Are covariant vectors representable as row vectors and contravariant as column vectors I would like to know what are the range of validity of the following statement:

Covariant vectors are representable as row vectors. Contravariant
  vectors are representable as column vectors.

For example we know that the gradient of a function is representable as row vector in ordinary space $ \mathbb{R}^3$ 
$\nabla f = \left [ \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z} \right ]$
and an ordinary vector is a column vector
$ \mathbf{x} = \left[ x_1, x_2, x_3 \right]^T$
I think that this continues to be valid in special relativity (Minkowski metric is flat), but I'm not sure about it in general relativity.
Can you provide me some examples?
 A: Yes, the statement holds true in general relativity as well. However, as we need to deal with tensors of higher and in particular mixed order, the rules of matrix multiplication (which is where the idea of the representation via row- and column-vectors comes from) are no longer sufficiently powerful:
Instead, the placement of the index determines if we are dealing with a contravariant (upper index) or a covariant (lower index) quantity.
Additionally, by convention an index which occurs in a product in both upper and lower position gets contracted, and equations must hold for all values of free indices.
If the given metric is non-Euclidean (which is already true in special relativity), mapping between co- and contravariant quantities is more involved than simple transposition and the actual values of the components in a given basis can change, eg:
$$
p^\mu = (p^0,+\vec p)\\
p_\mu = (p^0,-\vec p)
$$
and in general:
$$
p_\mu = g_{\mu\nu}p^\nu
$$
where $g_{\mu\nu}$ denotes the metric tensor and a sum $\nu=1\dots n$ is implied.
A: It is meaningful in general, though it is a matter of convention, not of truth. But it never leads to incorrect results if you make this convention.
This is thoroughly discussed in the entry
''How are matrices and tensors related?'' of Chapter B8: Quantum gravity
of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html
Note that in multivariate analysis one generally defines the gradient is the transpose of the (exterior) derivative, so ''gradient'' and ''derivative'' are slightly different notions. The transpose makes sense only given a metric, as it essentially consists in replacing raised/lowered indices by lowered/raised ones.
Thus unlike a covariant exterior derivative, a gradient is no longer covariant but contravariant (and hence a column vector).
A: As from my experience, I had very hard time to understand the "physical" difference between contra- & cov- things, I really understood them only when I read differential geometry and get involved with one-forms, even more, some authors (like Shuch) argue that it is wrong to say that Covectors are really vectors, they are different objects, they are one forms!
A: This is not a full answer, but rather an attempt to clear up some misconception about the gradient: In particular, in my opinion saying that the gradient is a covector doesn't make much sense.
There are two ways to interpret the concept of vectors and covectors:
The first one is to say there is only a single entity - the vector - which has covariant and contravariant components. This is inspired by classical tensor calculus: when doing calculations, we often do not care about the placement of the indices of a particular tensor - after all, we can always lower or raise them (ie go from column vectors to row vectors and vice versa) by contraction with the metric tensor.
If you take this point of view, differential and gradient are two names for the same entity. It is somewhat misleading to say that the gradient is a covector, as what we really mean is that the gradient is a vector whose covariant components are given by the partial derivatives (whereas its contravariant components are given by contraction of the covariant components with the inverse of the metric tensor).
The second point of view - which is the one I prefer - is that vectors (or, more precisely as we're doing differential geometry, tangent vectors) are distinct from covectors (aka 1-forms). However, the scalar product gives an isomorphism between tangent vectors and 1-forms. The gradient is the (pre-)image of the differential under this isomorphism and an actual vector.
