Sorry to be impatient but Ive still had no luck.
Hi I have a question here worth 17 marks in total. I am able to answer some of it however I am unsure where the marks are coming from:
a) State the definition of a group.
A group is a set with a binary operation which satisfies certain axioms.
A group must contain a set, binary operation and an identity element.
Im sure this might be worth 3 marks.
b) Suppose (G,~,e) is a group. Let a,b,c are elements of G and suppose that a~b=a~c
Prove that b=c stating where group axioms are being used.
All I can think of saying is that by the identity axiom and letting a=e on each side that infact b=c.
c) Suppose that n is a natural number and that A is a n x n matrix.
Set E(A)={X|X is ainvertible n x n matrix that commutes with A}
Prove that (E(A),.,I) is a group where . denotes matrix multiplication.
I would answer by the fact that by set definition X commutes with A so commutivity holds:
B and C are elements of E(A)
B.C=C.B
B, C and D are elements of E(A)
B.(C.D)=(B.C).D
And by identity axiom:
B is a element of E(A)
I.B=B=B.I
by multiplication by the identity matrix
Am I mostly correct?
I would appreciate any help or hints.
Cheers.
Commutivity is not in the group axioms. What did you have for the answer
to are (a), was it there? You will find the group axioms here.
E(A) inherits associativity from the group of all nxn matrices under multiplication.
closure: let X1 and X1 be in E(A), then
(X1.X2).A = X1.(X2.A) = X1.(A.X2) = (X1.A).X2 = (A.X1).X2 = A.(X1.X2)
(this uses associativity of matrix products, and that the elements of E(A)
commute with A).
Hence if X1 and X2 are in E(A) so is X1.X2.
E(A) inherits having an identity element from the group of all nxn matrices
(and that it commutes with A).
Now let X2=X1^-1, where X1 is a non-singular matrix which comutes with
A, and X2 is its bog standard matrix inverse, we need to show that X2 is
in E(A).
X2.A = X1^-1.A = X1^-1.X1.A.X1^-1
since X commutes with A so X1.A.X1^-1 = A.X1.X1^-1 = A.
But:
X1^-1.X1.A.X1^-1 = (X1^-1.X1).A.X1^-1=A.X1^-1
so X1^-1 commutes with A and so is in E(A).
RonL