>>8094314
https://en.wikipedia.org/wiki/Volume_form

>>8094314
Wedge product

>>8094314
https://en.wikipedia.org/wiki/Determinant

>>8094314
Here is an algorithm which will find the scalar value of a sequence of m vectors lying in an m-dimensional space (written off the top of my head):
Step 0: Assume our sequence of vectors is linearly independent (if they are not, then we can form a similar algorithm, which I will post if there is desire for it)
Step 1: Make a square matrix with our sequence of vectors
Step 2: Take the determinant of this matrix
Step 3: The absolute value of this scalar. That is the "scalar value" of our m-dimensional "parallelogram." In 3 dimensions this is volume, 2 is area, 4 is "hypervolume" etc.

Note that: 1 dimension is the trivial case. If we have a sequence of a single 1-vector then we simply have a scalar. This scalar is the value of its own length.

>>8094403
if the family is not linearly independent, the determinent (and the m-volume) is automatically 0.
example: parallelogram with two consecutive colinear sides
you don't need step 1

>>8094447
we are talking about what the volume/area/etc is though. that is not the same as the determinant.

I would argue that a matrix and a sequence of vectors are two entirely separate concept.

>>8094403
This is actually very easy to remember. But why is it that the determinants 'behave' like that giving area in 2d, volume in 3d,..?

>>8094530
because they determine how "spread out" the vectors are.
It just happens that that's exactly what surface/volume/other represent.

>>8094447
Without step 1, "the matrix" is undefined.

>>8094403
What's the general name for area,volume,higher dimension spaces? Supposing is called asdf, do you have an algorithm to calculate asdf's in any dimension? for example, calculating the area between to vectors in a 4 dimensions. l usually just take the magnitude from the cross product with a determinant, but l have no idea how to do this in 4d(l actually read cross product does not exist in 4d)

>>8094581
between to vectors -> between two** vectors, the w slipped off

>>8094564
it's very well defined.
It just has a determinant of 0.

>>8094581
you just take the determinant, i.e. put the vectors into a square matrix. the cross product and triple product are just special cases of the determinant in small dimension

the general name is hypervolume

>>8094581
The space taken up by an n-dimensional solid is referred to by a number called the "scalar value" of the solid. I actually said this in my post. It is a really weird name, but it is the only name I have ever come across. Maybe someone else has seen a different name?

>>8094555
this pretty much explains it. thinking of it like this explains how the cross product in 3 dimensions gives a vector with area of a parallelogram between the two arguments of the product.

>>8094674
hypervolume has been specifically used to refer to 4 dimensions from what I have read.

>>8094403
since i got several replies, I will post about what happens if the vectors do not form a basis of the dimension they rest in.

>>8096635
Given that we have a sequence of vectors, that sequence is not a basis if it linearly dependent. This means that we can remove vectors from the sequence until the sequence is linearly independent.

Let us have a sequence S. let v and u be vectors in S which are dependent on one another. If we remove either v or u from the sequence, then we may arrive at a linearly independent sequence. Let S \ {u} = A and let S \ {v} = B. ( \ denotes set difference).
WLOG, let A and B linearly independent now that their dependence relations of v and u have been remove.

Notice that the matrix formed by A and the matrix formed by B need not have the same determinant. (None of my assumptions made that a requisite.)

Visually, we can think of this as "seeing" the solid from different angles. Suppose you had 4 vectors in 3 dimensions. if two of those vectors are dependent on the remaining two vectors, then we may choose to remove different vectors and arrive have a different parallelogram embedded in 3D space.

In general, we do not try to make a determinant out of a sequence of vectors like this. It really almost makes more sense to say that there is no volume associated with this n-dimensional solid we have been toying with. You can consider this to be the intuitive reason why a linearly dependent square matrix has a determinant of zero.

Hope some anons found that interesting.

Find a higher dimension and we shall talk

>>8098010
How is high school treating you?

>>8098223
My diploma was only ever good for getting me off probation early.