Is there a general expression for the adjoint representation of $U(N)$ or $u(N)$? At least for low values of $N$ like $2$ or $3$ and such I would like to know if there are explicit matrices known giving the representation of $u(N)$ or $U(N)$ in the adjoint?  

(..a related query: Is it for the Lie group or the Lie algebra of U(N) that it is true that the weight vectors in the fundamental/vector representation can be taken to be N N-vectors such that all have weight/eigenvalue 1 under its Cartan and the ith of them has 1 in the ith place and 0 elsewhere and for the conjugate of the above representation its the same but now with (-1)?..I guess its for the u(N) since they are skew-Hermitian but would still like to know of a precise answer/proof..)
I've only recently started learning about group theory so it would be worth double checking the following. 

As I understand it, when we say something (for example a field, $X$) transforms under the adjoint rep of a group (say $U(N)$), the transformation can involve any matrix, $g$, belonging to the group $U(N)$. The adjoint rep doesn't put any constraints on the matrices we choose from $U(N)$, rather it tells us how to use those matrices in a transformation. When we say $X$ transforms in the adjoint rep of $U(N)$ it simply means it transforms as $X \rightarrow X' = gXg^{-1}$. The matrix $g$ can be any matrix belonging to $U(N)$. Note: $g^{-1}$ is the inverse of the matrix $g$. So when we say 'an object is in the adjoint representation of $U(N)$' this is a statement about how to act the matrices of $U(N)$ on an that object. It doesn't say anything about which matrices we choose from $U(N)$.
I don't know if it helps, but I think you can write
$$U(N)=SU(N)\times U(1)$$
usually (in physics) people consider semisimple groups, i.e. groups without U(1) factors (=tori). To give an explicit example, consider $U(2)=SU(2)\times U(1)$. The adjoint representation of the algebra can be defined by
$$(T_a)_{bc}=-if_{abc}$$
where $f_{abc}$ are the structure constants of the algebra. Perfoming an explicit calculation for $SU(2)\times U(1)$ and keeping in mind that the generator of $U(1)$ commutes with all other elements of the algebra we get the following generators for the adjoint representation:
$$
T_1=\begin{pmatrix}0&amp;0&amp;0&amp;0\\0&amp;0&amp;-i&amp;0\\0&amp;i&amp;0&amp;0\\0&amp;0&amp;0&amp;0\end{pmatrix}
T_2=\begin{pmatrix}0&amp;0&amp;i&amp;0\\0&amp;0&amp;0&amp;0\\-i&amp;0&amp;0&amp;0\\0&amp;0&amp;0&amp;0\end{pmatrix}
T_3=\begin{pmatrix}0&amp;-i&amp;0&amp;0\\i&amp;0&amp;0&amp;0\\0&amp;0&amp;0&amp;0\\0&amp;0&amp;0&amp;0\end{pmatrix}
T_4=0
$$
So you see, that with $T_1, T_2, T_3$ you can span a 3 dimensional real vector space, but with all four you can't since $T_4$ vanishes. This means, as I think, that while you can define an adjoint representation for $SU(2)$, you can't do so for $U(2)$ or $U(1)$.