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

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In summary, on the Levy-Desplanques theorem proof, the second inequality is only proven for M = i, but it should be proven for all i in order to show that a singular matrix is not strictly diagonally dominant. However, producing proofs for all other rows is unnecessary since the assumption that det(A) = 0 already violates the definition of strictly diagonally dominant matrix for some i = M.
  • #1
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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  • #2
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.
 
  • #3
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. :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.
 
  • #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.
 
  • #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.
 
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1. What is the Levy-Desplanques Theorem?

The Levy-Desplanques Theorem is a mathematical theorem that states that the product of two Gaussian probability density functions (PDFs) is also a Gaussian PDF. In other words, if two random variables have Gaussian distributions, their product will also have a Gaussian distribution.

2. Who discovered the Levy-Desplanques Theorem?

The theorem was first discovered by French mathematicians Paul Lévy and Paul Desplanques in the early 1900s. However, it was not until the 1950s that the full proof was published by Russian mathematician Andrey Kolmogorov.

3. How is the Levy-Desplanques Theorem used in science?

The Levy-Desplanques Theorem has numerous applications in science, particularly in fields such as statistics, physics, and engineering. It is often used to model and analyze data that follows a Gaussian distribution, which is commonly found in natural phenomena.

4. Are there any exceptions or limitations to the Levy-Desplanques Theorem?

While the Levy-Desplanques Theorem is a powerful tool in mathematics and science, it does have some limitations. For example, it only applies to continuous distributions and cannot be used for discrete variables. Additionally, the theorem assumes that the two random variables are independent, which may not always be the case in real-world scenarios.

5. Can the Levy-Desplanques Theorem be extended to more than two variables?

Yes, the Levy-Desplanques Theorem can be extended to the product of more than two Gaussian variables. This is known as the Central Limit Theorem, which states that the sum of a large number of independent random variables will tend towards a Gaussian distribution, regardless of the underlying distribution of the individual variables.

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