What is the meaning of Lax's definition of orientation of simplexes? I have some questions regarding simplexes after reading Lax's "Linear Algebra and its Applications", the chapter on determinants.
(1) I do not understand his definition of orientation. He says:

An ordered simplex $(0, a_1, \ldots,a_n)= S$ that is nondegenerate can have one
of two orientations: positive or negative. We call S positively
oriented if it can be deformed continuously and nondegenerately into
the standard ordered simplex $(0, e_1, \ldots,e_n)$, where $e_j$ is the $j$th
unit vector in the standard basis of $\mathbb{R}^n$. By such deformation we mean $n$
vector-valued continuous functions $a_j(t)$ of $t$, $0 \leq t \leq 1$, such that (i)
$S(t) = (0, a_1(t), \ldots , a_n(t))$ is nondegenerate for all $t$ and (ii)
$a_j(0) = a_j, a_j(1) = e_j$. Otherwise $S$ is called negatively oriented.

What does it mean to be deformed continuously and nondegenerately? I don't understand what kind of function $a_j(t)$ he has in mind, let alone what is $0 \leq t \leq 1$ doing in it. I feel I need a bit more thorough explanation about what he means by those functions.
(2) Then, he states that

The volume of a simplex S is given by the elementary formula $\text{Vol(S)}=\frac{1}{n} \cdot \text{Vol}_{n-1}\text{(Base)} \cdot \text{Altitude}$
(By base we mean any of the $(n-1)$-dimensional faces of $S$, and by altitude we mean the distance of the opposite vertex from the hyperplane that contains the base).

I would appreciate it if someone can elaborate a bit more on where this formula comes from, since it does not look obvious to me. I can see it is true in $2$ and $3$ dimensions, but how do we generalize it to $n$ dimensions?
 A: Here are some simple examples to explain the positive and
negative orientation thing.
Consider the simplex (triangle) $T$ with vertices $0=(0,0)$,
$a_1=(1,0)$ and $a_2=(1,1)$. This can be deformed to the
standard triangle by taking $a_1(t)=(1,0)$ and
$a_2(t)=(1-t,1)$. These $a_i(t)$ are vector valued functions of
$t$ (i.e. for each $t$ you get a vector in the plane) and they
are continuous (each component is continuous).
As $t$ varies from $0$ to $1$, the triangle deforms from $T$ to
the standard triangle. Everywhere along the way, it's a
"nondegenerate" triangle (i.e. it's not something stupid like
three points on a line). By the definition you quoted, this is
therefore a positively oriented triangle.
Take instead the triangle $T'$ with vertices $0=(0,0)$,
$a_1=(1,0)$ and $a_2=(1,-1)$. This one will be negatively
oriented. To see why, suppose you have a pair of continuous
functions $a_1(t)=(p(t),q(t))$ and $a_2(t)=(r(t),s(t))$ such that $0,a_1(0),a_2(0)$ is
the triangle $T'$ (i.e. $p(0)=1$, $q(0)=0$, $r(0)=1$, $s(0)=-1$) and $0,a_1(1),a_2(1)$ is your standard
triangle (i.e. $p(1)=1$, $q(1)=0$, $r(1)=0$, $s(1)=1$). Let $b_1(t)$ be the vector obtained from $a_1(t)$ by
rotating 90 degrees anticlockwise. The function
$f(t):=a_2(t)\cdot b_1(t)$ is continuous (because all the
components of all the vectors involved are
continuous). Moreover, $f(0)=(1,-1)\cdot(0,1)=-1$ and
$f(1)=(0,1)\cdot (0,1) = 1$. So $f$ is a continuous function on
$[0,1]$ taking negative and positive values, so it must be zero
for some $t\in(0,1)$. At this $t$, $a_2(t)$ is collinear with
$a_1(t)$ (because it's orthogonal to the vector orthogonal to
$a_1(t)$) so the triangle is "degenerate" (just three points on
a line). This shows that this triangle is negatively oriented.
Off the top of my head, I can't think of a quick way of proving
formulas for the area of a simplex.
A: This is a partial answer to the volume question. We know first that
we can measure lengths and angles of 2 or 3 dimensional objects using ruler and
protractor, and we define area and volume of the resulting objects
in a way that is useful and now familiar. Since we cannot step into
4+ dimensions with these instruments, we have to define length, angle, volume
in ways that accord with things we already know, that are interesting,
and that we hope to find useful.
One way to go from 3 to 4 dimensions is to begin by observing
the 3-fold symmetry of a unit cube
when rotated around the line connecting $(0,0,0)$ to $(1,1,1)$.
The part with vertices $(0,0,0), (1,0,0), (1,0,1), (1,1,0), (1,1,1)$
represents $1/3$ of the volume, because the rotation
$$
 A(x_1,x_2,x_3) = (x_2,x_3,x_1)
$$
has period 3,
preserves lengths, therefore angles, and therefore volume.
Therefore  that piece contains $1/3$ of the volume of the unit cube,
i.e., $1/3$, i.e. $1/3$ times the altitude times the base area.
Since that piece is seen to consist of pyramids with
square bases, we still have to convince ourselves that cutting them
in half so that they have triangular bases does not change the basic formula. Then a similar argument could
be made in 4 dimensions for a 4-fold symmetry and a
factor of $1/4$.
A different approach extrapolates what we know about calculus in
2 and 3 dimensions. In principle one feels that algebra comes before
calculus, but our brains allow us to reason somewhat circularly when
we are making up definitions. So observe that a simplex with base
$a_1,\ldots,a_{n-1}$ and "top" vertex $a_n$ can be sliced parallel
to the base, lets say at height $t$, with $0 \le t \le a$
where $a$ is the altitude. The slice is a rescaled version of the base,
with lengths decreased by a factor of $(1-t/a)$.
The cross section area ($n-1$-volume) we agree,
based on
experience with $n-1$ dimensions, is $(1-t/a)^{n-1}$ times the area
of the base. And based on our calculus experience we agree that the
$n$-volume could be taken to be
$$
 \int_0^a ({\rm slice\ area})\,dt
 =
 \int_0^a (1-t/a)^{n-1}({\rm base\ area})\,dt
 $$ $$
 =
 \Big[\frac{(1-t/a)^n}{-n/a}\Big]_0^a({\rm base\ area})
 =
  \frac{1}{n}({\rm base\ area})a.
$$
As I've emphasized, this proves nothing, but gives some hints for
understanding why the definitions are made as they are.
Ultimately, if you take the "most beautiful" approach,
and simply extrapolate
$$
 \frac{1}{2},\frac{1}{3},\ldots, \frac{1}{n}, \ldots 
$$
then these arguments are evidence that you have chosen well.
