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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.

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

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1yen
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1 Answers1

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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.
knucklebumpler
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  • thanks for the quick reply. Given a vector $x$, it is easy to see that $x = diag(x)(\sum_i e_i)$, which I think is what you are suggesting. Then $vol(x)=\prod x_i$. However, if $x_i=0$ for some i, then the volume is 0, but certainly there is some non-singular matrix $A$ such that $y=Ax$ and $y_i \neq 0$ for all i. This would suggest my determinant conjecture is wrong (it would be infinite). Or perhaps I am still not understanding something correctly. – 1yen Apr 08 '11 at 08:54
  • @1yen I'm not sure what $diag$ is in this context. In $\mathbb{R}^n$, a vector only has $n$-volume when $n = 1$ in which case we just call it length. So I'm not sure what $vol(x)$ is supposed to mean either. Typically, you will need $n$ independent vectors to span a parallelepiped with non-zero $n$-volume, no? What's the volume of the parallelepiped spanned by the vectors $e_i$. – knucklebumpler Apr 08 '11 at 08:59
  • I think I understand now...to consider a volume, you are correct, I need n lin. independent vectors. So I guess what I really want to say is $det(A)=vol(X)/vol(Y)$, where $X=[x_1,\ldots,,x_n]$ are n lin. independent vectors, and $Y=AX$. Thanks! – 1yen Apr 08 '11 at 09:19
  • @1yen, I'm not sure that you should accept this answer as there may be a better way to measure the volume that you're looking for. This is just a disappointment for your idea once you understand my answer. Somebody might know of a better way to do it. – knucklebumpler Apr 08 '11 at 09:21