Matrix generators for exceptional lie algebras The complex semisimple Lie algebra $\mathcal{L}$ includes the Cartan subalgebra $h_\alpha \in H$ and $e_\alpha$, where $\alpha$ are the roots. Why do they want to find the matrix representations for $ad(e_\alpha)$ instead of $e_\alpha$ to construct generators for the exceptional groups? My answer would be that, to write the generators in matrix form one should first take the map of ad from the lie algebra of $E_n$ to the lie algebra of $gl(n)$ and the map of Ad from the group $E_n$ to $GL(n)$, where $exp(tad(e_\alpha))=Ad(exp(te_\alpha))$, where the map Ad somehow is I? But this confuses me so I am not sure if that is the correct reason why.
 A: I don't know why this was downvoted, it seems fine to me.
A matrix representation is always a matrix representation of a linear map. Elements of an abstract Lie algebra are not in themselves linear maps so there is nothing to take a matrix representation of. 
Let's say someone gave you an abstract Lie algebra $\mathfrak{g}$ in terms of generators and relations and asked you to write down the matrix representation of the element $X \in \mathfrak{g}$. Then you could not answer it because the question is meaningless. 
By contrast, the map $ad(X)$ is a linear map: it maps elements of $\mathfrak{g}$ to elements of $\mathfrak{g}$ by the rule $Y \mapsto [X, Y]$. So it makes sense to ask for a matrix representation of $ad(X)$ w.r.t. a given basis of $\mathfrak{g}$. 
Similarly if you were given some map $\phi: \mathfrak{g} \to End(V)$ for some vector space $V$ we could ask about the matrix representation of $\phi(X)$ w.r.t. some basis of $V$. (Of course in practice this would only be interesting if $\phi$ is a Lie algebra morphism from $\mathfrak{g}$ to $\mathfrak{gl}(V)$, but this typically still leaves infinitely many choices on $V$'s of many different dimensions.)
Now in general Lie algebras do not come with a 'standard' map $\phi$ to the endomorphisms of some vector space and hence we cannot talk about 'the' matrix representation of a Lie algebra element - we first need to specify a way of transforming Lie algebra elements into linear maps. There are typically infinitely many choices but using $ad$ has the advantage that it is always available, and natural.
Now when you are looking at a Lie algebra that is defined as a set of linear maps with a certain property then we have a second 'natural' choice beyond ad of a way of assigning a linear map to each Lie algebra element because there the Lie algebra element is a linear map by construction. Examples are the Lie algebras of type $\mathfrak{gl}, \mathfrak{sl}, \mathfrak{so}$ etc. But in a sense they are the exception. Exceptional simple Lie algebras, but also non-semisimple Lie algbras and Lie algebras that are secretely $\mathfrak{sl}_n$ for some $n$ but you don't recognize it because it is defined in terms of cartan matrices rather than matrices don't automatically come with a 'prefered' choice of interpreting their elements as linear maps and hence the question of what is 'the' matrix representation of the element doesn't make much sense.
