Problem shown in the image Let V and W be finite - dimensional vector spaces and let T : V rightarrow W be linear as well as one - to - one and onto. Suppose that dim(V) = n and let (beta = {vi....vn} be a basis for V. Prove that {T(v1),..., T(vn)} is a basis for W. Let U : W rightarrow V be the unique linear map such that U(T(vi)) = Vi for all 1 Solution Part I) Since T is one-to-one and onto then in particular dim(W)=dim(V)=n Since { T(v1),...,T(vn) } contains n elements and dim(W)=n it suffices to prove it\'s linearly independent by theorem. Now suppose we a1,...,an such that a1T(v1)+...+anT(vn) = 0 Since T is linear it can be rewritten as T(a1v1+...+anvn) = 0 Since T is one-to-one => a1v1+...+anvn=0 Since {v1,...,vn} is a basis, then it\'s linearly independent, so a1=...=an=0 Hence the initial family is independent, and it\'s a basis of W Part II) Let x in V then since {v1,...,vn} is a basis of V we can find a1,...,an such that x = a1v1+...+anvn Now T(x) = T(a1v1+...+anvn) = a1T(v1)+...+anT(vn) So : U(T(x)) = a1U(T(v1))+...+an U(T(vn)) since both T and U are linear Now since U(T(vi)) = vi we have U(T(x)) = a1v1 + ... + anvn = x So U(T(x)) = x Similarly : Let y in W, then since {T(v1),...,T(vn)}is a basis for W, we can find a1,...,an such that y = a1T(v1)+...+anT(vn) So U(y) = a1U(T(v1))+...+anU(T(vn)) = a1v1+...+anvn => T(U(y)) = T(a1v1+...+anvn)=a1T(v1)+...+anT(vn) = y So T(U(y)) = y.

1. Problem shown in the image Let V and W be finite - dimensional vector spaces and let T : V
rightarrow W be linear as well as one - to - one and onto. Suppose that dim(V) = n and let (beta =
{vi....vn} be a basis for V. Prove that {T(v1),..., T(vn)} is a basis for W. Let U : W rightarrow V
be the unique linear map such that U(T(vi)) = Vi for all 1
Solution
Part I)
Since T is one-to-one and onto then in particular dim(W)=dim(V)=n
Since { T(v1),...,T(vn) } contains n elements and dim(W)=n it suffices to prove it's linearly
independent by theorem.
Now suppose we a1,...,an such that a1T(v1)+...+anT(vn) = 0
Since T is linear it can be rewritten as T(a1v1+...+anvn) = 0
Since T is one-to-one => a1v1+...+anvn=0
Since {v1,...,vn} is a basis, then it's linearly independent, so a1=...=an=0
Hence the initial family is independent, and it's a basis of W
Part II)
Let x in V then since {v1,...,vn} is a basis of V we can find a1,...,an such that x = a1v1+...+anvn
Now T(x) = T(a1v1+...+anvn) = a1T(v1)+...+anT(vn)
So : U(T(x)) = a1U(T(v1))+...+an U(T(vn)) since both T and U are linear
Now since U(T(vi)) = vi we have U(T(x)) = a1v1 + ... + anvn = x
So U(T(x)) = x
Similarly :
Let y in W, then since {T(v1),...,T(vn)}is a basis for W, we can find a1,...,an such that y =
a1T(v1)+...+anT(vn)
So U(y) = a1U(T(v1))+...+anU(T(vn)) = a1v1+...+anvn => T(U(y)) =
T(a1v1+...+anvn)=a1T(v1)+...+anT(vn) = y
So T(U(y)) = y