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Definition of the Determinant of a Matrix

  1. May 17, 2010 #1
    How could one derive a definition of the determinant from some of its basic properties such as det of product = product of dets or the determinant of a transpose is the determinant of the untransposed matrix?

    Upon instigating research on determiniants, all I've found are definitions that either only cover lower dimensional matrices or define the determinant to be some strange expression with little motivation.
     
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  3. May 18, 2010 #2

    lavinia

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    I think the determinant is the only mutilinear alternating function in the vector space - up to a constant. If you require the map to have a value of 1 on the identity matrix then you get the determinant.

    Try proving this - say by induction.
     
  4. May 18, 2010 #3
    To clarify, Lavinia means that the top exterior power of a vector space is one-dimensional. I.e. for an n-dimensional vector space V, the vector space of alternating, multilinear functions f:V^n->R is one-dimensional. This is only true for the top dimension.
     
  5. May 18, 2010 #4

    lavinia

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    Yes. this is the same statement.

    the determinant is a polynomial in the matrix entries. I wonder how this all translates into the properties of this polynomial.
     
  6. May 19, 2010 #5

    lavinia

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    How about this as another way to think of the determinant. Consider all functions on square matrices that are constant under conjugation.

    [tex] f(X) = f(AXA ^-1) [/tex]

    for all square matrices X and invertible square matrices ,A.

    For instance the trace is such a function.

    Define the determinant to be that function which equals the product of the diagonal entries on any diagonal matrix.
     
    Last edited: May 19, 2010
  7. May 20, 2010 #6
    Thanks for the responses. However, I have a question.
    Is the determinant uniquely determined by its multilinearity and the properties I mentioned above?
    I am hoping that the determinant can be made in to a type of analogue of the complex abosolute value with as few other conditions as possible (alternating multilinear is a bit too...esoteric for my tastes)
     
    Last edited: May 20, 2010
  8. May 20, 2010 #7

    lavinia

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    alternating multilinear on any n-vectors uniquely determines the determinant up to a constant.
     
  9. May 23, 2010 #8
    I like how it's covered in Linear Algebra Done Wrong, which is free and easy to find online. Once he gets to the definition of determinant I think the coverage is pretty standard but the motivation behind the definition is nicely included.
     
  10. May 23, 2010 #9
    An equivalent definition is that the determinant is the product of the eigenvalues of the linear transformation that corresponds to the matrix (ie., it is the factor of signed scale of n-paralleletopes embedded in the target vector space versus the domain). This definition is a bit more intuitive and easier to work with geometrically. If you're looking for a formula that refers directly to the entries of an arbitrary matrix representation of the linear transformation, then you are left with the alternating multilinear map, embodied in the Levi-Civita symbol or Laplace's expansion by minors (equivalent, just re-ordered).
    The relation between the two definitions is derived through basic geometric algebra (the wedge product).
     
    Last edited: May 23, 2010
  11. May 23, 2010 #10

    lurflurf

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    ^The product of eigenvalues is not equivalent or helpful is the matrix does not have any eigenvalues.
     
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