Question

Proof:

Let S ⊆ V be a subset of a vector space V over F. We have that S is linearly dependent if and only if there exist vectors v1, v2, . . . , vn ∈ S such that vi is a linear combination of v1, v2, . . . , vi−1, vi+1, . . . , vn for some 1 ≤ i ≤ n.

Answer 1

Let us first assume that there are vectors v₁, v₂, . . . , v_n ∈ S such that some v_i ,1 ≤ i ≤ n, is a linear combination of v₁, v₂, .,v_i-1,v_i+1, . . , v_n, then there exist scalars a₁,a₂,…,a_n(not all zero if v_i ≠0 ), such that v_i = a₁v₁+a₂v₂+…+a_i-1v_i-1 +a_i+1v_i+1 + ,…+a_nv_n . Then a₁v₁+a₂v₂+…+a_i-1v_i-1 +a_i+1v_i+1 + ,…+a_nv_n –v_i = 0. This means that a linear combination of v₁, v₂, . . . , v_n equals zero without the scalar coefficients of these vectors being all zero. Hence, by the definition of linear dependence, the vectors v₁, v₂, . . . , v_n ∈ S are linearly dependent. Therefore, the set S is also linearly dependent.

Now, let us assume that S is a linearly dependent set. This implies that there are vectors v₁, v₂, . . . , v_n ∈ S such that a linear combination of v₁, v₂, . . . , v_n, say a₁v₁+a₂v₂+…+a_nv_n = 0, without the scalar coefficients of these vectors being all zero. Further, if a_i≠ 0, then v_i = (1/a_i)( a₁v₁+a₂v₂+…+a_i-1v_i-1 +a_i+1v_i+1 + ,…+a_nv_n).Thus, v_i is a linear combination of v₁, v₂, .,v_i-1,v_i+1, . . , v_n.