Invarience Theorem

Statement:-  Any two bases of a finite dimensional vector space have same number of elements.

OR

The number of elements in a basis of a finite dimensional vector space is unique.

Proof:-  Let A={α₁,α₂,………………………..,α_m}

and              B={β₁,β₂,…………………………,β_m}

be two bases of a finite dimensional vector space V(F). Then

L(A)=V

If since β₁∊V

⇒          β is linear combination of α₁,α₂,……………………,α_m

∴   The set {β₁,α₁,α₂,……………………,α_m} is linearly dependent.

∴   α_i which is linear combination of its proceeding vectors  β₁,α₁,α₂,…………………..α_i-1

∵                              L(A)=V

∴   each vector in V is a linear combination of

α₁,α₂,…………,α_i-1,α_i,α_i+1,…………….α_m

and α_i is a linear combination of β₁,α₁,α₂,…………..,α_i-1 thus

each vector in V is a linear combination of

β₁,α₁,α₂,……..,α_i-1,α_i+1,………..α_m

The set S’={β₁,α₁,α₂,…………..,α_i-1,α_i+1,…………..α_m} spans V.

If β∊V

⇒  β is a linear combination of β₁,α₁,α₂,………..α_i-1,α_i+1,…………,α_m.

∴ The set {β₁,α₁,α₂,…………,α_i-1,α_i+1,…………α_m} is linearly dependent.

∵      ∃ α_i such that

α_i is a linear combination of β₁,β₂,…………..,α₁,α₂,.......,α_i-1

∵ each vector of V is a linear combination of

β₁,α₁,α₂,…………,α_i-1,α_i+1,……..α_j-1,α_j,α_j-1,………………α_m and α_j is linear combination of β₁,β₂,α₁,α₂,……………,α_i-1

∴         The set S’={β₁,α₁,α₂,……….,α_i-1,α_i+1,………..,α_j-1,α_j+1,……α_m} spans V.

In each steps consist of the exclusion 1.α type vector and inclusion of 1.β type vector and the resulting set generates V. All α type vectors can not be exhausted by β-type vector because in that case we obtain a spanning set which containing some β-type element only.

Thus a proper subset S generates V.

⇒ each vector in V is a linear combination of β-type vectors.

Consequently B will be linear combination which is contradiction.

∴                   m≮n

Similarly by changing of rules of bases we get

n≮m

Thus            m=n                                Proved