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Definition of Vector Space

Vector Space​

  • EE to be a non-null set and KK to be a scalar set
  • Vectors: elements of the set EE
  • ++ - Internal composition law, E×E→EE\times E \to E
  • .. - External composition law, K×E→EK\times E\to E
  • (E,+,.E,+,.) is said vector space of the vector set EE over the scalar field (K,+,K,+,) if and olf if the ten vector space axioms are verified

Axioms​

  • EE is closed with respect to the internal composition law: ∀u,v∈E:u+v∈E∀u, v ∈ E : u + v ∈ E
  • EE is closed with respect to the external composition law: ∀u∈Eand∀λ∈K:λu∈E∀u ∈ E and ∀λ ∈ K : λu ∈ E
  • Commutativity for the internal composition law: ∀u,v∈E:u+v=v+u∀u, v ∈ E : u + v = v + u
  • Associativity for the internal composition law: ∀u,v,w∈E×E:u+(v+w)=(u+v)+w∀u, v, w ∈ E × E : u + (v + w) = (u + v) + w
  • Neutral element for the internal composition law: ∀u∈E:∃!o∈E∣u+o=u∀u ∈ E : ∃!o ∈ E|u + o = u
  • Opposite element for the internal composition law: ∀u∈E:∃!−u∈E∣u+−u=o∀u ∈ E : ∃!−u ∈ E|u + −u = o
  • Associativity for the external composition law: ∀u∈E∀u ∈ E and ∀λ,μ∈K:λ(μu)=(λμ)u=λμu∀λ, μ ∈ K : λ(μu) = (λμ) u = λμu
  • Distributivity 1: ∀u,v∈E∀u, v ∈ E and ∀λ∈K:λ(u+v)=λu+λv∀λ ∈ K : λ (u + v) = λu + λv
  • Distributivity 2: ∀u∈Eand∀λ,μ∈K:(λ+μ)u=λu+μu∀u ∈ E and ∀λ, μ ∈ K : (λ + μ)u = λu + μu
  • Neutral elements for the external composition law: ∀u∈E:∃!1∈K∣1u=u∀u ∈ E : ∃!1 ∈ K |1u = u

Vector Subspace​

(E,+,.E,+,.) be a vector space, U⊂EU \subset E and U≠∅U \ne ∅ The triple (E,+,.E,+,.) is a vector subspace of (E,+,.E,+,.) if (U,+,.U,+,.) is a vector space over the same field KK with respect to both the composition laws

Proposition shows that we do not need to prove all 10 axioms, just need to prove closure of the two composition laws.

Null vector in Vector Spaces​

(E,+,.E,+,.) be a vector space over a field KK. Every vector subspace (U,+,.U,+,.) of (E,+,.E,+,.) contains the null vector. For every vector space (E,+,.E,+,.), at least two vector subspace exist

Intersection and Sum Spaces

Intersection Spaces​

If (U,+,.U,+,.) and (V,+,.V,+,.) are two vector subspace of (E,+,.E,+,.), then (U∩V,+,.U\cap V,+,.) is a vector subspace of (E,+,.E,+,.)

Sum Space​

(U,+,.U,+,.) and (V,+,.V,+,.) be a vector space of (E,+,.E,+,.). Sum subset is a set S=U+VS=U+V defined as S=U+V={w∈E∣∃u∈U,v∈V∣w=u+v}S=U+V=\{ w\in E |\exists u \in U, v \in V|w=u+v \}

Direct Sum​

(U,+,.U,+,.) and (V,+,.V,+,.) be two vector subspaces of (E,+,.E,+,.). If U∩V={o}U\cap V = \{ o\} the subset sum S=U+VS=U+V is indicated as S=U⊕VS=U⊕V and named subset direct sum

Linear dependence in nn dimensions

Basic Definition​

  • Linear combination of the nn vectors by means of nn scalars is the vector λ1v1+λ2v2+...+λnvn\lambda_1v_1+\lambda_2v_2+...+\lambda_nv_n
  • Said to be linear dependent if the null vector o can be expressed as linear combination by means of the scalars λ1,λ2,...,λn≠0,0,...,0\lambda_1,\lambda_2,...,\lambda_n\ne 0,0,...,0
  • Vectors are linearly independent if the null vector oo can be expressed as linear combination only by means of the scalars 0,0,...,00,0,...,0
  • Are linearly dependent if and only if at least one of them can be expressed as linear combination of the others
  • Would then check them as you would do in matrices

Linear Span

Set containing the totality of all the possibly linear combinations of the vectors, by means of nn scalars. L(v1,v2,...,vn)={λ1v1+λ2v2+...+λnn∣λ1,λ2,...,λn∈K}L (v_1 , v_2 , . . . , v_n ) = \{ λ_1 v_1 + λ_2 v_2 + . . . + λ_n n |λ_1 , λ_2 , . . . , λ_n ∈ K \}

Properties​

  1. Span L(v1,v2,...,vn)L(v_1,v_2,...,v_n) with the composition laws in a vector subspace of (E,+,.E,+,.)
  2. s∈Ns\in \N with s<ns<n. If ss vectors are linearly independent while each of the remaing n−sn-s vectors is linear combination of the linearly independent ss vectors, then L(v1,v2,...,vn)=L(v1,v2,...,vs)L (v_1 , v_2 , . . . , v_n ) = L (v_1 , v_2 , . . . , v_s )
  3. L(v1,v2,...,vn)=L(v(σ)1,v(σ)2,...,v(σ)no)L (v_1 , v_2 , . . . , v_n ) = L (v_{(σ)1} , v_{(σ)2} , . . . , v_{(σ)no })