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Linear Algebra - Determinant functions

  1. Aug 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Let R be the field of real numbers, and let D be a function on 2x2 matrices over R, with values in R, such that D(AB) = D(A)D(B) for all A, B. Suppose that D(I) != D ([0 1 1 0])

    Prove that
    a) D(0) = 0
    b) D(A) = 0 if A2= 0
    c) D(B) = -D(A) if B is obtained by interchanging the rows of A
    d) D(A) = 0 if one row of A is 0
    e) D(A) = 0 whenever A is singular

    2. Relevant equations



    3. The attempt at a solution
    I can prove a) by assuming A=B=0. But that also leaves the case that D(0) = 1 which I don't know how to disprove. I'm clueless on the rest of them.
     
  2. jcsd
  3. Aug 13, 2009 #2

    CompuChip

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    For the first one, try letting B unspecified (arbitrary).
    For the second one, use A = B.
     
  4. Aug 13, 2009 #3

    Dick

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    For the third one, use that D(I) is not equal to D(J) where I=[[1,0],[0,1]] and J=[[0,1],[1,0]]. If D(I)=1 then D(J) MUST be -1. Can you show that using I^2=J^2=I? Now use that for any matrix A, JA is A with the rows interchanged.
     
  5. Aug 13, 2009 #4

    CompuChip

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    That one's kinda brilliant :)
     
  6. Aug 13, 2009 #5

    Dick

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    Kinda fun is what it is. That's a cute problem.
     
  7. Aug 13, 2009 #6
    I proved e) and therefore automatically proved d) as follows.

    Let Q be the matrix [ [1 -1] [1 -1] ]. I have some singular function which will be of the form P = [ [X Y] [cX cY] ] where c is some arbitrary constant. Let L = [ [X Y] [0 0] ]

    Now Q times L is [ [X Y] [X Y] ] which is singular. Other singular matrices P can be made by "modifying" the bottom row of Q i.e. multiplying the bottom row of Q by some scalar c will give you any other singular function of the form P that can be made from L. So let us consider the most basic case.

    D(QL) = D(Q)D(L) .... Given
    but D(Q) = 0 since Q2 = 0
    Now, D(QL) = D(P) since QL = P
    So, D(QL) = 0 = D(P)

    Therefore, since c can take value zero also, the above result is true for ANY singular matrix, inluding L itself. Hence d) is also proved.
     
  8. Aug 13, 2009 #7
    Dick that was awesome!Thanks! Thanks CompuChip! Check my above post too.
     
  9. Aug 13, 2009 #8
    Here's part of a follow up question to that question (I proved the first part myself)

    D is an alternating 2-linear function on 2x2 matrices over some commutative ring K with identity. Using D(A) = det A D(I) show that det (AB) = (detA)(detB) without using computations with the entries.

    I proved that D(A) = det (A) D(I) for 2x2 matrices, but I can't seem to prove that second part.
     
  10. Aug 13, 2009 #9

    CompuChip

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    Can't you just write
    det(AB) = D(AB) / D(I)
    because D(I) is non-zero, and use what you know about D, in particular D(AB)?
     
  11. Aug 14, 2009 #10
    But for this follow up question, I cannot assume that D(AB) = D(A)D(B) and it is not given so.
     
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