Considering the collection of matrices (defined in the six dimensional tensor product space) $T_a = \frac{1}{2} \lambda_a \otimes \sigma_2$ and $T_i = \frac{1}{2} \lambda_i \otimes I$ where $\lambda_a$ are the $SU(3)$ generating Gell-Mann matrices for $a=1,3,4,6,8$, $\lambda_i$ are the Gell-Mann matrices for $i=2,5,7$, and $\sigma_2$ is the Pauli matrix. I'm asked to show that these generate a reducible representation of SU(3), and to reduce it.
There are a few things I can pick out from the matrices, mainly that $[T_i,T_j] = i\epsilon_{ijk} T_k$ for $\{i,j,k\} = \{2,5,7\}$, and $[T_a,T_b] = if_{abk} T_k$ for $\{a,b\} = \{1,3,4,6,8\}$ and $k=2,5,7$, where (I think) the $f_{abc}$ are undetermined.
The first comes from the fact that $\lambda_{2,5,7}$ generate an $SU(2)$ subalgebra of $SU(3)$. For the second, we know that $T_a$ are symmetric, so $[T_a,T_b]^T = -[T_a,T_b]$ and this means that if $[T_a,T_b] = i\epsilon_{abc} T_c$ ; then $T_c$ must be antisymmetric, and the antisymmetric $T_c$s are the $T_i$s.
Also, $[T_a,T_i] = i\epsilon_{aic} T_c$ for $c=1,3,4,6,8$ by the same idea as above, since $[T_a,T_i]^T = [T_a,T_i]$.
I'm struggling with using this information to reduce this representation of SU(3).
Gell-Mann matrices: https://en.wikipedia.org/wiki/Gell-Mann_matrices
Best Answer
You are almost there! Inspection of the structure constants f of SU(3) in the Gell-Mann basis confirms that they must vanish unless they contain an odd number of indices from the set $ \{2,5,7\}$, those of the imaginary generators. This is because the commutation relation includes an i on the r.h.s., so it must involve one or three imaginary generators.
So, indeed, as you observed, $[\lambda_i,\lambda_j] = i\epsilon_{ijk} \lambda_k$ ; $[\lambda_a,\lambda_b] = i2f_{abk} \lambda_k$ ; and $[\lambda_a,\lambda_i] = i2f_{aib} \lambda_b$.
Now, the imaginary $\sigma_2$s square to the identity, and therefore clearly preserve the Lie algebra of the λ/2s in transitioning to the Ts. That is, the all-imaginary Ts provide a representation of SU(3), as well.
The unitary group matrices $\exp (i\vec \theta\cdot \vec T)$ are thus real, sending real 6-vectors to real vectors. But you already know the irreducible 6 of SU(3) is complex--immediately visible from its Young tableau.
You can immediately see, by inspection, that the representation is reducible, if you appreciate $\sigma_2$ is equivalent to $\sigma_3$, diag(1,-1); so, if you moved the 2d matrices to the left of the direct product, the upper 3 components of the vectors in your 6d vector space never mix with the lower 3 ones under action of the group matrices. That is, your T matrices consist of an upper left 3×3 block acting on the upper 3 components and an identical but opposite lower right 3×3 block acting on the lower 3.
So what is the similarity transformation reducing T to this visibly reduced T'?
Consequently the unitary 6×6 matrix $\mathbb{1}_3\otimes S$ will provide the requisite similarity transformation.
Your representation reduces to a triplet and a mirror-image triplet!