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Looking for insight into what the Determinant means...

  1. Oct 30, 2015 #1
    In HS they just taught you the formula for it (the cofactor method) and a few other things, such as det(A) = 0 means A is singular.

    I finally reached Ch 5 of MIT OCW Intro to Linear Algebra, and I was really hoping that seeing how determinants are derived from first principles would give me some insight into what the determinant means...aside from whether or not the matrix is singular. I can follow the derivation, through the Big Formula, using the first three properties (det(I) = 1, row swap multiplies by -1, and linearity of determinants).

    I have seen how the [itex]A \times adj(A) = det(A)\times I[/itex]

    I have shown myself how the determinant for a 2x2 and 3x3 matrix arises purely from elimination, and considering the case if the last pivot is zero.

    Ie:

    [tex]\begin{bmatrix}a & b\\ 0 & d-\frac{bc}{a}\end{bmatrix}[/tex]

    and

    BDjXWcn.gif

    imgur link: http://i.imgur.com/BDjXWcn.gif

    What I've learned now just begs a further question, rather than, "why is the determinant important/ what does the determinant means?", I am now wondering, "what do the pivots, and the product of the pivots mean?" Aside of course from indicating whether or not the matrix is singular.
     
  2. jcsd
  3. Oct 30, 2015 #2

    Fredrik

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  4. Nov 4, 2015 #3
    The determinant can be interpreted geometrically. If A is an [itex]n\times n [/itex] matrix, then let [itex]r_1, r_2, \dotsm, r_n[/itex] be the n rows of A. The absolute value of the determinant of A would be the n-dimensional voume of the parallelotope corresponding to these n vectors. (Imagine one corner of the polytope at the origin, and n more corners located at [itex]r_i[/itex].) If the determinant is non-zero, then A has an inverse. Geometrically, this means there is a bijective map [itex]f:\mathbb{R}^n \to \mathbb{R}^n[/itex] such that f(A) = I. That is, there is a linear transformation (namely [itex]f(x)=A^{-1}x[/itex]) that maps the parallelotope associated with A to the unit parallelotope associated with the identity matrix I. In the case where the determinant (and hence volume) is 0, the parallelotope is degenerate. (In [itex]\mathbb{R}^3[/itex], one such case would be a box with height 0, so that it is really a flat square inside 3-dimensional space). As your intuition might suggest, any mapping that takes such a degenerate polytope to the unit parallelotope could not possibly be bijective.
     
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