Independent elements of matrices?

[perplexabot]

Hey all. I am currently reading an article and there is a paragraph that I am having a hard time understand. This is what the paragraph says:

"Since A_r = A_r^τ and A_i = -A_i^τ, we know that only the lower triangular (including the diagonal) elements of A_r are independent and only the strictly lower triangular (excluding the diagonal) elements of A_i are independent."

I don't exactly know what "independent elements" means in this case.

Are we talking about algebraic independence (because linear independence makes no sense to me in this case)? If yes, can someone please provide some insight into how it applies in this case? I read about algebraic independence on wiki, so I do have a general picture of what it is.

If you would like to refer to the article, here it is: http://www.ee.ucr.edu/~yhua/MILCOM_2013_Reprint.pdf
The paragraph is located under equation 9 of page 4 of the pdf.

Thank you PF.

 

[FactChecker]

If A_r and A_i have been derived by a process that forced them to be symetric and antisymetric, then A_r(i,j) ≡ A_r(j,i). for all (i,j) So once the elements above the diagonal are given, the elements below the diagonal are known. Same for the antisymetric matrix A_i, where A_i(i,j) ≡ -A_i(j,i) for all (i,l). For A_i we also know that A_i(i,i) ≡ 0.

So that gives a lot of equations that A_r and A_i must satisfy.

 

[perplexabot]

If A_r and A_i have been derived by a process that forced them to be symetric and antisymetric, then A_r(i,j) ≡ A_r(j,i). for all (i,j) So once the elements above the diagonal are given, the elements below the diagonal are known. Same for the antisymetric matrix A_i, where A_i(i,j) ≡ -A_i(j,i) for all (i,l). For A_i we also know that A_i(i,i) ≡ 0.

So that gives a lot of equations that A_r and A_i must satisfy.

OK, I think I get what you mean. So they are using less elements in order to describe the same set of equation(s). I wonder why the diagonal isn't included for A_i?

So this "independent" is just the everyday English definition, right?

 

[FactChecker]

OK, I think I get what you mean. So they are using less elements in order to describe the same set of equation(s). I wonder why the diagonal isn't included for A_i?

Because A_i(j,j) = -A_i(j,j) we know that A_i(j,j) =0. So the diagonal elements are not free.

So this "independent" is just the everyday English definition, right?

It also has the mathematical meaning. There are no equations forcing a relationship between them. They are all mathematically independent of each other.

 

[perplexabot]

Thank you for the quick responses and easy to understand explanation.

Because Ai(j,j) = -Ai(j,j) we know that Ai(j,j) =0. So the diagonal elements are not free.

I should have figured out that the diagonals are zeros, silly me (i didn't see it in your first post either). Thank you for that. Not exactly sure what you mean by "not free" tho?

There are no equations forcing a relationship between them.

What is meant by "them" here?

Apologies for my many questions, hopefully you can be patient with me.

I also have a question in this same article but about something different. You think I should open a new thread?

EDIT: I think I get why the diagonal pf Ai isn't included now. Since it is zero it really doesn't contribute to the equations, is this right?

 

[FactChecker]

Not exactly sure what you mean by "not free" tho?

(regarding the diagonal elements of Ai) Maybe I need to be more careful in my terminology. I meant that you can not assign any value you want to them. They must be 0.

What is meant by "them" here?

(regarding the elements below the diagonal) All the elements below the diagonal can take any value without any relationship to the other elements below the diagonal.

EDIT: I think I get why the diagonal pf Ai isn't included now. Since it is zero it really doesn't contribute to the equations, is this right?

The diagonal elements must be 0 for the equations to work. So they "contribute" in that sense. The definition of antisymetric forces the diagonal elements to be 0.

 

[perplexabot]

I got it! Thank you!