Prove that If a vector space is of dimension n and a set of vectors spans V, then that set of vectors must be linearly independent.
To prove that if a vector space is of dimension n and a set of vectors spans V, then that set of vectors must be linearly independent, we can use the definition and properties of vector spaces, dimension, and linear independence.
1. Let's start by assuming that V is a vector space of dimension n, and S is a set of vectors that spans V.
2. Since S spans V, every vector in V can be expressed as a linear combination of the vectors in S.
3. Let's assume that S is not linearly independent, which means there exist scalars c₁, c₂, ..., cₙ (not all zero) such that c₁v₁ + c₂v₂ + ... + cₙvₙ = 0, where v₁, v₂, ..., vₙ are vectors in S.
4. Rearranging the equation, we have -c₁v₁ - c₂v₂ - ... - cₙvₙ = 0.
5. Since V is a vector space, it satisfies the properties of closure under scalar multiplication and vector addition. Therefore, -c₁v₁ - c₂v₂ - ... - cₙvₙ = 0 is also a linear combination of the vectors in S.
6. By combining multiple linear combinations, we have (-c₁)v₁ + (-c₂)v₂ + ... + (-cₙ)vₙ = 0.
7. Since the scalars -c₁, -c₂, ..., -cₙ are not all zero (as assumed in step 3), this implies that we can express the zero vector as a nontrivial linear combination of the vectors in S, contradicting the assumption that S spans V.
8. Therefore, our assumption that S is not linearly independent is incorrect, and we can conclude that if a vector space is of dimension n and a set of vectors spans V, then that set of vectors must be linearly independent.
This proof shows that if a set of vectors spans a vector space of dimension n, then it must be linearly independent.