Linear Algebra: UNIT I: Vector Spaces
Vector Spaces - Theorem Part 8
Important Theorems for Engineering Maths or Mathematics - Vector Spaces - Theorem Part 8
Vector Spaces - Theorems Part 8
Theorem 25
Let V be a vector space
with dimension n.
(a) Any finite
generating set for V contains atleast n vectors, and a generating set for V
that contains exactly n vectors is a basis for V.
(b) Any linearly independent subset of V that
contains exactly n vectors is a basis for V.
(c) Every linearly
independent subset of V can be extended to a basis for V.
Proof:
Let
B be a basis for V.
(a)
Let G be a finite generating set for V. Then the subset H of G is a basis for
V. Therefore, H contains exactly n vectors. Since a subset of G contains n
vectors, G must contain atleast n vectors. If G contain exactly n vectors, then
H= G so that G is a basis for V.
(b)
Let L be a linearly independent subset of V containing exactly n vectors. By
replacement theorem, there is a subset H of B containing n−n=0 vectors such
that L ∪ H generates V. Thus H=
ϕ and L generates V. Since L is also linearly independent, L is a basis for V.
(c)
If L is a linearly independent subset of V containing m vectors, then there is
a subset H of B containing exactly n−m vectors such that L ∪ H generates V. Now L
∪ H contains atmost n
vectors, L ∪
H contains exactly n vectors and that L ∪
H is a basis for V.
Linear Algebra: UNIT I: Vector Spaces : Tag: maths, mathematics : - Vector Spaces - Theorem Part 8
Linear Algebra: UNIT I: Vector Spaces
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