# Consider two subsets of ℝ2 given as, S1 = {[1, -2], [3, 5]} and S2 = {[1, 1], [0, 0]}. Then,

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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.

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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,…, 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.