Linear Algebra: UNIT I: Vector Spaces
Vector Spaces: Theorems Part 1
Important Theorems for Engineering Maths or Mathematics - Vector Spaces: Theorems Part 1
Theorem 1: (Cancellation Law
for vector addition).
If
x, y and z are vectors in a vector space V such that x+z = y+z then x=y.
Proof:
There
exists a vector v in V such that z+v=0. (Property 4)
⇒ x=x+0 = x+(z+v) = (x +
z) + v
= (y + z)+v = y+ (z+v) = y + (0)= y..
Theorem 2
In any vector space V,
the following statements are true:
(a) 0x=0 for each x ∈ V.
(b) (−a) x = −(ax) = a
(−x) for each a ∈
F and each x ∈
V.
(c) a0 = 0 for each a ∈ F.
(d) ax = 0 either a = 0
or x = 0.
Proof:
(a)
Let us consider x ∈
V and 0 ∈ F.
It
follows that 0x + 0x = (0+0) x
= 0x
=
0x + 0
=
0 + 0x (Commutative law)
Hence
0x
=0 by the above previous theorem.
(b)
The vector − (ax) is the unique element of V such that ax + (−ax) = 0.
ax + (−
a) x = [a + (− a)] x
= 0x
=
0
(c)
a•0 = a0+ 0 (Additive identity)
=
a0 + [a0+ (− a0)] (Associative
law)
=
a(0+0) − a0 (Distributive law)
=
a0 − a0 (Identity law)
=
0 (Inverse)
(d)
If ax = 0 then a=0 or x=0.
Let
us assume that ax=0 and a≠0 such that x=0. Since a ∈
F then a−1 ∈ F.
ax =
0
a‒1(ax) = a‒1(0)
(a‒1a) x
= 0
1.x = 0
x=0
Theorem 3
The
vector 0 is unique. (ie) The identity is unique.
Proof:
Let
x ∈ V and e, e' ∈ V
x+e = V and x+e' = V
x+e
= x+e'
e = e' (By cancellation law)
Theorem 4
The vector y is unique
(ie) The inverse is unique
Proof:
Let
x ∈ V and y, y' ∈ V
x+y = 0 and x+y' = 0
x+y
= x+y'
y=y' (By cancellation law)
Linear Algebra: UNIT I: Vector Spaces : Tag: maths, mathematics : - Vector Spaces: Theorems Part 1
Linear Algebra: UNIT I: Vector Spaces
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Linear Algebra
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