Example 97 infinite sequences let k be a field a

Example 9.7: Infinite sequences
Let K be a field. A sequence of elements of K is an infinite list
( a
0
, a
1
, a
2
, a
3
... ) ,
where a
i
∈ K for all i . We also use the notation ( a
i
)
i ∈ N
, or occasionally ( a
i
) , to denote such a
sequence. Let Seq
K
be the set of sequences of elements of K . We add two sequences by adding
their i
th
elements:
( a
i
)
i ∈ N
+( b
i
)
i ∈ N
=( a
i
+ b
i
)
i ∈ N
.
We scale a sequence by scaling each of its elements:
k ( a
i
)
i ∈ N
=( ka
i
)
i ∈ N
.
Then Seq
K
is a vector space.
Example 9.8: Vector space of polynomials of unbounded degree
Let K be a field, and let P be the set of all polynomials (of any degree) with coefficients from K ,
i.e., expressions of the form
p ( x )= a
n
x
n
+ a
n − 1
x
n − 1
+ ... + a
1
x + a
0
,
where n ≥ 0 and a
0
, ... , a
n
∈ K . Addition and scalar multiplication of polynomials are defined in
the usual way . Then P is a vector space.

9.1. Definition of vector spaces 331
We conclude this section by deriving some initial consequences of the vector space axioms.
Proposition 9.9: Elementary consequences of the vector space axioms
In any vector space, the following are true:
(a) The additive unit is unique. In other words, whenever u + v = u , then v = 0 .
(b) Additive inverses are unique. In other words, whenever u + v = 0 , then v = − u .
(c) 0 u = 0 for all vectors u .
(d) The following cancellation law holds: if u + w = v + w , then u = v .
Proof. We prove the first three properties, and leave the last one as an exercise. Assume V is any vector
space over a field K .
(a) Consider arbitrary vectors u , v ∈ V and assume
u + v = u .
Applying the law (A1) (commutative law) to the left-hand side, we have
v + u = u .
Adding − u to both sides of the equation, we have
( v + u )+( − u )= u +( − u ) .
Applying the law (A2) (associative law) to the left-hand side, we have
v +( u +( − u ))= u +( − u ) .
Applying the law (A4) (additive inverse law) to both sides of the equation, we have
v + 0 = 0 .
Applying the law (A3) (additive unit law) to the left-hand side, we have
v = 0 .
This proves that whenever u + v = u , then v = 0 , or in other words, v = 0 is the only element acting
as an additive unit.
(b) Consider arbitrary vectors u , v ∈ V and assume
u + v = 0 .
Applying the law (A1) (commutative law) to the left-hand side, we have
v + u = 0 .

332 Vector spaces
Adding − u to both sides of the equation, we have
( v + u )+( − u )= 0 +( − u ) .
Applying the law (A2) (associative law) to the left-hand side, we have
v +( u +( − u ))= 0 +( − u ) .
Applying the law (A4) (additive inverse law) to the left-hand side, we have
v + 0 = 0 +( − u ) .
Applying the law (A1) (commutative law) to the right-hand side, we have
v + 0 = − u + 0 .
Applying the law (A3) (additive unit law) to both sides of the equation, we have
v = − u .
This proves that whenever u + v = 0 , then v = − u , or in other words, v = − u is the only element
acting as an additive inverse of u .
(c) First, note that the scalar 0 ∈ K satisfies the property 0 + 0 = 0, by property (A3) of the definition of
a field. Now let u ∈ V be any vector. Using the vector space law (SM2) (distributive law over scalar
addition) and 0 + 0 = 0, we have
0 u + 0 u =( 0 + 0 ) u = 0 u .
Next, we use a small trick: add − ( 0 u ) to both sides of the equation. This gives
( 0 u + 0 u )+( − ( 0 u )) = 0 u +( − ( 0 u )) .