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Volume Function

  1. Jul 1, 2011 #1
    I have a quick question. First let me give a definition.

    Let [tex] a_1, a_2, ..., a_k [/tex] be independent vectors in R^n. We define the k-dimensional parallelopiped [tex] \mathbb{P}(a_1, ..., a_k) [/tex] to be the set of all x in R^n such that [tex] x = c_1a_1 + \cdots + c_k a_k [/tex] for scalars c_i such that 0 <= c_i <= 1.

    Now let X be the n by k matrix whose i'th column vector is a_i. We define the k-dimensional volume of the parallelopiped [tex] \mathbb{P}(a_1, ..., a_k) [/tex] to be the number [tex] \sqrt{det[X^{tr}X]} [/tex].

    My question is, why do they define the volume-function this way? Where is the geometric motivation?

    All help would be greatly appreciated!
  2. jcsd
  3. Jul 2, 2011 #2

    Gib Z

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    Homework Helper

    I wish I could provide something closer to the generalization, but here's the motivation at least for k=2 (a parallelogram in N dimensional space):

    Draw two vectors, x and y, in the plane that spans your parallelogram. The area is the length of x, times the perpendicular height of the parallelogram. Call the projection of y, in the direction of x, p. It is known that [tex] p = \frac{ x \cdot y}{||x||^2} x [/tex] and we can see that the perpendicular height is [tex] ||y-p|| = \sqrt{ ||y||^2 - \frac{(x\cdot y)^2}{||x||^2} }[/tex] by squaring, expanding with the dot product and simplifying.

    Thus, [tex] A = ||x|| ||y-p|| = \sqrt{ ||x||^2 ||y||^2 - (x\cdot y)^2 } [/tex].

    By definition of the determinant of a 2x2 matrix and norm of a vector;

    [tex] A^2 = \det \left[ x\cdot x , x\cdot y ; x\cdot y, y\cdot y \right] [/tex]

    Note that we can express dot products in terms of matrix multiplication:[itex] x\cdot y = x^Ty [/itex]. Then if we let [itex] X = [ x , y ] [/itex] be the matrix with columns x and y, then we find that [tex] A^2 = \det(X^TX) [/tex].
  4. Jul 2, 2011 #3
    Excellent question!

    If you want geometric intuition about what you're doing, you want to read about Geometric Algebra. Hestenes' book "Clifford Algebra to Geometric Calculus" is a classic here, and he discusses this in detail, but I find it extremely dense, and not very accessible to beginners. Instead, look for Dorst et al's textbook, "Geometric Algebra for Computer Science". He does a great job of giving a robust geometric motivation for everything he does, and carefully explains nuances that beginners are likely to wonder about.

    Chapter 2 explains the wedge product [itex]a \wedge b \wedge \ldots[/itex] in great detail, and how it lets you build up k-dimensional "blades", which represent the subspace spanned by k vectors. Chapter 3 explains how you can go about actually measuring the magnitude of these blades (which turns out to be the k-volume of the parallelepiped).

    Dorst is definitely the book you're looking for -- if you're looking for a book.

    Here's the gist: with the k vectors you started with, you form the wedge product
    [tex]A_k = a_1 \wedge a_2 \wedge \ldots \wedge a_k[/tex]
    The result, [itex]A_k[/itex], compactly summarizes a ton of information about these vectors:
    • Its "attitude" tells you the subspace of [itex]\mathbb{R}^n[/itex] spanned by those [itex]k[/itex] vectors. (For k=2 and n=3, think about turning a plane around in different directions.)
    • Its "orientation" tells you how it compares to some reference orientation. (For instance, in 2D, it tells you whether clockwise angles or counterclockwise angles are taken to be "positive". In 3D, it tells you whether you have a right-handed or left-handed coordinate system.)
    • Its "magnitude" gives the volume of the parallelepiped which you defined earlier.

    Swapping any two factors in the wedge product reverses the sign. This means that if you use any vector twice, the result is zero. This is just what you want for a volume. (Think about the "parallelepiped" spanned by [itex]a_1[/itex] and [itex]2a_1[/itex].) In fact, this means it also vanishes if the vectors are linearly dependent. (Think about the "parallelepiped" spanned by [itex]a_1, a_2, (a_1+a_2)[/itex].)

    So, what you're looking for is the magnitude of [itex]A_k[/itex]. If it were just one vector, we'd find that by taking the dot product with itself:
    |a_1| = \sqrt{a_1 \cdot a_1}
    To generalize this to higher dimensions, we need something called the scalar product
    |A_k| = \sqrt{A_k \ast \tilde{A}_k}
    The tilde over the second factor means "reversion": don't worry too much about the details, because it turns out the calculation is exactly the same as the determinant you listed in your original post.

    So, to answer your question: the reason that form means the volume is because that's how you take the scalar product of a k-blade with itself.
  5. Jul 2, 2011 #4
    Off topic, but...

    Gib Z, your sig is hilarious. :)
  6. Jul 2, 2011 #5
    Thanks for the detailed reply both of you! It helped a GREAT deal.
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