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Determinant simple algebra

  1. Jul 28, 2011 #1
    Hi,

    if vector b * matrix K = 0 (bK=o) what methods can one use to show that the determinant of K is therefore also zero, without using eigenvalues.

    I have a feeling Im over complicating this.

    Knd regards

    Emma
     
  2. jcsd
  3. Jul 28, 2011 #2
    I assume b != 0. What else are you starting with? Are you allowed to use the fact that, for any square matrix K with determinant != 0, the equation Kx = y has a unique solution? If so, it is obvious that the solution to KTb = 0 is not unique, since KT0 = 0 also. Thus, the determinant of KT is 0, and since transposition doesn't change the determinant, the determinant of K must also be 0.
     
  4. Jul 28, 2011 #3
    Sorry thats correct that b!=0

    using for any square matrix K with determinant != 0, the equation Kx = y has a unique solution - is fine.

    Im almost understand where your heading with this - would you please however clarify two steps;

    you say KTb = 0 is not unique, since KT0 = 0 also. Thus, the determinant of KT is 0.

    How do you know the value of the determinant from that statement alone,

    Also is bK, the same as KTb, I noticed you switched the order of multiplication there.

    Thanks for your help
     
  5. Jul 28, 2011 #4
    Sorry, I should have written KTbT. I'm using the turnover rule for matrix multiplication: for any two matrices A and B, (AB)T = BTAT. In this case b is a row vector, a 1 x n matrix, and bT is a column vector, an n x 1 matrix. The only reason for doing it that way is that the uniqueness theorem is usually stated in terms of left multiplication and column vectors, but of course it also holds for right multiplication and row vectors.

    If the determinant of K (or KT) is not 0, the solution would be unique. Since the solution is not unique, the determinant must be 0.
     
  6. Jul 28, 2011 #5
    Thats great - I really appreciate your help!
     
  7. Jul 28, 2011 #6
    It can be proved directly using Kramer's rule. Write K=[k1;...;kn] (the kj are the rows) and let Kj be the matrix K with row j replaced with 0. Obviously det(Kj)=0 by row expansion.

    Also 0=b*K=b1*k1+...+bn*kn, so det(Kj)=det([k1;...;b1*k1+...+bn*kn;...;kn])=det([k1;...;bj*kj;...;kn])=bj*det(K). Choose any j such that bj<>0 and this shows that det(K)=0.
     
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