Linear Algebra: UNIT I: Vector Spaces
Vector Spaces: Theorems Part 3
Important Theorems for Engineering Maths or Mathematics - Vector Spaces: Theorems Part 3
Vector Spaces
Theorems Part 3
Theorem 8
If S is a non−empty
subset of a vector space V then the set W consisting of all linear combinations
of elements of S is a subpsace of V. (ie) (W=L(S) 
V and subspace)
Proof:
Given:
V
is a vector space over F.
S
is a non−empty subset of V.
W
= all linear combination of elements of S.
W
= L(S)
To
prove W is a subspace of V.
S≠ϕ; Let x∈S,
then 0 = 0.x ∈
W
0 ∈
W
Let
x, y ∈ W ⇒ x+y and cx are also linear
combination of elements of S.
x + y ∈
W and cx ∈
W.
W is a
subspace of V.
(ie)
If x, y ∈ W then x, y are linear
combinations of elements of S. So there exists elements u1, u2,
…. un and w1, w2 ...wm in S, such
that x=a1u1+ a2u2 + ... + anun,
and y=b1w1 + b2w2+ … + bmwm for some
choice of scalars a1, a2, ... an and b1, b2, ... bm.
Now
x+y = a1u1
+ a2u2 + …. +anun + b1w1 + b2w2+ … +bmwm
and
cx = (ca1)u1 +
(ca2)u2 + …. +(can)un
are linear combinations of elements of S.
W is a
subspace of V.
Note
L(S)
is the smallest subspace of V containing S. In other words if W is a subspace
of V there exists S
W then L (s) W.
Any
subspace of V that contains S must also contains L (S).
Linear Algebra: UNIT I: Vector Spaces : Tag: maths, mathematics : - Vector Spaces: Theorems Part 3
Linear Algebra: UNIT I: Vector Spaces
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