What is the QCD gluon field (mathematically)? In QCD, the gluon field is described as $A^a_\mu$. In the covariant derivative for the Lagrangian, it is multiplied by the Gell-Mann $SU(3)$ generator matrices $\lambda_a$ ($a=1..8$) as $\lambda_aA^a_\mu$. My question is, what is the mathematical structure of $A_\mu$ (for a given a)? Since $\mu=0..3$, it must have 4 components (I'm assuming $\mu$ does not mean $A(x_\mu)$), but the $\lambda$'s are $3\times 3$ matrices. Since $\lambda$ is a 3x3 matrix, and $A_\mu$ is a 1x4 vector, how can the two be "multiplied" together?
Is each element of $A_\mu$ a $3\times 3$ matrix? If not, then what?  Or does the notation $\lambda_aA^a_\mu$ mean something other than normal matrix multiplication is going on in this case?
What is the mathematical structure of $\lambda_aA^a_\mu$ (like, a 1x4 vector, or 3x3 matrix, or a 1x4 vector of 3x3 matrices)?
What does an element of $A_\mu$ represent physically?
 A: $A$ is a $4$-vector taking values in $3 \times 3$ matrices.  More precisely, it's a dual $4$-vector taking values in the Lie algebras $su(3)$, which is an 8-dimensional subspace of the 9-dimensional space of $3\times 3$ matrices.
If you pick coordinates $x^\mu$ on spacetime, you get a basis $dx^\mu$ for the dual vectors and you can expand $A = \sum_\mu A_\mu \otimes dx^\mu$.  Each component $A_\mu$ lives in $su(3)$, i.e., is a $3\times 3$ matrix. 
The symbol $\otimes$ is the tensor product, which is a kind of external multiplication, not the same thing as the usual internal multiplication.  In matrix terms, it means that each component of the 4-vector $A$ is a $3\times 3$ matrix.  (Equivalently, you can think that $A$ is a $3\times 3$ matrix, each entry of which is a $4$-vector.)
If you pick a basis $\lambda_a$ for $su(3)$, you can further expand each $A_\mu = \sum_a A^a_\mu \lambda_a$.  Now each component $A^a_\mu$ is just a real-valued function.  Combining these two, you can write
$$
A = \sum_\mu \sum_a A^a_\mu (\lambda_a \otimes dx^\mu).
$$
A: For a specified $a$, $A^\mu$ is a four-vector. Four-vectors are defined to be object that "transforms like four-vectors", i.e. with Lorentz transformations $A'^{\mu} = \Lambda^\mu_\nu A^\nu.$
In a suitable basis, you can write it out as a column vector:
$$ \left ( \begin{array}{c}A_0  \\ A_1 \\ A_2 \\ A_3 \end{array} \right ). $$
Different $a$'s imply different values for the entries of the four-vector.
I should also say that it is a vector field meaning that there will be a different numerical value of the same $A^\mu$ at each point in space (and time).
The $\lambda$s are just the generators of the $SU(3)$ group, so they are just the $3\times 3$ Gell-Mann matrices.
