Is Levy-Desplanques Theorem limited to specific values of i and M?

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Discussion Overview

The discussion revolves around the Levy-Desplanques theorem and its proof, specifically questioning whether the theorem's implications are limited to cases where the index i equals M. Participants explore the conditions under which a strictly diagonally dominant matrix is non-singular and the necessity of proving related inequalities for all indices.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions the validity of the proof provided for the Levy-Desplanques theorem, noting it only addresses the case where i equals M and suggesting that this limitation undermines the generality of the proof.
  • Another participant argues that to demonstrate a singular matrix is not strictly diagonally dominant, it suffices to show that at least one diagonal entry fails the property, rather than all entries.
  • A similar concern is reiterated by a participant who emphasizes the need for a proof that applies to all indices, not just the case where i equals M.
  • One participant uses an analogy about having all the money in the world to illustrate that finding a single counterexample (i.e., where i does not equal M) is sufficient to challenge the universality of the claim.
  • Another participant asserts that since the determinant condition already contradicts the definition of a strictly diagonally dominant matrix for some cases, further proofs for other indices may be unnecessary.

Areas of Agreement / Disagreement

Participants express disagreement regarding the necessity of proving the theorem for all indices. Some argue that proving it for one case is sufficient, while others maintain that a comprehensive proof is required for all indices to establish the theorem's validity.

Contextual Notes

Participants highlight limitations in the proof's scope, particularly its focus on specific cases and the implications of singular matrices in relation to diagonal dominance. There is an ongoing discussion about the definitions and assumptions underlying the theorem.

Calabi_Yau
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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.
 
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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.
 
Calabi_Yau said:
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.
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. :-p)

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.
 
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.
 
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.
 
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