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Linear Algebra proof (nonsingular matrices)

  1. Sep 14, 2006 #1
    Let A and B be n x n matrices and let C = AB. Prove that if B is singular then C must be singular.

    I have no idea how to prove this. I also don't understand how you can make such a claim without making some stipulations about A. I mean, if A were the 0 matrix, then C doesn't equal AB. And if A is singular, couldn't C also be singular? I was trying to prove this using row equivalence but I couldn't get there. Thanks
     
  2. jcsd
  3. Sep 14, 2006 #2

    AKG

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    What? C = AB by hypothesis, so if A = 0, then C = 0B = 0.
    Yes, but that has nothing to do with anything.

    Do you know what it means for a matrix to be singular?
     
  4. Sep 14, 2006 #3
    I think so. I think it means that it doesn't have an inverse. Doesn't it also mean that there is a 0 in the diagonal? I'm not good at writing proofs.
     
  5. Sep 14, 2006 #4
    ...in other words, if C=AB is invertible then B is invertible. that's how i would do it. if i had to do it exactly as stated i might use contradiction. suppose B is singular & AB is invertible, that is, [tex](AB)^{-1} = B^{-1}A^{-1}[/tex]. maybe it's easier that way. :confused:
     
  6. Sep 14, 2006 #5

    AKG

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    Singular means there's no inverse, correct. It doesn't mean there's a zero on the diagonal, and there are singular matrices with no zeroes on the diagonal.

    If B is singular, what can you say about the solutions to Bx = 0?
     
  7. Sep 14, 2006 #6
    the only solution is 0
     
  8. Sep 14, 2006 #7

    AKG

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    If B is non-singular, what can you say abou the solutions to Bx = 0?
     
  9. Sep 14, 2006 #8
    its zero? I might see where this is going
     
  10. Sep 15, 2006 #9

    AKG

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    I don't mean to confuse you too much. If B is non-singular, then Bx = 0 has only one solution, x=0, so post 8 is correct. If B is singular, then Bx = 0 has infinitely many non-zero solutions, so post 7 is incorrect. In fact, B is singular iff Bx = 0 has infinitely many non-zero solutions. This means that B is non-singular iff Bx = 0 has only the zero-solution. Don't you have any theorems like these?
     
  11. Sep 15, 2006 #10
    Yes, I actually misread post 5, I thought you had wrote nonsingular. I know the theorems. This is just the first course where I have to write proofs since 7th grade, also, I'm not particularly good at math and am taking linear algebra for mostly applications. (I don't deny that studying the proofs and theory will be a strong foundations for the applications.)

    So where do I start? a hint?
     
  12. Sep 15, 2006 #11

    HallsofIvy

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    You can also do this by looking at determinants:det(C)= det(AB)= det(A)det(B)
     
  13. May 1, 2010 #12

    tot

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    is it just me or is the math department lame.
    Why do we need so many contradicting words for the same thing
    correct me if I am wrong
    "non-singular"="One single trival solution"= "invertible"
    "singular" = "many solutions" ="not invertible"
     
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