Page 57 - DMTH502_LINEAR_ALGEBRA

Unit 1: Vector Space over Fields

Notes
If ( , )a b C then ( a , b ) C
b
and also ( a , b ) ( , ) = ( a a b b ) (0, 0)
,
a
Hence [(–a, –b) is the additive inverse, of (a, b)]
b
a
,
d
e
c
f
e
f
Also [( , )( , )]( , ) = [(ac bd bc ad ) ( , )]
= [(ac bd )e (bc ad ) ,(bc ad )e (ac bd ) ]
f
f
= [ (a ce df ) b (de cf ), (ce df ) a (de cf )]
b
= ( , )(a b ce df de cf )
,
= ( , )[( , )( , )]a b c d e f
Hence multiplication is associative in C.
Distributive laws also hold in C because,
a
,
b
f
c
d
( , )[( , ) ( , )] = ( , )(c e d ) f
e
b
a
= [ (a c e ) b (d f ), (c e ) a (d f )]
b
= [(ac bd ) (ac bf ),(bf ad ) (be af )]
,
,
= (ac bd bc ad ) (ae bf be af )
= ( , )( , ) ( , )( , )a b c d a b e f
Similarly, it can be proved that multiplication is distributive over addition in C from right too.
Multiplication is commutative in C because
a
d
c
b
,
( , )( , ) = (ac bd bc ad )
= (ca db cb da )
= ( , )( , )c d a b
Since (1, 0) C and also (1, 0) (a, b)
= (a, b) (1, 0) is multiplicative identity in C.
Multiplicative inverse for non-zero elements in C exists because if (a, b) is non-zero elements in
C then a and b are not zero at a time.
Let (c, d) be the multiplicative inverse of (a, b) then

c
b
d
( , )( , ) = (1, 0)
a
i.e. [(ac bd ),(bc ad )] = (1, 0)
so that ac – bd = 1, bc + ad = 0
a b
i.e., c = ,d
a 2 b 2 a 2 b 2

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