Let T:Rn-Rn be an invertible linear transformation, and let S and U be functions from Rn into Rn such that S(T(x))=x and U(T(x))=x for all x in Rn. Show that U(v)=S(v) for all v in Rn. This will show that T has a unique inverse.

Question

Let T:Rn->Rn be an invertible linear transformation, and let S and U be functions from Rn into Rn such that S(T(x))=x and U(T(x))=x for all x in Rn. Show that U(v)=S(v) for all v in Rn. This will show that T has a unique inverse.

anonymous · Answer

We are going to do a proof by contradiction.  Assume that $$S\ne U.$$Then there must exist at least one vector v in Rn such that:$$S(v)\ne U(v).$$ However, since T is an invertible linear transformation, there must exist a v' in Rn such that:$$T(v')=v.$$Therefore:$$S(v)=S(T(v'))=v'=U(T(v'))=U(v)$$which is a contradiction to S(v) not being equal to U(v).  Thus my initial statement was false, and U=S.

anonymous · Answer

for all x in Rn S(T(x))=U(T(x))
then SoT = UoT........let H be an inverse of T ( ToH = Id ) (right inverse )
Then SoToH = UoToH ====> So Id = Uo Id====> S=U