General question about representations of groups In my previous post I asked about the matrix representation of the cyclic group $C_n$ (not needed to answer this question). But it seems that I do not completely understand the following. Given a matrix representation $D$ such that $D(g)$ is a matrix for $g \in G$ we call $D$ often a representation and we identify a matrix representation with a representation if the underlying vector space $V$ has finite-dimension such that $\operatorname{GL}(V) \cong \operatorname{GL}(n,K)$ where $\dim(V)=n$ and $K$ the field. Now my question is: Given a matrix $D(g)$, can we always just assume that there exists some linear map $D$ working on some vector space $V$ and a basis for $V$ and that the choice of $V$ and its basis are arbitrary but such that the linear map $D$ gives the desired matrix $D(g)$? Because we can always write $D(g)=(D(a_1)_{A},\dotsc,D(a_n)_A)$ with A$=\{a_1,\dotsc,a_n\}$ the basis of $V$.
Hope this post is clear enough and thanks for your help!
 A: I am not sure if I understood your question correctly, but it seems to be a Linear Algebra question. It looks as if you are asking this: given a $n\times n$ matrix $A$ and a vector space $V$ over a field $K$ such that $\dim V=n$, is there always an endomorphism $f$ of $V$ and a basis $B=(a_1,\ldots,a_n)$ of $V$ such that the matrix of $f$ with respect to $B$ is $A$? Yes, there is. Actually, you can take any basis $(a_1,\ldots,a_n)$ of $V$ and consider the endomorphism $f$ of $V$ such that$$(\forall j\in\{1,2,\ldots,n\}):f(e_j)=\sum_{j=1}^na_{ij}e_j.$$
A: As written, the answer is no. If I arbitrarily assign a matrix to $g$ (or even each $g\in G$), there is no guarantee that the assignment $g\mapsto D(g)$ is a homomorphism $G\to GL(n,K)$.
However, given a homomorphism $D:G\to GL(n,K)$ you can certainly do what you are asking. Recall that given $A\in GL(n,K)$, one can define a linear transformation
$$L_A:K^n\to K^n$$
by $L_A(v)=A.v$. Now, let $V$ be an $n$ dimensional vector space and $\beta=\{v_1,\ldots,v_n\}$ any basis. Define an isomorphism $\phi:V\to K^n$ by $\phi(v_i)=e_i$. Then,
$$\phi^{-1}L_{D(g)}\phi\in GL(V)$$
and, writing the matrix for this element in the basis $\beta$ yields $$[\phi^{-1}L_{D(g)}\phi]_\beta=D(g).$$
