Page 151 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Show that M (R) is a ring with respect to addition and multiplication of matrices.
2
Solution: You can check that (M (R), +) is an abelian group. You can also verify the associative
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.
.
.
property for multiplication. We now show that A (B + C) = A B A- A C for A, B, C in M (R).
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a  a 12    b b 12  c  c 12  
A . (B + C) =  11    11    11  
 a 21 a 22    b 21 b 22   c 21 c 22 
a  a 12   b  c b  c 12 
=  11   11 11 12 
 a 21 a 22   b  c 21 b  c 22 
22
21
a  11 b  c 11  a 12 b  c 21  a 11 b  c 12  a 12 b  c 22 
11
22
12
21
=  a c c c c 

 21 b  11  a 22 b  21  a 21 b  12  a 22 b  22 
11
22
21
12
 a b  a c  a c 11 11  a c  a b  a b 22  a c 11 12  a c 
11
11
12
12 21
12 21
12
11
12 22
=  a c a c a b a c 

 a b  22 21  a c 21 11  22 21  a b  22 22  a c 21 12  22 22 
12
11
21
21
12 22 
12
12
11 11
11
11
=   a b  a b 21 a b  a b 22      a c  a c a c  a c 
11 12
11
12 21
12
 a b  a b 21 a b  a b 22   a c  a c a c  a c
12 22 
21 11
21 12
22
22 21
21
11
22
21
12
.
=  a  11 a 12   b 11 b 12     a  11 a 12   c 11 c 12  
.
 
 
 a 21 a 22   b 21 b 22   a 21 a 22   c 21 c 22 
= A.B + A.C
In the same way we can obtain the other distributive law, i.e., (A + B) . C = A. C + B . C  A, B,
C  M (R).
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Thus, M (R) is a ring under matrix addition and multiplication.
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Note Multiplication over M (R) is not commutative. So, we can’t say that the left
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distributive law implies the right distributive law in this case.
Example: Consider the class of all continuous real valued functions defined on the closed
interval [0, 1]. We denote this by C [0, 1]. If f and g are two continuous functions on [0, 1], we
define f + g and fg as
(f + g) (x) = f(x) + g(x) (i.e., pointwise addition)
and (f. g) (x) = f(x). g(x) (i.e., pointwise multiplication)
for every x  [0, 1]. From the Calculus course you know that the function f + g and fg are defined
and continuous on [0, 1], i.e., if f and g  C[0, 1], then both f + g and f .g are in C [0, 1]. Show that
C [0, 1] is a ring with respect to + and
Solution: Since addition in R is associative and commutative, so is addition in C [0, 1]. The
additive identity of C [0, 1] is the zero function. The additive inverse off  C [0, 1] is (–f), where
(–f)(x) = – f(A)  x  [0, 1]. See figure 14.1 for a visual interpretation of (- f). Thus, (C [0, 1], +) is
an abelian group. Again, since multiplication in R is associative, so is multiplication in C [0, 1].
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