Prove that a given matrix representation of a linear operator is, in fact, a representation I'm starting to learn about representation theory of groups, and I'm having trouble coming up with a technical proof. Here's the setup and the notation I'm using:
Let $g$ be an element of a group $G$. Given a vector space $V,$ let $T(g) \in Aut(V)$ be the linear operator assigned to $g$. But, this $T(g)$ is general, i.e. not assigned to a specific basis. Then, let $D(g)$ be the matrix representation of $T(g)$ with a particular basis.
Now, we define the matrix elements of $D(g)$ as follows:
\begin{equation}
T(g)e_i=e_kD(g)^k_i \,\,\,\,\,\,\,\,\,\,\,\,\,(*)
\end{equation}
where $k$ is the row index, $i$ is the column index, $e_k$ are the basis vectors, and summation over $k$ is implied. Note that $e_k$ is a row vector, if that maters. 
Now, I must show that $D$ is in fact a representation, by showing
(1) $D(E)=I$ (the identity $E \in G$ maps to the identity matrix) and
(2) $D(g_1g_2)=D(g_1)D(g_2)$. 
I think I have (1) mostly correct:
\begin{equation}
T(E)e_i=e_kD(E)^k_i \\
e_i=e_kD(E)^k_i
\end{equation}
and since $e_i$ is a basis vector, it cannot be written as a linear combination of any of the other basis vectors. Thus $D(E)^k_i=0$ for $i \neq k$ and $D(E)^i_i=1$. Thus $D(E)=I$.
For part (2), I am struggling — especially with implied summations and indices flying around. Here is what I have so far:
\begin{equation}
T(g_1g_2)e_i=e_kD(g_1g_2)^k_i \\
T(g_1)T(g_2)e_i=e_kD(g_1g_2)^k_i \\
T(g_1)e_kD(g_2)^k_i=e_kD(g_1g_2)^k_i
\end{equation}
Now I have summations over $k$ implied on both sides. I want to perhaps apply definition $(*)$ to $T(g_1)e_k$ on the left hand side, but I am unsure about how to handle the summation over $k$ if I do so. Any suggestions? Or another approach entirely?
 A: We have a group $G$, a finite dimensional vector space $V$ over a field $\mathbb F$ and a group homomorphism $T\colon G\to \mathrm{Aut}(V)$. This setup guarantees us that $T(g_1g_2)=T(g_1)\circ T(g_2)$ and $T(e)=\mathrm{id}_V$.
Now we pick a particular basis $v_1, \dots, v_n$ of $V$, in other words we have a linear isomorphism $\varphi\colon V\to \mathbb{F}^n$. This allows us to define:
$$D\colon G\to \mathrm{Aut}(\mathbb F^n),~D(g) = \varphi\circ T(g)\circ \varphi^{-1}$$
It is clear that each $D(g)$ is an automorphism (it's a composition of isomorphisms) and that $D$ is a group homomorphism:
$$D(g_1)\circ D(g_2) = (\varphi \circ T(g_1) \circ \varphi^{-1}) \circ (\varphi \circ T(g_2) \circ \varphi^{-1}) = \varphi \circ T(g_1g_2) \circ \varphi^{-1} = D(g_1g_2)$$
(We don't need to check that it preserves the identity, this is automatic from elementary group theory).
Edit: I put the construction in a more explicit way.
Take a linear map $f\colon V\to V$. A vector $f(v_i)$ can be uniquely written in terms of basis as:
$$f(v_i) = a_i^1 v_1 + \dots + a_i^n v_n$$
We gather these $a_i^j$ numbers into a matrix. (This is what basically the map $\varphi$ does in the definition of $D(g)$).
This assignment has the following properties:


*

*the map $\mathrm{id}_V$ is represented by the identity matrix $I_n$,

*the map $f\circ g$ is represented by the product of matrices $a^i_jb^j_k$.

