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Cross Product - Determinant form

  1. Mar 28, 2010 #1
    Did someone just realize that taking the determinant of a specific matrix gives you the cross product formula, or is there is a specific conceptual reason why it works?
  2. jcsd
  3. Mar 28, 2010 #2
    One definition of the determinant is as the unique (up to a constant) totally antisymmetric multilinear function on n vectors in n dimensions. Thinking of n-1 of the arguments (rows or columns) held fixed makes this a linear function on the remaining argument which is thus an element of the dual space and so (via the inner product) can be identified with an element of the space. In the case of n=3, n-1=2 and so this process identifies a pair of elements of the space with a third. This defines a product operation which is antisymmetric and linear in both arguments and is orthogonal to either of them - which (up to a constant) are the defining properties of the cross product.
  4. Mar 28, 2010 #3
    Sorry for my ignorance, but I got lost after you mentioned dual space and space. Could you explain what you mean?
  5. Mar 28, 2010 #4
    By "space" I just meant a vector space. The set of 3D vectors is one example but we can also consider higher dimensional spaces like R^n (or even infinite dimensional ones as well), and the "dual space" is a set of linear functions of vectors in the space - that is number-valued functions with the property that f(av+bw)=af(v)+bf(w). (But that's just a fancy term which you don't need to know for what follows).

    It can be shown that for a finite dimensional real vector space with an inner product (like the dot product in R^3) any real linear function is of the form f(v)=u.v for some vector u. So if we consider the determinant of a 3x3 matrix as a function of its first column then this must be a dot product with some vector which depends on the other two columns. (By direct calculation it is the dot product with the vector whose components are 2x2 determinants, but you seemed to be asking for a more conceptual explanation). Since the determinant is linear in each column and antisymmetric (changes sign when any two rows or colums are exchanged) the vector whose dot product with the first column gives the determinant depends linearly on each of the other two columns and changes sign when they are reversed. Also since the determinant is zero when columns repeat, this combination of the last two columns is orthogonal to both of them (since the corresponding function of the first column is zero when it matches either of the other two columns). These conditions are enough to determine the cross product.

    [More generally, in an n-dimensional space there is a generalization of the cross product defined not for pairs but for groups of n-1 vectors by taking as components the signed minor determinants which multiply by the entries of the first column to get the overall determinant of the nxn matrix with the given n-1 vectors as its remaining columns]
  6. Mar 29, 2010 #5


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    Welcome to PF!

    Hi alQpr! Welcome to PF! :wink:
    I think gsingh2011 :smile: is asking a much simper question …

    a determinant is usually a number, and is derived from a transformation matrix …

    so how does what should be a number turn out to be a vector (technically, a pseudovector :rolleyes:), and what is the transformation that it's derived from?

    I can't think of a good answer to any of this, I can only think of more problems :redface:

    … for example, the determinant of a product of two transformations is the product of the determinants, but what would the first "product" mean in this case? :confused:

    Does anybody have a good answer to this? :smile:
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