intuitionvector-spaces
I go by Vector. It's a mathematical term, represented by an arrow with both direction and magnitude. Vector! That's me, because I commit crimes with both direction and magnitude. Oh yeah!
For Euclidean vectors, it's natural to think of vectors having direction and magnitude. What about vectors in abstract vector spaces? Is it possible to (sensibly) define notions of direction and magnitude for all vectors?
(If not, what is an example which highlights why it's not always possible?)
There are several misconceptions in the OP about both mathematicians' and physicists' use of the word "vector", and even about what scalars and tensors are. To keep this a concise overview I'll be linking to fuller explanations.
Firstly, anything you've heard about magnitude and direction was just an attempt to help schoolchildren avoid certain fallacies without having to explain the entire concept of a vector space to them. The aim is to make sure they understand that, for example, a particle's momentum points a certain way but its amount of energy doesn't.
In general, vectors are not tuples. Admittedly some sets of tuples satisfy the axioms of a vector space if you define arithmetic the usual way, but vectors are so much more general than that case, as examples discussed above show. What is true in general is that, if a vector space $V$ has a basis of the form $\left\{e_i|i\in I \right\}$, then each vector in $V$ is expressible as a linear combination of the $e_i$. Depending on the details, this "linear combination" might be a sum or an integral. Armed with this, the coefficients used can provide a tuple representation of vectors (although in some cases you need infinitely many numbers), but the vector is an independent object. The map is not the territory. In fact, making a terrain look different by creating a new map that's rotated relative to an old one is a special case of what you'll sometimes here called a basis change. Since you're familiar with $\mathbb{R}^n$, I'll give a simple example. The vectors $\left(\begin{array}{c} 1\\ 0 \end{array}\right),\,\left(\begin{array}{c} 0\\ 1 \end{array}\right)$ comprise a basis of $\mathbb{R}^2$, but I can rotate a 2D map by an angle $\theta$ because $\left(\begin{array}{c} \cos\theta\\ \sin\theta \end{array}\right),\,\left(\begin{array}{c} \sin\theta\\ -\cos\theta \end{array}\right)$ comprise a basis too.
I should also point out in passing that, while in some contexts the word "basis" simply means a choice of $\left\{e_i|i\in I \right\}$ for which this can be done, the proper definition requires that the linear combination need only use finitely many of the $e_i$. Many vector spaces of interest that do not have finite dimension nonetheless meet some additional technical conditions that allow the less strict meaning of "basis" to be useful. However, the famous statement that two bases of a vector space have the same cardinality refers to the finite-combinations-only definition.
So that's what mathematicians mean by vector spaces. A vector space is always "over" a field of scalars. Just as a vector is defined as an element of a vector space which in turn has a long definition, a scalar is defined as an element of a field which in turn has a long definition.
Or is it? Let's talk about what physicists really mean when they discuss vectors. On the one hand, they know about all the mathematics I mentioned above. On the other hand, they also want to describe nature in terms of quantities that transform in certain convenient ways when we switch coordinate systems, to exemplify "symmetries". This leads them to define "vector" in a stricter way. For example, one thing schoolchildren aren't told is that, although angular momentum has a magnitude and direction, it's not a vector because of the way it transforms under reflections. The distinction in $\mathbb{R}^3$ between vectors and axial vectors takes some explaining. The confusion is understandable. Position and momentum are "in" $\mathbb{R}^3$ and are vectors; angular momentum is "in" $\mathbb{R}^3$ is an axial vector. The reason is simply that none of these things are really "in" a famous set of tuples, because they're not tuples at all; they're quantities that admit a tuple representation. That's one similarity axial vectors have with "true" vectors.
With the development of differential geometry, we realised there is a more elegant way to talk about all this. Instead of distinguishing between true vectors and axial vectors, we can distinguish between contravariant and covariant vectors, provided our "both types count" definition of vector means "rank one tensor". quantity $T^{\alpha_1\cdots\alpha_p}_{\beta_1\cdots\beta_q}$ with $p,\,q$ non-negative integers is called a tensor of rank $p+q$ and order (or type) $\left(p,\,q\right)$ iff a coordinate transformation of spacetime from $x^\mu$ to $x^{'\nu}$ obeys $$T^{'\alpha_1\cdots\alpha_p}_{\beta_1\cdots\beta_q}=\sum_{\gamma_1\cdots\gamma_p \delta_1\cdots\delta_q}\frac{\partial x^{'\alpha_1}}{\partial x^{\gamma_1}}\cdots\frac{\partial x^{'\alpha_p}}{\partial x^{\gamma_p}}\frac{\partial x^{\delta_1}}{\partial x^{'\beta_1}}\cdots\frac{\partial x^{\delta_q}}{\partial x^{'\beta_q}}T^{\gamma_1\cdots\gamma_p}_{\delta_1\cdots\delta_q}.$$(We never actually write the summation sign; we take for granted that any index that appears twice, once as a subscript and once as a superscript, is summed over all possible values. In relativity, there is one such value for each spacetime dimension.) A tensor of rank $0$ is a scalar, and is unchanged under coordinate transformations. A tensor of positive rank is called covariant if $p=0$, contravariant if $q=0$ and mixed otherwise. Mixed tensors have $p\geq 1$ and $q\geq 1$, so have rank $\geq 2$.
Something that looks like a tensor by virtue of its indices may not transform the right way to actually be a tensor. (Of course, if there are no indices at all something would "look like a scalar", but might not be one.) Here are three important examples.
The direction of a vector can be formalized in a couple of different ways.
In Euclidean space, one can define a direction vector to be a vector $\vec u$ such that $|\vec u| = 1$. So, for example, the direction vectors in the plane are those vectors based at the origin whose tip is on the unit circle, and therefore the direction vectors are in one-to-one correspondence with angles chosen in the interval $[0,2\pi)$, where the direction vector corresponding to the angle $\theta \in [0,2\pi)$ is $\vec u = \langle \cos \theta, \sin \theta \rangle$. Next, given an arbitrary nonzero vector $\vec v \ne \vec 0$ in Euclidean space, one can define the direction of $\vec v$ to be the direction vector $$\vec u = \frac{1}{|\vec v|} \vec v = \biggl\langle \frac{v_1}{\sqrt{v_1^2+\cdots+v_n^2}},..., \frac{v_n}{\sqrt{v_1^2+\cdots+v_n^2}} \biggr\rangle $$ And now it's clear what it means to say that two vectors $\vec v$ and $\vec w$ have the same direction: it means that $\frac{1}{|\vec v|} \vec v = \frac{1}{|\vec w|} \vec w$.
Another way to formalize direction, which works in any vector space $V$ regardless of whether it has a norm, goes like this. Consider the set of nonzero vectors $V - \{\vec 0\}$. Define an equivalence relation on $V - \{\vec 0\}$, where $\vec u, \vec v \in V-\{\vec 0\}$ are equivalent if there exists a scalar $r > 0$ such that $r \vec u = \vec v$. One can formally define the direction of a nonzero vector to be its equivalence class, under this equivalence relation. So to say that two nonzero vectors $\vec v,\vec w$ have the same direction means that they are equivalent. In fact you can even formally define the direction of $\vec v$ to be its equivalence class, namely the subset of all vectors $\{r \vec v \mid r > 0\}$; geometrically, this is just the open ray parallel to $\vec v$.
Best Answer
Summary
There's a reasonable definition of direction in abstract vector spaces, but that doesn't always include "orientation." To talk about magnitudes as lengths, you really need extra structure provided by a norm into an ordered field.
Direction
In abstract vector spaces, you can link a weak idea of "direction" with a vector directly by just defining:
That is, each $1$-dimensional subspace could be thought of as a class of vectors in the same direction. Notice, though, that this scheme has you think of the zero vector as being "in all directions," and maybe philosophically then it has no direction at all :)
Some folks might also include a component of orientation when they're thinking about "direction," so we should discuss that too. As far as I can tell, this necessitates $F$ to be an ordered field $F$ so that you can establish a dichotomy of what is positive and what is negative. (You don't have to have orientation if you're happy with the definition of direction above, but I think it's worth discussing.)
Given a direction $\langle v \rangle$ (and an ordering on your field of course), one could say that the elements of $A=\{\lambda v\mid \lambda>0\}$ are mutually oriented in the same way, and $B=\{\lambda v\mid \lambda<0\}$ as oriented in "the other" way, and they are oriented oppositely to those things in $A$. This is a problem for finite fields, which can't be ordered. In characteristic $2$ fields, for example, $v=-v$ for all vectors, so the dichotomy doesn't exist at all, there.
Magnitude
To talk about lengths in an abstract vetor space, you really need an extra structure called a norm. This gives you a way to measure how "long" vectors are. Being able to compare lengths of vectors with a norm again only makes sense when you are working with a norm into an ordered field, so that you can distinguish which magnitude is greater.
On the other hand, if you're just happy to have some sort of scalar for each vector, then there are generalizations of norms into nonordered fields that would work. You just couldn't interpret their values as lengths. Really, our geometric intuition about length is all bound up in ordered fields.