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This question already has an answer here:

The Poor Man’s Introduction to Tensors - by Justin C. Feng. https://web2.ph.utexas.edu/~jcfeng/notes/Tensors_Poor_Man.pdf

On page 3, "Before I can tell you what a tensor is, I must tell you what a vector really is; in fact, you will later see that a vector is a type of tensor.
A vector is simply a directional derivative. Before you write me off as a nut, take a look at the directional derivative of some scalar function $f(x^i)$"

$$\nu.\nabla(f(x^j))=\nu^i\dfrac{\partial}{\partial{x^i}}f(x^j)$$

To me this is still ok, the author later removes the function itself and claim,

$$\nu.\nabla=\nu^i\dfrac{\partial}{\partial{x^i}}$$
currently i see $\nu.\nabla$ as some form of operator.

But my trouble arises when on page 4, equation 19, author then gives equation, stating the following "Some people get rid of the function f altogether, and write the following:"

$$\nu=\nu^i\dfrac{\partial}{\partial{x^i}}$$ and $$\nu^i = \nu(x^i)$$

My problem is that seem that this definition of vector seems very un-intuitive to me,maybe because i am acquainted with all the old definition. It is as if vector were a operator. Can anyone give me the intuition behind this? Or can someone help me make its sense in term of old definitions of vector i have like in simple Euclidean geometry.

Edit: Just to be exact from the possible duplicate, My problem with intuition is exactly this new operator kind of definition As compared to n-tuple that i have been seeing all along. Exactly how do they connect if they do so, Or if they are different or if one generalizes the other.

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marked as duplicate by Henning Makholm, user91500, Travis Willse, Watson, Daniel W. Farlow Jul 8 '16 at 16:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Regarding your edit: You have probably learned that a vector is an $n$-tuple that transforms in a certain way during change of coordinates. If you take a look at $\partial/\partial x^\mu$, this transforms as $\partial/\partial y^\mu=\frac{\partial x^\nu}{\partial y^\mu}\partial/\partial x^\nu$. From this you can derive the components of "differential operator vectors" transform the same way as your $n$-tuples do. $\endgroup$ – Bence Racskó Jul 8 '16 at 10:00
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In euclidean geometry you have a so called affine structure, which essentially states that points are separated by translation vectors. An affine space is a set of points $A$ together with a vector space $V$ and an action of $V$ on $A$ as an abelian group: If $p\in A$ is a point and $v\in V$ is a vector then $p+v\in A$ is another point of $A$, and we can view $v$ as a vector that points from $p$ to $p+v$.

The reason we don't use affine spaces most of the time, is that if you fix a designated point, which we shall call the origin, $O$, then $A$ and $V$ canbe identified. The point $p$ in $A$ is identified with the vector $v\in V$ for which $p=O+v$. The nature of the abelian group action is defined in such a way to make this identification unique.

Why am I saying all this? Because if we want to define a vector at a point, rather than one that points between points (remember, in physics, the electric field at $x$, $E(x)$ is a vector that is located at $x$ but it doesn't point to any other point, it just has a magnitude and direction), the affine structure will enable that:

If you want to have a vector at $p$, you can take a straight curve $\gamma(t)=p+tv$ $(p\in A,\ v\in V)$, and take the derivative $$ \left.\frac{d}{dt}(p+tv)\right|_{t=0}=\lim_{h\rightarrow0}\frac{(p+hv)-p}{h}=v, $$ however by taking a look at the last limit, you can see the vector points from $p$ to $p+hv$ where $h$ tends to zero. The vector is at $p$.

Tensor calculus on the other hand, even if implicit, takes place generally on a differentiable manifold, a generalization of curves and surfaces to higher dimensions. A differentiable manifold is usually not an affine space, so we cannot define vectors the same way I did above. We have to use some other way to grasp the intuitive concept of a direction and magnitude at a point, without relying on this affine structure.

There are multiple ways to do this, one is to take an equivalence class of curves at a point, two curves being equivalent if they go through the point both and also their derivatives in any chart also coincide. This grasps the point of a "velocity" at a point. Another way to do it is to take first-order differential operators that act on scalar functions. This characterizes the concept of a vector by being the "velocity" with which you differentiate functions in a certain direction.

In the end, it doesn't actually matter. A recurring point in modern mathematics is that it doesn't so much matter what an object is, aside from what it can do and what its properties are. It however is useful to take vectors to be differential operators whatever definition you use, if for nothing else because it can reduce notational clutter. You can write $X(f)$ for the directional derivative of $f$ in the direction of $X$ instead of $d_Xf$, and you can write the commutator of two vector fields $X$ and $Y$ as $$ [X,Y](f)=XY(f)-YX(f) $$ instead of $$ [X,Y]=d^{-1}[d_X,d_Y] $$ for example.

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I am pretty sure most people as they are learning mathematics think of a vector as an arrow in space. This is useful but it isn't strictly correct.

What is a vector then? A vector is an element of a vector space.

So the answer to your question is actually fairly simple. An operator can be a vector by being an element of a vector space.

I would recommend that you prove that the appropriate set of directional derivative operators can be turned into a vector space.

It would also be worthwhile for your understanding to see if you can associate each operator with a vector in $\Bbb R^n$ and then make that connection rigorous by constructing an isomorphism between the two vector spaces.

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    $\begingroup$ This explains why one can call the set of certain differential operators on scalar functions a "vector space" (and, consequently, how its elements can be called "vectors" in an abstract sense) -- but not how it is useful to use those operators in place of the geometric vectors we already know and love (and which also form an abstract vector space). Note that the OP is quoting from physics notes, and physicists use "vectors" with a narrower meaning than mathematicians do. $\endgroup$ – Henning Makholm Jul 8 '16 at 9:35
  • $\begingroup$ I'm not sure that the OP is asking why it is useful to use those operators? Although I am certain that will be a question he will ask! $\endgroup$ – Bernard W Jul 8 '16 at 9:44
  • $\begingroup$ x @Bernard: Well, the point is not just that they form a vector space, or that it is a vector space that's useful in itself, but specifically that these operators correspond uniquely to geometric vectors, and that tensor calculus prefers to use them to represent ordinary geometric vectors instead of geometric vectors themselves. $\endgroup$ – Henning Makholm Jul 8 '16 at 9:47
  • $\begingroup$ Exactly the point! thanks! "tensor calculus prefers to use them to represent ordinary geometric vectors instead of geometric vectors themselves" $\endgroup$ – Mann Jul 8 '16 at 9:52
  • $\begingroup$ @HenningMakholm I see your point. I think I had the same thing in mind when I suggested that the OP should try to construct the isomorphism between the two spaces and I just took for granted that once that is done the 'connection' the OP was asking for should be obvious. $\endgroup$ – Bernard W Jul 8 '16 at 9:52

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