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Levy-Desplanques Theorem

  1. Dec 31, 2013 #1
    On the Levy-Desplanques theorem proof: http://planetmath.org/levydesplanquestheorem, they only prove the second inequality for M = i. What about if i ≠ M? e.g. if we are doing it for the first line on a singular matriz and M ≠ 1 we can't get to the second inequality.

    I thought that to prove: A strictly diagonally dominant matrix is non-singular (1)

    You had to prove: A singular matrix is not strictly diagonally dominant (2).

    Howver, they only prove (2) for i = M, whereas it should be for all i!

    What am I missing here? I can't understand how proving for only i 0 M constitutes a proof, and I can't prove it for all i.
  2. jcsd
  3. Dec 31, 2013 #2


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    To show that a singular matrix is not strictly diagonally dominant, you have to show that any single diagonal entry fails the property, not that all diagonal entries fail the property.
  4. Jan 1, 2014 #3
    Think of it this way:

    I make a claim, say, "I HAVE ALL THE MONEY IN THE WORLD!!! BWAHAHAHA!!" (Caps for dramatic silliness and excuse to link to a picture of Neil Patrick Harris as Dr. Horrible. I am clearly justified. :tongue:)

    If you have one cent, I don't have all the money in the world. I can have all but that one cent, but I still don't have all of it. Thus, by finding one case where it isn't true, the whole statement isn't true. Same thing here.
  5. Jan 1, 2014 #4
    Yes, it provides with a counterexample, for i = M, which proves it isn't true for all i, but I thought we had to prove it is wrong for ALL i.
  6. Jan 3, 2014 #5
    Since for some i = M the assumption that det(A) = 0 already violates the definition of strictly diagonally dominant matrix — that requires the definition be true for all i — then there is no point in producing proofs for all other rows.
    Last edited: Jan 3, 2014
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