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Showing det(C) = det(A)det(B)

  1. May 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Let http://imageload.dk/files/370495b6ec270c79fbf66c176f26e442.JPG [Broken] and

    http://imageload.dk/files/6ec41abfcdc8d231593c4a5ee8ee4fe7.JPG [Broken]

    where 0 is the j x k-zeromatrix.

    Show that det(C) = det(A)det(B)


    2. Relevant equations
    ??


    3. The attempt at a solution

    Got no clue. So I could really use a clue or some help here :)



    Regards
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 19, 2009 #2
    Hint: Use induction on the dimension of [tex] B [/tex], and expand the determinant along the bottom row.
     
  4. May 20, 2009 #3
    Hmmm, not totally sure what you mean.
    Induction, isn't that the thing with:
    1 + 2 + 3 + 4 + ... + n = (n)(n+1)/2 ?

    Not sure how I use that on the dimension of B tbh :?
     
  5. May 20, 2009 #4
    You can also use the definition of a determinant, i.e. the expression involving a summaton over permutations. Then it is trivial.
     
  6. May 20, 2009 #5

    C.E

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    Are you familiar with elementary matrices?
     
  7. May 20, 2009 #6
    You mean:
    det(A) = a1jA1j+...+ajjAjj
    det(B) = b1kB1k+...+bkkBkk

    And then when I multiply det(A) and det(B), I get the same as if I were to do the determinant of C, which will be det(A)*det(B)-0*0=det(A)*det(B) ?

    Or that won't work because they don't have the same dimensions ?
     
  8. May 20, 2009 #7
    A matrix that does the same to an already existing matrix, as if you were to row-operate ?
     
  9. May 20, 2009 #8
    I mean this: If A is an [itex]n\times n[/itex] matrix then, by definition, we have:

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

    where the summation is over all permutations [itex]\pi[/itex] of the numbers [itex]1\ldots n[/itex]
     
  10. May 20, 2009 #9
    Ok, that way...

    But how does that help me :?
    I can make det(A) and det(B) into a linear transformation, but if I do the same on C, won't I just get det(C) = A*B ?
     
  11. May 20, 2009 #10
    Consider some arbitrary permutation [itex]\pi[/itex]. Then consider the contribution to the determinant:

    [tex]\operatorname{sign}(\pi)C_{1,\pi(1)}C_{2,\pi(2)}\ldots C_{n,\pi(n)}[/tex]

    Where [itex]C_{r,s}[/itex] means the element in the rth row and sth column of C. And n =j+k is the dimension of C.

    Now unless the permutation [itex]\pi[/itex] permutes the numbers tranging from 1 to j amongst themselves, you will get zero. So, this means that such permutations [itex]\pi[/itex] can be decomposed as a product of permutations

    [tex]\pi=\rho\sigma[/tex]

    where [itex]\rho[/itex] permutes the numbers ranging from 1 to j and acts as the identity on the numbers ranging from j+1 to j+k and [itex]\sigma[/itex] is a permutation that permutes the numbers ranging from j+1 to j+k while acting as the identity on the numbers ranging from 1 to j.

    Then, if you use that the sign of a product of permutations is the product of the signs of the permutations, you are done.
     
  12. May 20, 2009 #11
    Hmmm, I'm really sorry. But I'm not sure I understand you here :?
    It sounds a bit confusing in my ears - even though it's probably not :S
     
  13. May 20, 2009 #12
    If you are going to choose elements from the first j rows of C, then unless you choose them from submatrix A you'll get zero, right? So, the values you must choose from rows 1 to j are restricted in the range from 1 to j. Therefore the permutation [itex]\pi[/tex] will map the range 1 to j to itself.
     
  14. May 21, 2009 #13
    Induction is a proof technique that takes advantage of an important property of the integers, namely, well-ordering. It's often compared to pushing down a stack of dominoes. What you do is show that your result holds for some base case (for your exercise, this would be the case where [tex] B [/tex] is a one-by-one matrix), and then show that if it holds for the case corresponding to a natural number [tex] n [/tex] (in your case, [tex] B [/tex] would be an [tex] n \times n [/tex] matrix), then it necessarily holds for [tex] n + 1 [/tex]. Carrying this out in your case isn't difficult, if you're familiar with this kind of argument.

    However, I actually like Count Iblis's idea better here. Although they're easy to crank out, I try to avoid induction proofs where possible, because they typically don't confer a very intuitive grasp of the theorem upon the reader (or writer). The proof using permutations gives an immediate and intuitive idea of why this result should be true.
     
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