Statement:-There exists a basis for each finite dimensional vector space.

Proof:- Let V be vector space & let S={ α₁,α₂,……………..,α_m } be a finite subset of V such that L(S)=V. If S is linearly independent then it is a basis of V.

If S is linearly dependent then ∃ a vector α_i ∊ S

Which can be expressed as a linear combination of the proceeding vectors α₁,α₂,…………………..,α_i-1

If we remove this vector 𝛼_I from S and the remaining set we denoted by S₁ , then

                   S₁={α₁,α₂,…………….,α_i-1,α_i+1,…………….,α_m}

Which also generates V.

Since if α∊V

Therefore α can be expressed as a linear combination of element of S. Thus

∃            a₁,a₂,…………………….a_m such that

α=a₁α₁+a₂α₂+………..+a_i-1α_i-1+a_i+1α_i+1+………………….+a_mα_m …(1)

∵    α_i = linear combination of α₁,α₂,……………….,α_i-1

⇨   α_i=b₁α₁+b₂α₂+………………….+b_i-1α_i-1…………………………(2)

Putting α_I in (1)

α=a₁α₁+a₂α₂+……..+a_i-1α_i-1 + {a_ib₁α₁+…………..a_ib_i-1α_i-1}+a_i+1α_i+1 +………………+a_mα_m

α= (a₁+a_ib₁)α₁+(a₂+a_ib₂)α₂+…………..+(a_i-1+a_ib_i-1)α_i-1+…………… a_i+1α_i+1+………………………a_mα_m

α = linear combination of S₁

L(S₁)=V

If S₁ is linearly dependent, then it will be basis of V.

If S₁ is linearly dependent then proceeding as above we get a set of (m-2) vector which generates V. Continue this process after finite number steps we get a subset of S.

Which generate V and so that will be basis of a vector space V. At the most we may be left with a single generating V, which is clearly linearly independent and therefore it will be a basis.

Thus, a basis for each finite-dimensional vector space.

This theorem can also be expressed as if a finite set S of non-zero vectors spans a finite dimensional vector space. A subset of S which form a basis of V.