Vector Spaces: Theorems Part 6

Vector Spaces

Theorems Part 6

Theorem 15

Let V be a vector space and S  V is linearly independent set. Prove that every vector in the span of S can be uniquely written as linear combination of vectors of S.

Proof:

Let v ∈ span (s).

Then v = α₁v₁ + α₂v₂ + …. + α_nv_n = ⁿΣ_i=1 α_iv_i

Suppose that v= β₁v₁ + β₂v₂ + …. + β_nv_n = ⁿΣ_i=1 β_iv_i

Now 0 = v−v

 = ⁿΣ_i=1 α_iv_i − ⁿΣ_i=1 β_iv_i

 = (α₁ − β₁)v₁ + (α₂− β₂)v₂ +...+ (α_nβ_n)v_n = 0

where v₁, v₂, …. v_n ∈ S

Since S is linearly independent, a_i – β_i = 0 for all i.

a_i = β_i

Theorem 16

Let S be a linearly independent subset of a vector space V, and let v be a vector in V that is not in S. Prove that S ∪ {V} is linearly dependent if and only if v ∈ L(S).

Proof:

If S ∪ {v} is linearly dependent, then there are vectors u₁, u₂, u₃ ... u_n in S ∪ {v} such that a₁u₁+ а₂u₂ + а₃u₃ + …. +a_nu_n = 0 for some non zero scalars a₁, a₂, a₃ ... a_n. Because S is linearly independent, one of the u_i's say u₁ equals v.

Thus a₁v + a₂u₂ + a₃u₃ + … +a_nu_n = 0 and so

 v = a₁^‒1 [ − α₂u₂ − α₃u₃ − α₄u₄ …..  – α_nu_n ]

 v = − [a₁⁻¹a₂]u₂ − [a₁⁻¹a₃]u₃− [a₁⁻¹a₄]u₄ …..  − [a₁⁻¹a_n]u_n

Since v is a linear combination of u₂, u₃ ….. u_n which are in S, we have v ∈ L(s).

Conversely, let v ∈ L(S). Then there exist vectors v₁, v₂, v₃, ... v_m in S and scalars b₁, b₂, b₃ ... b_m such that

 v = b₁v₁ + b₂v₂ + b₃v₃ + ... + b_mV_m.

Hence 0=b₁v₁ + b₂v₂ + b₃v₃ + ... + b_mv_m + (− 1) v.

Since v ≠ v; for i = 1, 2, 3,... m the coefficient of v in this linear combination is non zero, and so the set

{v₁, v₂, v₃ ... v_m, v} is linearly dependent.

  { S ∪ {v} } is linearly dependent.

Theorem 17

A set S={v₁, v₂, v₃, ... v_k }; k ≥2 is linearly dependent if and only if at least one of the vectors v_j can be written as a linear combination of the other vectors in S.

Proof:

Let S be a linearly dependent set. Then there exist scalars endent set Then there exist s c₁, c₂, c₃ ... c_k (not all zero) such that

 c₁v₁ + c₂v₂ + c₃v₃ + …. + c_kv_k = 0. Because one of the coefficients must be non zero, no generality is lost by assuming c₁ ≠ 0. Then solving for v₁ as a linear combination of the other vector produces c₁v₁ = ‒ c₂v₂ ‒ c₃v₃ ‒ c₃v₃ ‒ …. ‒ c_kv_k

 v₁= ‒ (c₂/c₁)v₂ ‒ (c₃/c₁)v₃‒ (c₄/c₁)v_{4 ……} ‒ (c_k/c₁)v_k

Conversely, suppose the vector v₁ in S is a linear combination of the other vectors.

(ie) v₁ = c₂v₂ + c₃v₃ + …. + c_kv_k

Then the equation − v₁+ c₂v₂ + c₃v₃ + …. + c_kv_k = 0 has atleast one coefficient, −1, that is non zero, and conclude that S is linearly dependent.

Definition:

Minimal generating set: Let V be a vector space over F and S contained in V then S is called minimal generating set for V if

(i) L(S) = V.

(ii) No proper subset of S will generate.