Algebraic Homomorphisms and Singular Transformations in Mathematics
An exploration of Cayley's theorem analog for groups in algebra, showcasing the isomorphism of algebras with unit elements over fields to subalgebras of some vector space. Additionally, a discussion on characterizing singular transformations in finite-dimensional vector spaces over fields. The explanations are supported with proofs and logical reasoning.
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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)
