Representation of SU(3) generators Let's discuss about $SU(3)$. I understand that the most important representations (relevant to physics) are the defining and the adjoint. In the defining representation  of $SU(3)$; namely $\mathbf{3}$,  the Gell-Mann matrices are used to represent the generators
$$
\left[T^{A}\right]_{ij} = \dfrac{1}{2}\lambda^{A},
$$
where $T^A$ are the generators and $\lambda^A$ the Gell-Mann matrices. In adjoint representation, on the other hand, an $\mathbf{8}$, the generators are represented by matrices according to
$$
\left[ T_{i} \right]_{jk} = -if_{ijk},
$$ 
where $f_{ijk}$ are the structure constants.
My question is this, how can one represent the generators in the $\mathbf{10}$ of $SU(3)$, which corresponds to a symmetric tensor with 3 upper or lower indices (or for that matter how to represent the $\mathbf{6}$ with two symmetric indices). What is the general procedure to represent the generators in an arbitrary representation?
 A: The generic, abstract answer to your question is in Coleman's excellent book. However, I have not seen these 10×10 explicit matrices around, as they are not exactly low-lying fruit. I have not stumbled upon them, but it may not be impossible they are in somebody's thesis. 
In principle, you might Kronecker-multiply 3 triplets, and shed off the two octets and the singlet in the Clebsch-Gordan decomposition, ${\bf 3}\otimes{\bf 3}\otimes{\bf 3}={\bf 10} \oplus {\bf 8} \oplus {\bf 8} \oplus{\bf 1}$, having worked out the Casimirs of all of those, as in the WP article which has a somewhat better basis than Gell-Mann's. The 10 is the D(3,0) in that notation, and both its quadratic and cubic Casimirs are equal to 6. In practice, it might take you about 1-2 days to work them out.
Alternatively, you might slug through the WP article and tease them out in a baryon decuplet basis, where the 10-vector they act on is the obvious $(\Delta^{++},\Delta^{+},\Delta^{0},\Delta^{-}, \Sigma^{* +},\Sigma^{* 0},\Sigma^{* -}, \Xi^{* 0},\Xi^{* -}, \Omega^- )$. The normalized Cartan 
subalgebra generators for $I_3$ and hypercharge (B+S), then, are the predictable diagonal ones, $F_3$=diag(3/2,1/2,-1/2,-3/2, 1,0,-1, 1/2,-1/2, 0)/$\sqrt{15}$ and $F_8$=diag(1,1,1,1, 0,0,0, -1,-1, -2)/$\sqrt{20}$, etc.
Try $U_+= F_6+iF_7$ first!
Edit: Indeed, Richard Shurtleff 
arXiv:0908.3864 has produced a mathematica notebook which produces these generators and checks their algebra.
