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Problem with proof

  1. Mar 13, 2007 #1
    Im stuck on this proof.

    Let A and B be nxn matricies such that AB is singular. Prove that either A or B is singular.

    Sooooo, here we go.

    Let M = AB where is M is the given singular matrix.

    Becuase M is singular then

    Mx=0 has an infinite amount of solutions.

    Let J be one of the non zero solutions



    this is where I get stuck.
    If knew that B was singular I think I could prove M is singular but Im having trouble from this way around.
    Any ideas?
  2. jcsd
  3. Mar 13, 2007 #2

    matt grime

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    Stop proving this and prove the converse - if A and B are invertible, then so is AB.
  4. Mar 13, 2007 #3

    matt grime

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    You aren't asked to prove this (though it is clearly true). You are told that M is singular, and asked to show that either A or B is too (or both).
  5. Mar 13, 2007 #4


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    You could just use determinants: det(AB)= det(A)det(B)= det(M). Since M is singular, det(M)= what? What does that tell you about det(A) or det(B)?
  6. Mar 13, 2007 #5

    Let A and B be invertible then M must be inverble also.

    But since M is singular then A or B must be singular.

    Is that it? By contradiction?
  7. Mar 13, 2007 #6

    so det(M)=0 so det(A)=0 or det(B)=0

    nonsingular matrix cannot equal zero so A or B must be singular.

  8. Mar 14, 2007 #7
    You've got to use determinants. That way you just map your matrices to set of real numbers R, and everything is just much easier ther!
  9. Mar 14, 2007 #8


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    Something like that, yes. You know that M = AB is singular, so, det(AB) = detA detB = 0 implies det A = 0 or det B = 0 ("or", of course, includes the case where both equal zero, too).
  10. Mar 14, 2007 #9
    Nonsense. It's equally easy to simply find the inverse of AB directly.
  11. Mar 14, 2007 #10
    Yes, but we're saying here that AB is singular here, so it doesn't have an inverse. Plus, finding inverse involves finding the determinant first...
  12. Mar 14, 2007 #11
    Using determinant gives the easiest proof.
  13. Mar 14, 2007 #12


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    There are (at least) two easy methods:

    1. Use determinants. det(AB) = det(A)det(B). If AB is singular, det(AB) is zero. det(A) and det(B) are just real numbers, so if their product is zero, what do you know?
    2. Prove the contrapositive (not the converse), that is, prove that if A and B are non-singular, then AB is non-singular. So if A and B have inverses, A-1 and B-1, find a matrix that should be the inverse of AB, and prove that the matrix you found really is the inverse.
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