# Thread: Isomorphism and dimensional vector spaces

1. ## Isomorphism and dimensional vector spaces

1. Let U,V be finite dimensional vector spaces,with bases B,C respectively.
Prove:a linear map f:U->V is an isomorphism iff [f]_C^B is invertible.(I'm so sorry but I honestly couldn't get Latex to work properly,slightest change to "[f]" and it would cause an unknown latex error)

2. Let T:M_n(R)->T:M_n(R),T(A)=AC,C$\displaystyle \in$ M_n(R).
Prove that it is an isomorphism iff C is invertible.

Again so sorry for the inconvenience and any help would be appreciated.

2. I presume that $\displaystyle [f]_C^B$ means the matrix representation of f using those bases.

An isomorphism is, by definition, both one to one and onto. If the vector y is in V then there exist a unique x in U such that f(x)= y. That says that, for any y in v, there exist a unique x such that $\displaystyle [f]_C^Bx= y$. It follows immediately that kernel of f is just {0} and from that that the matrix is invertible.

3. for problem 2, suppose C is invertible, and let A be any 2x2 real matrix. then T(AC^-1) = (AC^-1)C = A(C^-1C) = AI = A, so

A has the pre-image AC^-1, thus T is onto.

suppose T(K) = KC = 0. then (KC)C^-1 = 0C^-1 = 0. but (KC)C^-1 = K, whence K = 0, so T is 1-1. thus T is an isomorphism

(the proof that T is R-linear is rather straight-forward).

on the other hand, suppose T is an isomorphism. since all isomorphisms are invertible maps, T^-1 exists. since T is onto,

I is in the domain of T^-1. let B = T^-1(I).

then BC = T(B) = T(T^-1(I)) = I, so B = C^-1, C is invertible.