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 R^{n} consists of n vectors.

Let S be a subset of a vector space R^{n}, the **basis is a linearly independent spanning set** for R^{n}.

A set of vectors {v_{1}, v_{2},…, v_{p}} in a vector space V is said to be **linearly independent** if the vector equation c_{1}v_{1} + c_{2}v_{2 }+…+ c_{p}v_{p} = 0 has only one trivial solution **c**_{1} = 0, c_{2} = 0,…, c_{p} = 0;

The set is said to be **linearly dependent** if there exists weights **c**_{1}, c_{2},…, c_{p} not all 0, such that c_{1}v_{1} + c_{2}v_{2 }+…+ c_{p}v_{p} = 0

__Calculation:__

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

Let S_{1} = (v_{1}, v_{2}) and the vector equation be av_{1} + bv_{2} = 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 S_{2} = (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.**