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,………………………..,αm}
and B={β1,β2,…………………………,βm}
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,……………………,αm
∴
The set {β1,α1,α2,……………………,αm}
is linearly dependent.
∴
αi which is linear combination of its proceeding vectors β1,α1,α2,…………………..αi-1
∵ L(A)=V
∴
each vector in V is a linear combination of
α1,α2,…………,αi-1,αi,αi+1,…………….αm
and αi
is a linear combination of β1,α1,α2,…………..,αi-1
thus
each vector
in V is a linear combination of
β1,α1,α2,……..,αi-1,αi+1,………..αm
The set
S’={β1,α1,α2,…………..,αi-1,αi+1,…………..αm}
spans V.
If β∊V
⇒
β is a linear combination of β1,α1,α2,………..αi-1,αi+1,…………,αm.
∴ The set {β1,α1,α2,…………,αi-1,αi+1,…………αm}
is linearly dependent.
∵ ∃
αi such that
αi is a linear combination of β1,β2,…………..,α1,α2,.......,αi-1
∵
each vector of V is a linear combination of
β1,α1,α2,…………,αi-1,αi+1,……..αj-1,αj,αj-1,………………αm
and αj is linear combination of β1,β2,α1,α2,……………,αi-1
∴ The set S’={β1,α1,α2,……….,αi-1,αi+1,………..,αj-1,αj+1,……αm}
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
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