[Math] Matrix Determinants and parallelopiped volumes

determinantlinear algebramatrices

I recently read the very interesting discussion on matrix determinants here: What's an intuitive way to think about the determinant?

I was hoping to ask a question related to this topic, and decided to start a new question (I hope this is ok!).

Matrix determinants can be thought of as the coefficient by which an oriented volume changes under the linear mapping of that matrix.

For example, consider $Ax=y$. We can consider the entries of $x$ and $y$ as the vertices of an $n$-d parallelopiped. Then $vol(x)$ is the volume of that parallelopiped, and $vol(y)=|det(A)|vol(x)$.

My question truly has 2 parts. The first is whether my example above is a correct interpretation. And the second is, if $x \in \mathbb{R}^n$, how do I compute $vol(x)$?

This interests me, because if finding the "volume" of a vector is easy, than we can use that to find the determinant, i.e.

$|det(A)|=\frac{vol(y)}{vol(x)}$

Best Answer

Well, you can calculate the volume of the parallelepiped spanned by the standard basis vectors $e_i$ with $1$ in the $i$'th coordinate and $0$ elsewhere, right?

So given a parallelepiped spanned by $n$ vectors $v_i$, what's the matrix which maps $e_i$ on $v_i$?

The solution to this will give you yet another interpretation of how the determinate measures volumes.

EDIT: I reread your question and there are some things that I want to clear up.

  1. The expression $vol(x)$ is meaningless for a vector.
  2. This means that the expression $Ax = y \Rightarrow |y| = |det(A)|vol(x)$ is meaningless as well.
  3. It takes $n$ vectors to specify a parallelepiped with $n$-volume. The volume will be zero only if all of the vectors are independent. A square has zero 3-volume for instance.
  4. That parallelepiped can be defined as $P(v_1, \ldots, v_n) = \{x: x = c_1e_1 + \ldots + c_ne_n, 0 \leq c_1, \ldots, c_n \leq 1 \}$
  5. The determinate measure oriented $n$-volume. It does not measure oriented $n-d$ volume.
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