Page 62 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Solution:
R
(i) V 1 ( , ) is an abelian group as (R, +) is a field.
V 2 (a b ) a b R and , a b . R

V 3 ( ) a a a , R and a . R

a
V 4 ( a ) ( ) , , R and a . R
V 5 1 . a a . 1 a , 1 R and a . R
Hence R is a vector space over R.

C
(ii) V 1 ( , ) is an abelian group because C is a field
V (u ) u C and , u C
2
(using left distributive law of multiplication over addition in C.)

V . ( )u u , u , C and u . C
3
(using right distributive law in C)

u
V ( u ) ( ) , , C and u C
4
(associative law of multiplication in C)
V 1 . u u for 1 C for u . C
5
Hence C is a vector space over the field C.

Example 33: A field K can be regarded as a vector space over any subfield H or K.
Solution: We consider K as a set of vectors. Let us regard the elements of the satisfied H as scalars.

Let addition of vectors be the composition in the field K. Let us define the scalar multiplication
as follows:
If a H and K, a is the product of these two elements in the field K.

V Since K is a field, therefore (K, +) is an abelian group.
1
V 2 ( a ) a a a H and , , . K

This is a consequence of the left distributive law in K because
a , , K (because H < K and a H)

V 3 (a b ) a b , a b H and . K This is due to the right distributive law in K.

V ( ) ( a b ) , a b H and . K This result is due to associativity of multiplication in K.
a
b
4
V 5 . 1 . K where 1 is the unity of the subfield H. But H K and as such 1 is also the
unity of the field K.
Hence K is a vector space over H.

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