We are used to $\mathbb R$ being an ordered field: through the construction of real numbers we know that either $a\leq, a=b$ or $a\geq b$.Thus we're able to use real numbers to say something is bigger, better or faster than something else. It is also well-known, at some level, that $\mathbb R^d$, the vector space whose elements are $d$-length tuples of real numbers, is not ordered. This has momentous consequences: you're no longer able to say who's the better man when there are multiple scores measuring multiple factors, or what's best for society when each person has a scale from better to worse. You can hope the true scoreable thing falls within a one-dimensional manifold and perform some kind of 'dimensionality reduction' (averaging is a crude method; regression and SVD are better, but still reliant on ad hoc hypotheses). But this is no longer "the math". It's methodology, and it's questionable.
A vector space can be given some structures that are weaker than ordering but produce useful mathematics. First, you can define a distance $\mathrm{dist}\colon \mathbb R^d \times \mathbb R^d \to \mathbb R$ -- a function that is scalable ($\mathrm{dist}(\alpha \dot a, \alpha \dot b) =|\alpha|\cdot \mathrm{dist}(\mathbb a,\mathbb b) $ for a scalar $\alpha$), sub-additive ($\mathrm{dist}(\mathbb a,c) \leq \mathrm{dist}(\mathbb a,\mathbb b) + \mathrm{dist}(b,c)$) and point-separating ($\mathrm{dist}(a,a)=0$). Equivalently, you can define a norm $\|\cdot\|$ that is scalable, sub-additive, and point-separating, such that $\mathrm{dist}(a,b) = \|a-b\|$.
Note that these axioms imply that both distance and norm are always positive (except for the trivial case $\|(0,\cdots,0)\|=0$). In other words, they give our vector field a sense of size, but not order: you can try to rank vectors by their norm, but $\|(0,-1)\|=(0,1)\|$ even though one is "positive" and the other is "negative". To introduce this idea of an orientation on a vector space we need to develop some additional concepts.
A multilinear map (or tensor) of order $k$ on $\mathbb V$ is a function $\omega\colon \mathbb V^k \to \mathbb R$ that is linear on each of its component functions $\omega_i\colon \mathbb V \to \mathbb R$. The set of all tensors of order $k$ on $\mathbb V$ forms a vector space by inheriting the linear structure of its component functions, as the following examples will show.
Rank 1 tensors are linear functionals (or covectors), that is, vectors in the dual space $\mathbb V^\ast$ of linear maps $v^\ast\colon \mathbb V \to \mathbb R$. They can be represented by $(1\times \dim \mathbb V)$-dimensional row matrices as follows: given a basis $(\mathbf e_1,\cdots,\mathbf e_n)$ of $\mathbb V$, we define $\mathbf a = (a_1, \cdots, a_{\dim \mathbb V})$ as the basis vector of $\omega$, given by $$ \mathbf a = \begin{bmatrix} \omega(\mathbf e_1) & \cdots & \omega(\mathbf e_{\dim \mathbb V})^\top \end{bmatrix} $$ Then, if $\mathbf{u} = \sum_{i=1}^m u_i e_i$, we define $$ \omega(\mathbf u) = \omega\left(\sum_i u_i \mathbf e_i\right) = \sum_i \omega(u_i \mathbf e_i) = \sum_i u_i \omega(\mathbf e_i)=\sum_i a_i x_i = \mathbf a^\top \mathbf u $$ We verify that $\mathbf a$ is indeed the basis of the vector space of all covectors by noting that $\alpha(\omega \mathbf u)= \mathbf a^\top(\alpha \mathbf u)=\alpha(\mathbf a^\top \mathbf u)$ and $\omega(\mathbf u) + \omega(\mathbf v) = \mathbf a^\top\mathbf u + \mathbf a^\top\mathbf v=\mathbf a^\top (\mathbf u + \mathbf v)$ for all $\mathbf u, \mathbf v \in \mathbb V$.
Rank 2 tensors (bilinear maps) have a similar matrix correspondence. This can be seen by arranging the tensor mapping across the basis vectors of $\mathbb V$ as the rows of a matrix:
$$[A]{ij} = \omega(ei,e_j) $$
Let, then, $\mathbf u = \sum_i u_i e_i, v = \sum_i v_i e_i$, then $$\omega(\mathbf u,\mathbf v) = \sum_{i} [A]{ij} ui v_i = \mathbf u ^\top A \mathbf v$$ The same arguments as before can be used to show that $A$ is a basis for the vector space of all rank 2 tensors.
Being a bilinear form, rank-2 tensors admit the familiar matrix representation given by $$ \begin{bmatrix} -- & \mathbf u & -- \end{bmatrix} \begin{bmatrix} \omega(\mathbf e_1,\mathbf e_1)&\cdots& \omega(\mathbf e_1 ,\mathbf e_{\dim \mathbb V})\ \vdots & \ddots & \vdots \ \omega(\mathbf e_{\dim \mathbb V},\mathbf e_1)& \cdots &\omega(\mathbf e_m,\mathbf e_m) \end{bmatrix} \begin{bmatrix} | \ \mathbf v \ | \end{bmatrix} $$ The rows of the matrix $A$ are, then, the component linear maps corresponding to the basis vectors of $\mathbb V$ and form together a basis for the space of rank 2 tensors.
Tensors of higher rank can be similarly given bases by structures of the form $$ [A]{i1,i_2,\cdots,i_k} = \omega(\mathbf e_{i_1},\mathbf e_{i_2},\cdots, \mathbf e_{i_k}) $$ which could conceivably be represented as ``$k$-dimensional matrices'' (although this is hard to visualize even in three dimensions).
Tensors of rank $k$ are said to be $k$-forms if they are alternating -- that is, if swapping the position of two parameters changes its sign.
$1$-forms are just covectors (which are trivially alternating), while $2$-forms are bilinear maps $\omega\colon \mathbb V\times \mathbb V \to \mathbb R$, such that $$ \omega(q,p) = -\omega(p,q) $$
For $k=3$, in turn, $\omega(p,q,r) = -\omega(q,p,r) = \omega(q,r,p) = -\omega(p,r,q) = \omega(r,p,q) = -\omega(r,q,p)$. The set of $k$-forms on the vector space $\mathbb V$ (i.e. such that each of its arguments is a vector in $\mathbb V$) is denoted $\Omega^k(\mathbb V)$, and is itself a vector space with some restrictions on its basis vectors.
Through a $k$-form, any $k$ vectors can be given an orientation, either positive or negative. If $k$ matches the dimension of $\mathbb V$, then the sign of $\omega(\mathbf e_1, \cdots, \mathbf e_n)$ is the orientation of the vector space.
As a bonus: for each $k\in \mathbb N$, the order-$k$ volume element $\det_{(k)} \in \Omega^ k(\mathbb V)$ (with $\dim \mathbb V = k$) is the unique alternating tensor such that $$ \det_{(k)}(\mathbf e_1, \cdots, \mathbf e_n) = 1 $$ (where $(\mathbf e_1,\cdots, \mathbf e_n)$ is the canonical basis of $\mathbb V$).
Because $\det_{(k)}$ is alternating, it defines an oriented volume alongside an orientation given by $\det_{(k)}(\mathbf e_1, \cdots, \mathbf e_n)$.
The familiar determinant of a $(m\times m)$ matrix is then given by $\det A = det_{(m)} (\mathbf a_1, \cdots, \mathbf a_m) $ where $\mathbf a_i$ is the $i$-th column of $A$
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