Proving the Equivalence of Linear Independence and Basis for the Span of Vectors in R^n

Understanding the relationship between linear independence and the basis of a vector space is fundamental in linear algebra. Specifically, we aim to prove that any set of vectors in (mathbb{R}^n) forms a basis for its span if and only if the set is linearly independent. This article will explore the two main parts of this proof in detail.

Part 1: Linear Independence Implies Basis

Let (S {mathbf{v_1}, mathbf{v_2}, ldots, mathbf{v_k}}) be a set of vectors in (mathbb{R}^n) that is linearly independent. The span of (S), denoted as (text{span}(S)), is the set of all linear combinations of the vectors in (S).

A set of vectors is called a basis for a vector space if it is both linearly independent and spans the vector space. Since (S) is linearly independent, any vector in (text{span}(S)) can be expressed as a linear combination of the vectors in (S). This implies that (S) spans its own span. Therefore, (S) forms a basis for (text{span}(S)).

Part 2: Basis Implies Linear Independence

Assume (S) is a basis for its span, meaning (S) spans (text{span}(S)) and is a basis for it. If (S) spans (text{span}(S)), then any vector in (text{span}(S)) can be written as a linear combination of the vectors in (S).

Suppose (S) is not linearly independent. This means there exists at least one non-trivial linear combination of the vectors in (S) that equals the zero vector: n[c_1 mathbf{v_1} c_2 mathbf{v_2} ldots c_k mathbf{v_k} mathbf{0},]nwith at least one (c_i eq 0).

If such a relation exists, you can express one vector, say (mathbf{v_j}), in terms of the others: n[mathbf{v_j} -frac{c_1}{c_j} mathbf{v_1} - ldots - frac{c_{j-1}}{c_j} mathbf{v_{j-1}} - frac{c_{k-1}}{c_j} mathbf{v_{k-1}} - ldots - frac{c_k}{c_j} mathbf{v_k},]nwhich shows that (mathbf{v_j}) can be expressed as a linear combination of the other vectors in (S). This contradicts the assumption that (S) is a basis since a basis must be linearly independent.

Therefore, (S) must be linearly independent.

Conclusion

We have demonstrated that:

if a set of vectors is linearly independent, then it forms a basis for its span. if a set of vectors forms a basis for its span, then it is linearly independent.

Thus, we conclude that a set of vectors in (mathbb{R}^n) forms a basis for its span if and only if the set is linearly independent.

This equivalence is a cornerstone of linear algebra, providing a clear understanding of how vector independence and spanning sets are interconnected. For further reading, consider exploring related topics such as vector spaces, linear transformations, and matrix representations.