Are adjoint representation matrices the generators of the Adjoint representation? My question is, are the (Lie Algebra) adjoint representation matrices in general the generators of the (Lie Group) Adjoint representation (i.e. treating the Adjoint representation as a matrix Lie Group)?
My justification is that $ad(\xi) = T_e (Ad(\xi))$.
Therefore we treat the Adjoint representation as its own matrix Lie Group (living in the Lie Algebra vector space, i.e. it acts on Lie Algebra elements like $\xi$ in the above statement). Therefore, since it is a matrix Lie Group, we can in general define a Lie Algebra representation on it by its generating matrices (with the Lie Bracket represented by the usual matrix commutator)
The reason I ask this is because I was confused why the adjoint representation of su(2) (with the basis given below) could seemingly represent the Lie Bracket either by matrix-vector multiplication (with a vector in the Lie Algebra), or by their usual matrix commutator. (e.g. $[T_1,T_2]=T_3$ could be represented by $T_1*(0 1 0)= (0 0 1)$ or represented by $T_1*T_2-T_2*T_1 = T_3$)
My solution is therefore that adjoint is defined so that the matrix-vector multiplication represents the Lie Bracket but, since (by different reasoning) the basis of the adjoint are also 'raw' elements of the Lie Algebra of the Adjoint matrix Lie Group, they can also be represented by generators of the Adjoint matrix Lie Group (and therefore represent the Lie Bracket by the usual commutator).
$$ T_1 = \begin{bmatrix} 0&0&0\\0&0&-1\\0&1&0\end{bmatrix}, T_2 = \begin{bmatrix} 0&0&1\\0&0&0\\-1&0&0\end{bmatrix}, T_3 = \begin{bmatrix} 0&-1&0\\1&0&0\\0&0&0\end{bmatrix}$$
 A: To avoid extended discussion in the comments I wanted to put a more full answer. I'll try to pull in a few points on other questions you have asked as well so it's all in one place. In short, yes.
First things first, there are a lot of brackets and commutators flying around so we have to be careful what we're doing.
A Lie algebra representation is a Lie algebra homomorphism $\rho:\mathfrak{g}\to \mathrm{End}(V)$ where $V$ is some vector space. The homomorphism part means that $\rho([X,Y]) = [\rho(X),\rho(Y)]$ where the bracket on the left is whatever bracket we have defined on $\mathfrak{g}$ and the one on the right is the canonical bracket given by the commutator on $\mathrm{End}(V)$. Picking a basis of $V$ allows one to write $\rho(X)$ as a $n\times n $ matrix ($n = \dim V$). If $\rho$ is faithful (aka injective) this is unique for each $X \in \mathfrak{g}$ so we can identify $\mathfrak{g}$ as a subset (indeed Lie subalgebra) of $\mathrm{End}(V)$. As I have said elsewhere, the homomorphism part of the definition of a representation guarantees that the set of matrices have the same Lie bracket relations as $\mathfrak{g}$ does.
There is a special representation where $V = \mathfrak{g}$ called the adjoint representation $ \mathrm{ad}:\mathfrak{g}\to\mathrm{End}(\mathfrak{g})$. Which is defined as follows: $\mathrm{ad}(X)$ is the linear map on $\mathfrak{g}$ with $\mathrm{ad}(X)(Y) = [X,Y]$. It is easy to see that this is a homomorphism i.e. that $\mathrm{ad}([X,Y]) = [\mathrm{ad}(X), \mathrm{ad}(Y)]$. Indeed this is just the Jacobi identity rearranged. For a semisimple Lie algebra, the adjoint representation is faithful and so we can represent the Lie algebra by matrices in this case. Note now we are representing elements of $\mathfrak{g}$ both as $\dim\mathfrak{g}\times\dim\mathfrak{g}$ matrices and as $\dim\mathfrak{g} \times 1$ vectors simultaneously. So we have our original element $X$, its adjoint image $T_X = ad(X)$ viewed as a big matrix and it viewed as a vector $v_X$ to make that big matrix make sense. Then $[X,Y] = T_X*v_Y$ under the appropriate identifications.
Tying this to the Lie group story, the Adjoint group is defined as the image under the Adjoint representation of any Lie group with Lie algebra $\mathfrak{g}$. Thus it is automatically linear (i.e. a matrix group) as a subgroup of $GL(\mathfrak{g})$. Then you can generate the the Adjoint action using your chosen basis of $\mathfrak{g}$ viewed as $\dim\mathfrak{g} \times \dim\mathfrak{g}$ matrices if you so desire.
More generally, this works any time there is a valid Lie group representation. Note a Lie algebra representation of $\mathfrak{g}$ doesn't imply that a specific corresponding Lie group $G$ has a valid representation. The adjoint representation is special in that every Lie group has an Adjoint representation. For other representations of $\mathfrak{g}$ not all Lie groups with Lie algebra $\mathfrak{g}$ also have a representation there. E.g. the special orthogonal group has no spin representations and the adjoint group often doesn't act on the original "vector" representations.
