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What is the determinant of a matrix?

  1. Jul 26, 2012 #1
    Okay so I'm a first year engineering student and I'm taking linear algebra.

    I understand how to take determinants of nxn matrices, and I know how to do co-factor expansion. But I still don't understand what a determinant is.

    I don't like learning algorithms on how to do certain things in math without knowing why it works. I hope someone can answer the "Why?" question for me.

    To keep is simple as possible: What the hell is a determinant?
     
  2. jcsd
  3. Jul 26, 2012 #2

    rbj

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    best to read about it:

    http://en.wikipedia.org/wiki/Determinant

    or in your textbook.

    the determinant is an operator that maps a square matrix to a number. sometimes this class of operator is called a "functional". this number sorta-kinda roughly corresponds to what might be considered a "magnitude" of the matrix. sorta like the length of a vector, but this magnitude is raised to the nth power where n is the number of rows or columns of the square matrix.
     
  4. Jul 26, 2012 #3

    tiny-tim

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    Hi KingKai! :smile:

    It's how much larger the coordinate system is.

    The matrix transforms from one set of coordinates to another.

    The determinant is the ratio of the "volumes" of a "unit box" before and after the transformation.

    (and if you're doing an integral, eg ∫∫∫∫ f(w,x,y,z) dwdxdydz, once you've transformed the function and the limits, you also have to multiply by the determinant of the "Jacobian" matrix)
     
  5. Jul 26, 2012 #4

    rbj

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    that is an excellent answer, tiny. and a fact i hadn't thunked of before. but you don't need both the ratio of volumes and a unit box, do you? if you have any box (that has some volume) and transform it to another box by multiplying each corner coordinates by the transforming matrix, the volume of the resulting box is the volume of the box you start with times the determinant of the transforming matrix. ain't that so?
     
  6. Jul 26, 2012 #5
    This is wayy above my level of understanding. I only just learned how to do an integral like a month ago lol.

    τ∏αηκ∫ anyways haha.

    - ℝγαη
     
  7. Jul 26, 2012 #6

    rbj

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    King, leave the "Jacobian" thing alone for a couple of years.

    do you understand that if you multiply an n×n matrix with a length-n vector (otherwise known as a 1×n matrix or "column vector"), what you get as a result (or "product") is another length-n vector. you got that, right?

    now, imagine in 3-dimensional space (so n=3), and you have a box anywhere in this "3-space". there are eight corners of this box and each corner has coordinates of some ( x, y, z) values that you can represent as a vector. specifically as a column vector:

    [tex] \begin{bmatrix}
    x_1 \\
    y_1 \\
    z_1
    \end{bmatrix}[/tex]

    and you map each corner of that box to a corresponding corner of another box by use of some 3×3 square matrix. it's the same matrix used to transform each corner of the first box to the new box. the new box will have a volume equal to the volume of the first box times the determinant of the mapping matrix.

    does that make sense to you?
     
    Last edited: Jul 26, 2012
  8. Jul 26, 2012 #7

    tiny-tim

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    hi rbj! :smile:

    yes, you're right of course, but it usually crops up when dealing with things like dwdxdydz, so i decided it would be easier to understand if i kept to boxes :wink:

    on the other hand :rolleyes:
    Hi Ryan! :smile:

    ok, just consider a 1x1x1 box at the origin (so its volume is 1)

    a 3x3 matrix transforms it into a parallelepiped (a partially-collapsed box with sloping faces), and the volume of the new box is the determinant :wink:

    (and if the box has been turned inside-out, then the determinant is negative)

    same for a general n x n matrix, only less easy to visualise! :biggrin:
     
  9. Jul 26, 2012 #8
    I kinda sorta get it, I'll probably have to do an example of this "volume change" problem.

    Math is really starting to get weird lol...

    thanks a bunch rbj and tiny tim.

    You both get a check mark √ √
     
  10. Jul 26, 2012 #9
    These are great explanations. Can someone so the same for the trace of tensor?
     
  11. Jul 27, 2012 #10

    tiny-tim

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    hi cosmik debris! :smile:

    so far as i know, the trace of a tensor has no particular physical visualisation

    it is an invariant … it is the same under any unitary change of basis … so it does turn up in formulas like the einstein's field equations

    but in eg the moment of inertia tensor, the trace is the sum of the principal moments of inertia, which does not seem to have any physical application
     
  12. Jul 29, 2012 #11
    Thanks.
     
  13. Aug 8, 2012 #12
    Thank you. I never understood what it meant, but now I do. But how do you go from that to computing it? Also, why do books so rarely cover the definition of it?
     
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