Download Vector Spaces

Dr. Neal, WKU
MATH 307
Vector Spaces
A vector space over ℜ is an algebraic structure in which we can add elements and
perform scalar multiplication. There are 10 axioms that we assume and that allow us to
do these algebraic operations. Five of the axioms are for addition and five are for scalar
multiplication.
Axioms of Addition
1.
2.
3.
4.
Closure: If u , v ∈ V , then u + v is another element in V .
Associative: For all u , v , w ∈ V , (u + v) + w = u + (v + w) .
Commutative: For all u , v ∈ V , u + v = v + u .

0
V
Additive
Identity:
There
exists
a
“0”
element
in
,
denoted
by
, such that


€
0 + v = v = v + 0 for
€ all v ∈ V .
5. Additive Inverse:
€ For every v ∈ V , there exists an element −v ∈ V such that

v + (−v) = 0 = −v + v .
€
Note: Properties 1, 2, 4, and 5 are the axioms of an algebraic structure called a group.
€ be non-empty. By adding€Axiom 3, we obtain a commutative
By Axiom 4, a group must
group. So a vector space, with its addition operation, is a commutative group.
The set of integers Z and the set of rational numbers Q are commutative groups
under normal addition. But the set of natural numbers ℵ is not a group because there
are no additive inverses in ℵ and there is no zero element.
The following theorem holds for any group V , in particular for a vector space:
Theorem 3.1. (a) The additive identity in V is unique. (b) Every element’s additive
inverse is unique.
Proof. (a) Suppose u is another “zero” vector such
  that u + v = v = v + u for all v ∈ V .
Because 0 also has this property, we have u = u + 0 = 0 .

(b) Suppose v + w = 0 = w + v for some w . Then


w = 0 + w = (−v + v) + w = −v + (v + w) = −v + 0 = −v .
€
Axioms of Scalar Multiplication
1.
2.
3.
4.
5.
Closure: If v ∈ V and c is any scalar (i.e., real number), then c v ∈ V .
Associative: c(d v) = (c d)v for all v ∈ V and all c , d ∈ ℜ .
Scalar Distributive: c (u + v) = cu + cv for all u , v ∈ V and c ∈ ℜ .
Vector Distributive: (c + d)v = c v + d v for all v ∈ V and all c , d ∈ ℜ .
€
Multiplicative
Identity: For the scalar 1, 1v = v for all€v ∈ V .
€
€
€
€
€
Note: The set of rational numbers Q ,€although a commutative
group, is not a vector
€
space over ℜ for it is not closed under scalar multiplication: For c = π and v = 2 , then
c v ∉ Q . However, the set of real numbers ℜ and the set of complex numbers C are
vector spaces over ℜ .
€
Dr. Neal, WKU
The set of all m × n matrices Mm, n is a vector space over ℜ with the m × n zero matrix
n
being the additive identity element. In particular, n -dimensional space R is a vector
space over ℜ . Also, the set of all polynomials Pn with real coefficients and having
degree ≤ n is a vector space over ℜ .
Theorem 3.2. Let V be a vector space.
(a)
(b)
(c)
(d)

0 v = 0 for all v ∈ V .
c 0 = 0 for all scalars c .

If c v = 0 , then c = 0 or v = 0 .
For every
vector v in V , −1v is the additive inverse − v .
€
Proof. (a) First, 0 v = (0 + 0)v = 0 v + 0v . Now add the additive inverse −(0v) to both
sides:


0 = 0v + −(0v) = (0v + 0v ) + −(0v) = 0v + (0v + −(0v)) = 0v + 0 = 0v .
€

 



(b) First, c 0 = c(0 + 0) = c 0 + c 0 . Now add the additive inverse −(c 0) to both sides to


obtain
0
=
c
0.
€

(c) Suppose
. If c = 0, then we are done. So suppose
c
v
=
0
€ c ≠ 0. We now must show

that v = 0 . But by property (b) we have
 1 1
1 
0 = 0 = (c v) =  c v = 1v = v
c 
c
c
(d) We will show that −1v behaves as the additive inverse: For every v ∈ V we have


(−1v) + v = −1v + 1v = (−1 + 1) v = 0v = 0 and v + (−1v) = 1v + (−1v) = (1+ −1)v = 0v = 0 .
€