You mean "infinitessimal generator," i.e. generator of the corresponding Lie algebras. Anyhow, the unitary group $U(n)$ is generated as a group by $SU(n)$ and $U(1)\subset U(n)$, which is the subset of diagonal matrices with $e^{i\theta}$ in the top left corner and 1's elsewhere. By differentiating, your ninth generator is probably the matrix with an $i$ in the top left corner and zeroes elsewhere.
For a quick reference, look at the first few paragraphs of http://en.wikipedia.org/wiki/Unitary_group.
Edit: I don't have enough reputation to comment, but the number of generators of the $U(n)$ is infinite. In fact this is already true for the circle $U(1)$ (any finite collection of elements generates a countable subset under the group laws). In the interest of using the language correctly, you are looking at a basis for the Lie algebra. The dimension means dimension as a real manifold, which is also the dimension of the lie algebra as a real vector space (even though its a real subspace of complex matrices).
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'?
- You can easily prove that
$$
S\equiv ( \mathbb{1} +i\sigma_1)/\sqrt{2} , \qquad S^\dagger S=\mathbb{1} ~,\qquad S^\dagger \sigma_2 S=\sigma_3.
$$
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!
Best Answer
The Dynkin diagram of $SU(4)$ has 3 nodes, which means that it carries three elements in it's Cartan subalgebra. Consider those as three possible choices for $J_z$. Each of those can be attached with a pair of raising/lowering operators $J_{\pm}$ -- to create one $SU(2)$ algebra each.
EDIT:
BTW, a simple way to see this is the following: Consider the generators on the diagonal (you'll have $N-1$ independent ones, due to the tracelessness condition) as the $J_z$. Then, the corresponding $J_\pm$ are those generators that are one step off-diagonal above/below, and there are exactly $N-1$ of them.
I presume that argument can be translated to a statement about groups, but I don't think I'm equipped to do that.