How can I find the (nine) generators of $\text{U}(3)$?
I have started out by considering $\text{SU}(3)$, which is an 8-dimensional subgroup of $\text{U}(3)$. The generators of $\text{SU}(3)$ that are often used by particle physicists are the Gell-Mann matrices $\lambda^a$ (for $a = 1,2,\ldots,8$), given by:
\begin{alignat*}{4}
\lambda^1 &= \left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{array} \right) \,, \quad &\lambda^2 &= \left( \begin{array}{ccc}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0
\end{array} \right) \,, \quad &\lambda^3 &= \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 0
\end{array} \right) \\
\lambda^4 &= \left( \begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{array} \right) \,, \quad &\lambda^5 &= \left( \begin{array}{ccc}
0 & 0 & -i \\
0 & 0 & 0 \\
i & 0 & 0
\end{array} \right) \,, \quad &\lambda^6 &= \left( \begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array} \right) \\
\lambda^7 &= \left( \begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & -i \\
0 & i & 0
\end{array} \right) \,, \quad &\lambda^8 &= \frac{1}{\sqrt{3}} \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -2
\end{array} \right) \,,
\end{alignat*}
and satisfying the commutation relation
\begin{equation*}
[\lambda^a,\lambda^b] = 2 i f^{abc} \lambda^c \,,
\end{equation*}
where $f^{abc}$ are the completely antisymmetric structure constants of $\text{SU}(3)$. Each element of $\text{SU}(3)$ can be written as $\exp (i \theta^j \lambda^j/2)$, where $\theta^j$ are real numbers and a sum over the index $j$ is implied. Furthermore, the Gell-Mann matrices are traceless, Hermitian, and obey the relation $\text{Tr}(\lambda^i \lambda^j) = 2\delta^{ij}$.
As the Gell-Mann matrices have nice properties, I would like to extend this set to obtain a full set of generators for $\text{U}(3)$. But how can I find this 'ninth' generator?
Any help would be appreciated.
Best Answer
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).