vacinammo288

2022-04-30

I've been away from math far too long, and now my memory plays tricks on me when I try to recall simple facts.

I know that $det({v}_{1},\dots ,{v}_{n})$ is the oriented volume of the simplex determined by the origin and the vectors ${v}_{1},\dots ,{v}_{n}$ (up to a constant factor depending on the dimension $n$).

But I fell into utter confusion when I tried to make up why, especially when I tried to prove that this formula works for a shifted simplex, too.

What I'm sure is that det is an alternating bi-linear form that fits naturally with an oriented volume function.

What I've found on the net is dependent on what definition is used, and prone to circular reasoning.

Explicit question: how and why is the oriented volume of a simplex related to the determinant of its vectors?

What I've tried: proved it for $n=1,n=2,n=3$, and I've seen that this is not the way to go.

I know that $det({v}_{1},\dots ,{v}_{n})$ is the oriented volume of the simplex determined by the origin and the vectors ${v}_{1},\dots ,{v}_{n}$ (up to a constant factor depending on the dimension $n$).

But I fell into utter confusion when I tried to make up why, especially when I tried to prove that this formula works for a shifted simplex, too.

What I'm sure is that det is an alternating bi-linear form that fits naturally with an oriented volume function.

What I've found on the net is dependent on what definition is used, and prone to circular reasoning.

Explicit question: how and why is the oriented volume of a simplex related to the determinant of its vectors?

What I've tried: proved it for $n=1,n=2,n=3$, and I've seen that this is not the way to go.

bailaretzy33

Beginner2022-05-01Added 15 answers

I'm only addressing one point in your question, so this is not a complete answer, but this is a bit too much to put in comments.

A simplex in $n$ dimensions has $n+1$ vertices. You choose any one of these, it doesn't matter which, and you then have $n$ vector differences between the other $n$ vertices and your chosen vertex. You put those vectors in the columns of your matrix.

If one vertex is the origin the easy way is to choose that vertex, and then the difference vectors are simply the position vectors of the other $n$ vertices. But those vectors never measure the volume by virtue of their use as position vectors; they measure the volume by virtue of incorporating the lengths of the edges of the simplex and the angles between those edges and the various hyperplanes spanned by other edges. And it is the difference vectors that capture those properties.

So if you start with a simplex with vertices at $0,{v}_{1},{v}_{2},\dots ,{v}_{n}$ the volume is $det({v}_{1},\dots ,{v}_{n})$ but it may be more intuitive to think of it as $det({v}_{1}-0,\dots ,{v}_{n}-0).$ If you then shift all the vertices to $u,({v}_{1}+u),\dots ,({v}_{n}+u),$ the volume is $det(({v}_{1}+u)-u,\dots ,({v}_{n}+u)-u)=det({v}_{1},\dots ,{v}_{n}),$ that is, the volume does not change. (Even the way we calculate it does not really change except that taking the difference of vectors in each column is non-trivial.)

A simplex in $n$ dimensions has $n+1$ vertices. You choose any one of these, it doesn't matter which, and you then have $n$ vector differences between the other $n$ vertices and your chosen vertex. You put those vectors in the columns of your matrix.

If one vertex is the origin the easy way is to choose that vertex, and then the difference vectors are simply the position vectors of the other $n$ vertices. But those vectors never measure the volume by virtue of their use as position vectors; they measure the volume by virtue of incorporating the lengths of the edges of the simplex and the angles between those edges and the various hyperplanes spanned by other edges. And it is the difference vectors that capture those properties.

So if you start with a simplex with vertices at $0,{v}_{1},{v}_{2},\dots ,{v}_{n}$ the volume is $det({v}_{1},\dots ,{v}_{n})$ but it may be more intuitive to think of it as $det({v}_{1}-0,\dots ,{v}_{n}-0).$ If you then shift all the vertices to $u,({v}_{1}+u),\dots ,({v}_{n}+u),$ the volume is $det(({v}_{1}+u)-u,\dots ,({v}_{n}+u)-u)=det({v}_{1},\dots ,{v}_{n}),$ that is, the volume does not change. (Even the way we calculate it does not really change except that taking the difference of vectors in each column is non-trivial.)

Brianna Sims

Beginner2022-05-02Added 19 answers

Here is a simple conceptual way of doing it. The determinant function is invariant under the operation of adding one row to another, it is multiplied by $c$ when multiplying a row by $c$, and changes sign when two rows are exchanged, finally the determinant of the identity is 1. Now it is easy to show that there is at most one such function with those properties. (Do row reduction.)

Now the volume, suitably modified to be signed has all these properties. The first adding a multiple of one vector to another is a shear transform. Draw a picture of parallelograms, two dimensions suffice cause only two vectors are involved. Multiplying a vector by $c$, also multiples the volume by $c$, taking care of signs. The conclusion is that the two functions are equal.

Now the volume, suitably modified to be signed has all these properties. The first adding a multiple of one vector to another is a shear transform. Draw a picture of parallelograms, two dimensions suffice cause only two vectors are involved. Multiplying a vector by $c$, also multiples the volume by $c$, taking care of signs. The conclusion is that the two functions are equal.

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