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

[Calabi_Yau]

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.

 

[Office_Shredder]

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.

 

[Mandelbroth]

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. )

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.

 

[Calabi_Yau]

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.

 

[Xiwi]

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.