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Why the determinant of a matrix is equal to its transpose

  1. Jun 8, 2009 #1
    1. The problem statement, all variables and given/known data
    I don't understand why the determinant of a matrix is equal to its transpose....how is this possible?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 8, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: determinants

    If you mean the determinant of a matrix is equal to the determinant of the transpose, why do you think that's so impossible? Try and prove it.
     
  4. Jun 9, 2009 #3
    Re: determinants

    ohhh it works o.o
    can i see a rigorous proof of this?
     
  5. Jun 9, 2009 #4

    Mark44

    Staff: Mentor

    Re: determinants

    Yes, if you have functioning eyes and someone shows you such a proof.
     
  6. Jun 9, 2009 #5
    Re: determinants

    lol thats funny. What i meant is May you show me such a proof, please?
     
  7. Jun 9, 2009 #6

    Mark44

    Staff: Mentor

    Re: determinants

    I might, if you prove it to yourself with a 2 x 2 matrix first.
    A = [a b; c d] (this is in row-major order).
     
  8. Jun 9, 2009 #7
    Re: determinants

    A = [a b; c d]
    det (A) = ad-bc

    A transpose = [a c; b d]
    det A transpose = ad - bc = det A
     
  9. Jun 9, 2009 #8
    Re: determinants

    In this case, a visual interpretation and the fact that you can take determinants along and row/column is usually a fair explanation.

    For a possible proof, you can prove the determinant inductively for matrices of size n. Then you take the idea of permutations to swap rows/columns to show that it does not change the determinant. It should be straight forward from then on.
     
  10. Jun 9, 2009 #9
    Re: determinants

    The proof is trivial:

    If A is an n by n matrix, then:

    [tex]\det(A) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{i,\pi(i)}[/tex] (1)

    The determinant of the transpose can thus be written as:

    [tex]\det(A^{T}) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{\pi(i),i}[/tex]

    So, to prove that the determinant of the transpose is the same, we have move the permutation from the second index to the first in (1). We can do this as follows.

    In the product, it doesn't matter in which order the matrix elements are multiplied:

    [tex]\prod_{i=1}^{n}A_{i,\pi(i)} = \prod_{i=1}^{n}A_{\pi^{-1}(i),i}[/tex]


    Using that the sign of a permutation is the same as the sign of its inverse, gives:


    [tex]\det(A) =\sum_{\pi}\operatorname{sign}(\pi^{-1})\prod_{i=1}^{n}A_{\pi^{-1}(i),i}[/tex]


    The set of all inverse permutations is the same as the set of all permuations, so we can write this as


    [tex]\det(A) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{\pi(i),i}[/tex]
     
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