How does the matrix for $ad(x)$ compare to $x \in L$ for a lie algebra $L$? How does the matrix for $ad(x)$ compare to $x \in L$ for a lie algebra $L$?
I understand how to compute the matrix of $ad(x)$ given a basis for $L$, but for some reason some things that are supposed to be obvious to me just arn't. Like for example, $L$ is solvable if and only if $ad(L)$ is solvable. Could somebody give me some general insight into how the matrix representations of $x$ and $ad(x)$ relate to eachother? Perhaps in the cases when $x$ is upper triangular, $L$ is solvable, and stuff like that.
I realize this is an open ended question, but I often get my best answers when I ask questions as such!
Thanks in advance you guys are the best!!
 A: In general, there is no reason why $L$ should consist of matrices and there is no need to understand a relation between matrix representations to understand the result that you quote. The point is that $ad:L\to ad(L)$ is a homomorphism of Lie algebras which is surjective by definition. Moreover, the kernel of this map is by definition the center $\mathfrak z(L)$ of $L$. This means that $ad(L)\cong L/\mathfrak z(L)$ as a Lie algebra. Now quotients of solvable Lie algebras are always solvable, so if $L$ is solvable then also $ad(L)$ is solvable.
Conversely, there is a general result that if you have a Lie algebra $L$ and a solvalble ideal $K\subset L$ such that $L/K$ is solvable then $L$ is solvable. Since $\mathfrak z(L)$ is abelian and thus solvable this shows that solvability of $ad(L)$ implies solvability of $L$.
A: Andreas Cap's answer is the best. I just want to show you one striking example for a Lie algebra $L$ given as matrices where there is very little relation between an element $x$ (as matrix) and the matrix of its adjoint map $ad(x) \in \mathfrak{gl}(L)$:
Namely, choose your favourite very high natural number $n$ and your favourite very complicated matrix $x \in M_n(\mathbb C)$, and let $L$ be the one-dimensional Lie algebra spanned by $x$, which you can thus view as a subalgebra of $\mathfrak{gl}_n(\mathbb C)$. Like every one-dimensional Lie algebra, $L$ is abelian, meaning that $ad(x) = 0 \in M_{n^2}(\mathbb C)$. And you chose your favourite crazy matrix for $x$, which could be anything, maybe lower triangular, maybe semisimple, maybe nilpotent of degree $17$, maybe skew-symmetric, whatever!
I leave it to you to come up with Lie algebras of dimension $2, 3, 4, ..., 28$ which are made up of the funniest and mutually most different matrices (some orthogonal, some upper triangular, some with only imaginary eigenvalues, ...), where still $ad(x)=0$ for all of them.
So you see that in general, there is little reason to expect a strong relation between some Lie algebra given as matrices and the matrices representing $ad(x)$. Which should make it all the more remarkable that in some special cases, one can make statements. For example if $L$ is a semisimple Lie subalgebra of some $\mathfrak{gl}_n$ (i.e. its elements $x \in L$ are given as $n \times n$-matrices) then we have that $x$ is semisimple iff $ad(x)$ is semisimple, and $x$ is nilpotent iff $ad(x)$ is nilpotent. For a highly nontrivial proof of this nontrivial theorem (which moreover needs the ground field to have characteristic $0$), see e.g. Bourbaki's treatise one Lie Groups and Algebras, chapter I, §6 no. 3.
