Algebraic Theorems and Transformations in Mathematics

Name : M.Vijaya Shankari. Department: Mathematics College: HKRH College,Uthamapalayam Subject : Algebra ll

1)Analog of a cayleys theorem for groups: Statement:- If A is an algebra with unit element over F. Then, A is isometric to a sub algebra of A(V) for some vector space V over F. Proof:- Since A is an algebra over F. It is a vector space over F. We use V=A to prove this theorem. If a A, let Ta: A---- A be defined by VTa= Va for all v A

Claim:- Tais a linear transformation on V(=A) Let v1,v2 A 1)( v1+ v2) Ta= ( v1+ v2)a=v1a+v2a=v1Ta+v2Ta 2)( v)Ta= ( v)a= (va)= (vTa) Hence Tais a linear transformation on A Now, Consider the mapping : A--- A(V) defined by a =Ta for any a A Claim 2:- is an isomorphism of A into A(V) if a,b A and , F Then for all v A vT( a+ b)= v( a+ b)= v( a)+v( b)= (av)+ (bv)= (vTa)+ (vTb)= v( Ta)+v( Tb)=v( Ta+ Tb)

Hence, We have vT(a+b)= v(Ta+Tb) Hence, T( a+ b)= Ta+ Tb Therefore, ( a+ b) = T( a+ b)= Ta+ Tb ( a+ b) = (a )+ (b ) Therefore, is the vector space homomorphism of A into A(V) For a,b A vTab=v(ab)=(va)b=(vTa)b=(vTa)Tb= v(TaTb)

Therefore, Tab= TaTb Therefore, (ab) = Tab= TaTb=(a )(b ) Therefore is a ring homomorphism of A. Hence we have proved that is an algebra homomorphism of A into A(V) TO DETERMINE THE KERNEL OF Let a A be in the kernel of , Then a =0 Implies that Ta=0

Therefore vTa=0 for all v V Now V=A and A has a unit element e Hence, eTa=0 Implies that 0= eTa=ea=a Implies that a=0 The kernel of must therefore merely consist of zero alone. Hence is an isomorphism of A into A(V)

2)Characterisation of singular transformation: Statement:- If V is a finite dimensional over F, Then T A(V) is singular iff there exist a v 0 in V such that vT=0 Proof:- Assume that T is singular Since T is singular, there is an S 0 in A(V) such that ST=TS=0

Since S0, there is an element w V such that wS0 Let v=ws then v 0 and vT=(wS)T=w(ST)=w(0)=0 Hence we have produced a nonzero vector v in V, Which is annihilated by T Conversely, Assume that there exist a v 0 in V with vT=0 To prove: T is not invertible. Suppose that T is invertible

Then T-1exists Therefore, (vT)T-1=0T-1=0 Implies that (vT)T-1=0 Implies that v(TT-1) =0 Implies that v(1)=0 Implies that v=0 This is a contradiction to v 0 Hence T is not invertible

3)Characterisation of regular transformation: Statement: If V is finite dimensional over F, Then T A(V) is regular iff T maps V onto V Proof:- If T is regular, then given v V, V = (vT-1)T Therefore, vT = v Hence, T is onto

Conversely, Assume that T is onto To prove that T is regular Suppose that T is not regular, There exist a vector v1 0 in V such that v1T=0 We can fill out from v1to a basis v1,v2,...,vnof V, Then any element in vT is a linear combination of the elements w1 =v1T, w2=v2T,..., wn=vnT. Since w1=0, vT is spanned by the (n-1) elements w2,...,wn Therefore, dim(vT) (n-1) < n=dim v But then vT must be different from V. i.e., T is not onto which is a =><= Therefore, T must be regular

4)If F is a characteristics root of T A(V), Then for any polynomial q(X) is a characteristic root of q(T) Proof:- Suppose F is a characteristics root of T There is a nonzero vector v in V such that vT= v vT2= (vT)T=( v)T= (vT)= ( v)= 2v

Continuing in this way we get vTk= kv for all positive integer k If q(x)= 0xm+ 1xm+1+ 2xm+2+...+ mfor i F Then, q(T)= 0Tm+ 1Tm+1+...+ m(1) Therefore, v(q(T))=v( 0Tm+ 1Tm+1+...+ m(1)) = 0(vTm)+ 1(vTm+1)+...+ m(v) = 0( mv)+...+ m(v)

=v(0m+1m-1+...+m) Therefore, v(q(T)) = (q( ))v Implies that v(q(T)-q( ))=0 Therefore q( ) is a characteristic root of q(T)