1v1 + c2v2 +…+ cpvp = 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.