How to construct matrix representations of particle multiplets? Similarly to how the SU(2)-triplet W-bosons can be written as a vector with three components $\big(W^1, W^2, W^3\big)^T$ or as a 2x2 matrix by contracting with the Pauli matrices $\big(\textbf{W} = W^i \frac{\sigma^i}{2}\big)$ , I am trying to figure out how to do this for other representations and groups. Specifically, I am trying to construct the matrix representation of an SU(3) sextet/sextuplet (this is a flavor SU(3) not an extension of SU(2)$_L$).
It is clear to me that for multiplets in the adjoint representation (e.g. an SU(2) triplet or an SU(3) octet) we simply contract with the respective group generators. But how can this be done in general?
Thank you very much in advance!
 A: An n-dimensional representation is basically a set of n-dimensional vectors, as  for your 3d vectors of your example, along with the $n\times n$ matrices (satisfying the Lie algebra) acting on them.
Such matrices, in any representation, normally form an orthonormal basis set, so they parameterize the adjoint representation--which can therefore be written as a vector in the space of representation matrices, as you illustrate for su(2). (In your example, the representation matrices need not be in the fundamental, as you chose.)
However, in other representations, like the 6 of su(3) you chose, n does not coincide with the dimension of the algebra, here 8, and so representation matrices are not a good basis to represent your 6-vectors, and you normally need to stick to vectors, not matrices.
You do know what the representation matrices of the 6 look like, in su(3), right?
Since the 6 is a symmetric representation, you might also use the Jordan-Schwinger realization in terms of bosons oscillators for it. Just plug in the Gell-Mann matrices for M, just as the article does for the Pauli matrices...

Edit addressing your comment
Indeed,  after your comment, I finally appreciated what you are asking about.  The 6 of your reference is, indeed, a sporadic construction whose name I do not know (as yet), and which does not appear to generalize, except for the symmetric irrep of the su(n) s, of dimension n(n+1)/2, coincidentally the dimension of symmetric $n\times n$  matrices!
So, for your sextet, indeed, you may package your 6-vector in the symmetric 3×3 traceful (!) matrix
$$
S\equiv \begin{pmatrix} a&b&c\\ b&d&e\\ c&e&f \end{pmatrix} \\ = b\lambda_1+c\lambda_4+e\lambda_6+{a+d+f\over 3} 1\!\! 1+ {a-d\over 2}\lambda_3 + {a+d-2f\over 2\sqrt 3}\lambda_8,
$$
which is therefore not in the algebra of su(3), by dint of the trace: it is in the algebra of u(3), not quite spanning it, as explained above. (It has forfeited the imaginary generators $\lambda_2,\lambda_5,\lambda_7$, an su(2) subalgebra.)  Cf. the footnote.
Nevertheless, it transforms as three triplet columns from the left under SU(3)L; in addition, the transposed (not hermitian conjugated!!) group element on the right, to preserve symmetry, $USU^T$... you are not in Kansas anymore: So there is no suitable commutator expression similar to that for the adjoint transformation!)$^\natural$.
The construction appears coincidental, but given the J-S construction, above, may well generalize on occasion to some higher symmetric reps, sporadically. I know of no refs for it, but if I find out more, I'll add  such.

$\natural$ 
The above 6 parameters of S relate to the parameters of the symmetrized triplets' tensor product $\tfrac{1}{2} ({\mathbf v}\otimes {\mathbf w}^T +  {\mathbf w}\otimes {\mathbf v}^T   )$ by the evident $a=v_1w_1$ ,  $d=v_2 w_2$ ,  $f=v_3 w_3$ ,  $b=(v_2 w_1+ v_1 w_2)/2$ ,  $c=(v_3 w_1+ v_1 w_3)/2$ ,  $e=(v_3 w_2+ v_2 w_3)/2$ . You may work out the obvious generalization for n >3.  
