Intuition of vector of $n$ dimensions. Actually I was reading Lectures of Physics by Feynman where he introuduced gradient theorem. While I was googling about this theorem, I came across vector of $n$ dimensions. I'm having a great problem in understanding this. What does it mean, when we say a vector has $n$ components? What is the direction? I really need a good intuition.
 A: The most familiar $n$-dimensional vector space that look like a 'typical' vector space is undoubtedly $\Bbb R^n = \{(x_1, x_2, \ldots, x_n) : x_i \in \Bbb R\}$, the space of of all tuples consisting of $n$ real numbers.
As a hands-on simplified stand-in for $\Bbb R^n$, your best bet is probably $3$-space, $\Bbb R^3$. In $\Bbb R^3$, each component of a vector represents a distinct 'direction' (as well as a length), and all of our 'pure' directions, like $(0, 0, 1)$, are at right angles to all other pure directions, like $(1, 0, 0)$. Any given vector, say $(3, -1, 2)$, specifies something pointing $3$ units in the $x_1$ direction, backwards $1$ unit in the $x_2$ direction, and $2$ units in the $x_3$ direction.
In $\Bbb R^n$, the same line of reasoning is perfectly good, except instead of only having $3$ 'pure' directions, we have $n$ of them, all still at right angles to one another. With the vector $(2, 5, -2, 1, -5) \in \Bbb R^5$, we can decompose it into 'pure' directions, just like we did in $\Bbb R^3$; we just have more of them now.
If all of your vectors will always be from some Euclidean space $\Bbb R^n$, then congratulations, you are officially working with the nicest space I can think of! This need not be the case though; there are plenty of different kinds of $n$-dimensional vector spaces, and even infinite-dimensional spaces that show up throughout math and physics. Most of them are not as nice as $\Bbb R^n$; 'direction' may not mean what it does in $\Bbb R^n$, we may not be given a natural way to speak of 'length', or the 'length' may look completely unlike the length of a line-segment, to name some basic differences.
However, a solid understanding of Linear Algebra will enable you to understand the similarities and differences among spaces, and work reasonably comfortably in almost any vector space you can dream up.
A: Mathematically a vector isn't really an arrow, just an element of a set, called vector space, whose elements can basically be added together and multiplied with a number the usual way.
There are no geometric concepts in a vector space, except for colinearity. We can always say, that if $V$ is a vector space over $\mathbb{R}$ (or any other field for that matter, but let us just look at the reals) then if $x,y\in V$ then those two vectors are colinear iff $y=\alpha x$, where $\alpha\in\mathbb{R}$ is some appropriate number.
Any other concept of geometry comes from additional structures you introduce on your vector space, such as a norm, an inner product, a symplectic form, etc. Depending on what structure you impose, the geometry induced might be like euclidean geometry (inner product) or completely alien (symplectic form for example).
The dimension of a vector space is basically just the amount of "degrees of freedom" in the vector space, nothing about directions.
Since you spoke of rigid bodies, let me use them as an example. In classical mechanics, the so called configuration space of a rigid body is the set $\mathbb{R}^3\times SO(3)$, where $\mathbb{R}^3$ is the usual euclidean 3-space, and $SO(3)$ is something called the special orthogonal group. This is basically the set of all rotations of the euclidean 3-space. It is not a vector space, but it is a set that can be continuously parametrized by a triple of numbers, so it is similiar enough that I can use it as an example.
Usually this representation of $SO(3)$ is done by specifying three angles. Therefore, a rigid body's state at any time can be specified by giving the 6-tuple $$ (x(t),y(t),z(t),\alpha(t),\beta(t),\gamma(t)). $$
The first three of these refer to the rigid body's position in space, the second three refers to the rigid body's orientation in space.
As you can see, a rigid body has six degrees of freedom, so one can be described by tuples of 6 numbers. These also carry geometric data, since it gives both spatial position and orientation. There is absolutely no sense in assigning directions or length to this tuple though (and since the addition of two such tuples is meaningless, they are not even vectors really).
EDIT:
Here is an expansion of my comment, in response to yours, since the character limit on comments prevents me to post a meaningful answer.
Imagine the equation for a circle of radius $a$ in the plane. This usually takes the form $$ x^2+y^2=a^2, $$ however if I write $$ r=a, $$ this equation also means the same thing.
Two different equations describe the same thing!
To describe a point on the circle of radius $a$, I need to specify an $x$ and $y$ value that obeys my first equation. But a circle is what we call a one-dimensional manifold, it is parametrizable by one parameter. The equation $x^2+y^2=a^2$ provides a constrait on the parameters $x$ and $y$, that reduces the degrees of freedom in specifying a point on the circle to one parameter.
This one parameter can be an angle for example $\varphi$.
The reason why the abstract formalism exists in linear algebra is to be able to express geometric data in a way that is invariant of any representation of said data.
In my example, both $x^2+y^2=a^2$ and $r=a$ is a concrete representation of a circle.
When we actually calculate something, we use representations, that depend on a particular coordinate system. When we want to formulate something, we can do so abstractly, that directly means something geometrical. And by geometrical, I don't mean euclidean specifically, so it doesn't necessary have anything to do with actual angles or lengths or anything.
When a real vector space, $V$ is for example, 6 dimensional, it means that an element $x$ of $V$, when assigned to a representation, will have 6 independent components. Six independent number is needed to actually give the data encoded in the single vector, $x$. The degrees of freedom in the $V$ space is 6. The space $V$ can be an euclidean vector space, in which case, the six data represents multidimensional "directions" and lengths, yes. But it can carry any other kinds of data that needs six numbers to be specified.
